0^0=1

[DonH]

Well, according to my computer Basic program it does.
So, if you have a lot of nothing and multiply it by itself no times, you
get "one" - what?
At least this is consistent with the zero power of any integer, as all
give 1.
Can any of this be proved?
(Zero, while a mathematical symbol, is not a number, but denotes the absence
of number?)

[Virgil]

In article <iskkr.6000$%E2....@viwinnwfe01.internal.bigpond.com>,

x^0 = 1 for any x other than 0
0^x = 0 for any x other than 0

Virtually the only reason to define ANY value for 0^0 is so that
calculators and computers do not throw a fit when they try to evaluate
it.
--

[Pubkeybreaker]

On Apr 20, 5:47 pm, Virgil <vir...@ligriv.com> wrote:
> In article <iskkr.6000$%E2.2...@viwinnwfe01.internal.bigpond.com>,

>
>  "DonH" <donlhumphr...@bigpond.com> wrote:
> > Well, according to my computer Basic program it does.
> >    So, if you have a lot of nothing and multiply it by itself no times, you
> > get "one" - what?
> >    At least this is consistent with the zero power of any integer, as all
> > give 1.
> >    Can any of this be proved?
> > (Zero, while a mathematical symbol, is not a number, but denotes the absence
> > of number?)
>
> x^0 = 1 for any x other than 0
> 0^x = 0 for any x other than 0
>
> Virtually the only reason to define ANY value for 0^0 is so that
> calculators and computers do not throw a fit when they try to evaluate
> it.
> --

Consider the number of mappings from a set of b elements to a set of
c elements.

Now let b = c = 0

[Tonico]

Or, since we're dealing with real or complex numbers, consider lim x^x
= lim e^{x log(x)} , when x --> 0 ...this is a good reason to define
0^0 = 1 .

Tonio

[Richard Tobin]

In article <fc51d462-6ab6-4087...@fv28g2000vbb.googlegroups.com>,

Tonico <Toni...@yahoo.com> wrote:

>Or, since we're dealing with real or complex numbers, consider lim x^x
>= lim e^{x log(x)} , when x --> 0 ...this is a good reason to define
>0^0 = 1 .

But equally one could consider the limit of 0^x.

For integers the set mapping argument seems compelling, but when
dealing with the reals it seems better to leave it undefined or define
it explicitly if needed.

-- Richard

[DonH]

"Richard Tobin" <ric...@cogsci.ed.ac.uk> wrote in message
news:jmsvh1$nml$1...@matchbox.inf.ed.ac.uk...

# Yes, 0^0 may be an arbitrary device to accommodate a computer, as Barron's
Maths Study Dictionary excludes zero(^0) (pg.62) when defining the "zero
exponent".
Which still leaves, eg. 5^0 = 1.
If I have five apples and multiply them by themselves zero times, do I
end up with one apple?
Also, does 5^0 = 5 * 1/5?
Has x^0 = 1 ever been proved? Or, is it merely a convenient dogma?

[LudovicoVan]

"DonH" <donlhu...@bigpond.com> wrote in message
news:zKpkr.6017$%E2....@viwinnwfe01.internal.bigpond.com...

> Has x^0 = 1 ever been proved? Or, is it merely a convenient dogma?

Some things get proved, some things get defined.

-LV

[Peter Webb]

"DonH" <donlhu...@bigpond.com> wrote in message
news:zKpkr.6017$%E2....@viwinnwfe01.internal.bigpond.com...

x^a * x^b = x^(a+b)
x^a * x^0 = x^(a+0) = x^a
x^0 = 1

[LudovicoVan]

"DonH" <donlhu...@bigpond.com> wrote in message

news:iskkr.6000$%E2....@viwinnwfe01.internal.bigpond.com...

> (Zero, while a mathematical symbol, is not a number, but denotes the
> absence of number?)

Of course zero, in modern math, is a number.

-LV

[Dave]

Your last statement assumes that x^a is defined and non-zero. But if x = 0 and a > 0, then x^a = 0, and if x = 0 and a < 0, then x^a is undefined. Thus, your "proof" breaks down in the case under discussion: 0^0.

Dave

[Peter Webb]

"Dave" <dave_an...@juno.com> wrote in message
news:26122529.1302.1334983483051.JavaMail.geo-discussion-forums@ynlp3...

I was responding to the last line of the previous post, "Has x^0 = 1 ever
been proved?". I am quite aware that there are restrictions on x and a. You
didn't even manage to list them all correctly; if x<0 and a is irrational
then x^a is undefined. (If you want to nit-pick, get it correct).

The OP clearly knows very little maths, and I just wanted to show him why
x^0 is defined as being equal to 1 in a simple way.

[David C. Ullrich]

That's far from the truth!

Consider power series. For example,

exp(x) = sum_{n=0}^infinity x^n/n!

If we didn't define 0^0 then the left side of that equation
would be undefined when x = 0. That would be bad;
exp is equal to its power series _except_ when x = 0?

[Richard Tobin]

In article <iskkr.6000$%E2....@viwinnwfe01.internal.bigpond.com>,
DonH <donlhu...@bigpond.com> wrote:

> So, if you have a lot of nothing and multiply it by itself no times, you
>get "one" - what?

To calculate 2^3, you don't "have a lot of two and multiply it by
itself three times".

On the other hand, you might say that you start with one, and multiply
it my two three times. So to get 0^0, you would start with one and
multiply it by zero no times.

-- Richard

[G. A. Edgar]

> x^0 = 1 for any x other than 0
> 0^x = 0 for any x other than 0

Woh! 0^(-1)=0 That will prove a lot of interesting things for me!

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

[DonH]

"Peter Webb" <r.peter...@gmail.com> wrote in message
news:jmtm3v$dhq$1...@news.albasani.net...

# If x^(a+0) = x^a, then the implication is that x^0 = 0; the x^0 is
dedundant.

[Frederick Williams]

DonH wrote:
>

>
> # If x^(a+0) = x^a, then the implication is that x^0 = 0; the x^0 is
> dedundant.

x^a = x^(a+0) = x^a x^0 rather implies that x^0 = 1.

--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting

[DonH]

"Frederick Williams" <freddyw...@btinternet.com> wrote in message
news:4F92FE5B...@btinternet.com...

> DonH wrote:
>>
>
>>
>> # If x^(a+0) = x^a, then the implication is that x^0 = 0; the x^0 is
>> dedundant.
>
> x^a = x^(a+0) = x^a x^0 rather implies that x^0 = 1.

# Does it? Maths' Order of Operations states that parenthesis is evaluated
first, which reduces (a+0) to a; leaving only x^a. Hence, x^0 = 0.

[Frederick Williams]

DonH wrote:
>
> "Frederick Williams" <freddyw...@btinternet.com> wrote in message
> news:4F92FE5B...@btinternet.com...
> > DonH wrote:
> >>
> >
> >>
> >> # If x^(a+0) = x^a, then the implication is that x^0 = 0; the x^0 is
> >> dedundant.
> >
> > x^a = x^(a+0) = x^a x^0 rather implies that x^0 = 1.
>
> # Does it? Maths' Order of Operations states that parenthesis is evaluated
> first, which reduces (a+0) to a; leaving only x^a. Hence, x^0 = 0.

p^(q+r) = p^q p^r

[Bart Goddard]

"DonH" <donlhu...@bigpond.com> wrote in
news:ZsDkr.6045$%E2....@viwinnwfe01.internal.bigpond.com:

>
> "Frederick Williams" <freddyw...@btinternet.com> wrote in message
> news:4F92FE5B...@btinternet.com...
>> DonH wrote:
>>>
>>
>>>
>>> # If x^(a+0) = x^a, then the implication is that x^0 = 0; the x^0
>>> is dedundant.
>>
>> x^a = x^(a+0) = x^a x^0 rather implies that x^0 = 1.
>
> # Does it? Maths' Order of Operations states that parenthesis is
> evaluated first, which reduces (a+0) to a; leaving only x^a. Hence,
> x^0 = 0.

One wonders how you get from x^a = x^a to the conclusion that
x^0 = 0.


--
Cheerfully resisting change since 1959.

[Jamerino Ball-sack]

Lern hau two uze Wikipedia, you fucking imbecaile.


http://en.wikipedia.org/wiki/0%5E0#Zero_to_the_zero_power

[Dave]

Whether x^a is complex or undefined when x < 0 and a is irrational depends on whether you say that sqrt(-1) is undefined or i, doesn't it?

Dave

[Pubkeybreaker]

On Apr 22, 2:22 am, Dave <dave_and_da...@juno.com> wrote:
> On Saturday, April 21, 2012 2:03:23 AM UTC-5, Peter Webb wrote:

> > "Dave" <dave_and_da...@juno.com> wrote in message

> >news:26122529.1302.1334983483051.JavaMail.geo-discussion-forums@ynlp3...
> > > On Friday, April 20, 2012 11:25:42 PM UTC-5, Peter Webb wrote:

> > >> "DonH" <donlhumphr...@bigpond.com> wrote in message
> > >>news:zKpkr.6017$%E2....@viwinnwfe01.internal.bigpond.com...
> > >> > "Richard Tobin" <rich...@cogsci.ed.ac.uk> wrote in message
> > >> >news:jmsvh1$nml$1...@matchbox.inf.ed.ac.uk...
> > >> >> In article
> > >> >> <fc51d462-6ab6-4087-97c1-63bda0170...@fv28g2000vbb.googlegroups.com>,

> Dave- Hide quoted text -

No, it doesn't.

[Frederick Williams]

Pubkeybreaker wrote:
>
> On Apr 22, 2:22 am, Dave <dave_and_da...@juno.com> wrote:

> >
> > Whether x^a is complex or undefined when x < 0 and a is irrational depends on whether you say that sqrt(-1) is undefined or i, doesn't it?

>

> No, it doesn't.

Well, I'm not so sure. If "you say" that sqrt(-1) is undefined, then
that implies that reals only are being considered; and in that case, if
x^a were (non-real) complex, then it would be undefined.

[Frederick Williams]

Dave wrote:

>
> Whether x^a is complex or undefined when x < 0 and a is irrational depends on whether you say that sqrt(-1) is undefined or i, doesn't it?

You're muddled. Towards an unmuddling: do you want your disjuncts to
correspond like this

... complex or undefined ...
undefined or i ...

(as you wrote), or like this

... complex or undefined ...
i or undefined ...

?

[Frederick Williams]

DonH wrote:

> (Zero, while a mathematical symbol, is not a number, but denotes the absence
> of number?)

Strictly speaking, if zero is a mathematical symbol, then it is not a
number. But the man in the street speaks (where numerals and numbers
are concerned) of symbols and things symbolized as if they were the
same.

It is understandable that the ancients found zero hard to understand,
but consider: you have eleven pounds in your bank account and you take
eleven pounds out. Your statement will read

... £11 ...
... £0 ...

that '0' doesn't indicate absence of number, it indicates absence of
money.

[Pubkeybreaker]

On Apr 22, 9:34 am, Frederick Williams <freddywilli...@btinternet.com>
wrote:

> Pubkeybreaker wrote:
>
> > On Apr 22, 2:22 am, Dave <dave_and_da...@juno.com> wrote:
>
> > > Whether x^a is complex or undefined when x < 0 and a is irrational depends on whether you say that sqrt(-1) is undefined or i, doesn't it?
>
> > No, it doesn't.
>
> Well, I'm not so sure.  If "you say" that sqrt(-1) is undefined, then
> that implies that reals only are being considered; and in that case, if
> x^a were (non-real) complex, then it would be undefined.

x^a is undefined EVEN WHEN sqrt(-1) IS defined.

(hint: x^a for a irrational has infinitely many possible values)

[Pfs...@aol.com]

On Sat, 21 Apr 2012 07:22:22 +1000, "DonH" <donlhu...@bigpond.com>

wrote:

>Well, according to my computer Basic program it does.

> So, if you have a lot of nothing and multiply it by itself no times, you
>get "one" - what?

> At least this is consistent with the zero power of any integer, as all
>give 1.
> Can any of this be proved?

>(Zero, while a mathematical symbol, is not a number, but denotes the absence
>of number?)
>

A definition of convenience --not a mathematical fact

[Tonico]

Well, I'm not sure: since we can put x^a = e^{a Log(x)} , with Log =
complex logarithm (Log Z = log|Z| + i*arg(z), with log = the usual
real logarithm function) it's all a matter of chosing some branch of
this function (many times "taking away" the non-positive real axis and
t), and thus we have a nice function: for irrational a we'd have that
its argument is 0 (or another integer multiple of 2Pi*i, if another
branch is chosen).

Tonio

[Frederick Williams]

Pubkeybreaker wrote:

>
> x^a is undefined EVEN WHEN sqrt(-1) IS defined.
>
> (hint: x^a for a irrational has infinitely many possible values)

I think I was taught (but it was a long time ago and I might have
misremembered) that

u^v = exp(v log u) by definition (and agreeing with certain special
sub-cases previously defined or
deduced)

with the principal branch of the logarithm being chosen. (Not that that
helps with 0^0.)

[Wally W.]

How is a definition in math *not* a mathematical fact?

Aren't definitions more basic than axioms?

[Tonico]

On Apr 22, 5:57 pm, Pfss...@aol.com wrote:
> On Sat, 21 Apr 2012 07:22:22 +1000, "DonH" <donlhumphr...@bigpond.com>

Err...can you name one "mathematical fact" that does not stem from
some definition?

Tonio

[DonH]

"Bart Goddard" <godd...@netscape.net> wrote in message
news:XnsA03CAF9BF5D61go...@74.209.136.98...

# Maybe. But I'd claim x^0 is actually a nonsense, as it tries to use a
non-number (which is what zero is) in a calculation. Zero merely denotes
the absence of number.

x^a * x^b = x^(a+b)

x^a * x^0 = x^(a+0) = x^a
x^0 = 1
... is Webb's original statement. Is x^0 a true algebraic expression, or
invalid?
Perhaps the situation can be resolved by dealing with actual entities.
Say, there are six apples.
This can be represented by 1+1+1+1+1+1, or 2+2+2, or 3+3. Or, even 2^2
+2.
But does 6^0 = 1?
That is, you multiply all six by itself no times, and you get one apple?
Or, do you regard "all six" as an aggregate? In which case, "one" has
special significance?
6 x 1 = 6. But 6 x 0 = 0... unless the original six remain; then 6 x 0 =
6
On the other hand, if you start with no apples, then 0 x 6 = 000000 = 0.
0^0 = 1? If you have "a" nothing, then you still have the original
"nothing", as one "entity"?
Or 0^0 = 0. Fair enough. But serve you right for using 0 in calculating.
PS: If a computer says 0^0=1, lest it go crazy, then it is giving the wrong
answer- and your faith in computers is diminished.

[Tonico]

On Apr 22, 10:11 pm, "DonH" <donlhumphr...@bigpond.com> wrote:
> "Bart Goddard" <goddar...@netscape.net> wrote in message
>
> news:XnsA03CAF9BF5D61go...@74.209.136.98...
>
>
>
>
>
> > "DonH" <donlhumphr...@bigpond.com> wrote in
> >news:ZsDkr.6045$%E2....@viwinnwfe01.internal.bigpond.com:
>
> >> "Frederick Williams" <freddywilli...@btinternet.com> wrote in message

> >>news:4F92FE5B...@btinternet.com...
> >>> DonH wrote:
>
> >>>> #  If  x^(a+0) = x^a, then the implication is that x^0 = 0; the x^0
> >>>> is dedundant.
>
> >>> x^a = x^(a+0) = x^a x^0 rather implies that x^0 = 1.
>
> >> # Does it?  Maths' Order of Operations states that parenthesis is
> >> evaluated first, which reduces (a+0) to a; leaving only x^a.  Hence,
> >> x^0 = 0.
>
> > One wonders how you get from x^a = x^a to the conclusion that
> > x^0 = 0.
>
> > --
> > Cheerfully resisting change since 1959.
>
> # Maybe.  But I'd claim x^0 is actually a nonsense, as it tries to use a
> non-number (which is what zero is) in a calculation.  Zero merely denotes
> the absence of number.

*** This, as your personal opinion, can be of some interest to some
people, but mathematicswise it is nonsense. ****

>    x^a * x^b = x^(a+b)
>    x^a * x^0 = x^(a+0) = x^a
>    x^0 = 1
>   ... is Webb's original statement.  Is x^0 a true algebraic expression, or
> invalid?

*** x^0 is an algebraic expression, and who knows what you meant by
"invalid" in this context...***

>   Perhaps the situation can be resolved by dealing with actual entities.
>   Say, there are six apples.
>   This can be represented by 1+1+1+1+1+1, or 2+2+2, or 3+3.  Or, even 2^2
> +2.
>   But does 6^0 = 1?

*** Yes, 6^0 = 1 because we, mathematicians, have agreen on this. It's
adefinition. ****

>   That is, you multiply all six by itself no times, and you get one apple?
>   Or, do you regard "all six" as an aggregate?  In which case, "one" has
> special significance?
>   6 x 1 = 6.   But 6 x 0 = 0... unless the original six remain; then 6 x 0 =
> 6
>   On the other hand, if you start with no apples, then 0 x 6 = 000000 = 0.
>   0^0 = 1?

*** You were already given a good deal of different opinions,
approaches and contexts in which one can decide 0^0 = 1 for some
particular case, sometimes one just decides to leave that expression
undefined, etc.

Tonio ***

 If you have "a" nothing, then you still have the original
> "nothing", as one "entity"?
>   Or 0^0 = 0.  Fair enough.  But serve you right for using 0 in calculating.
> PS: If a computer says 0^0=1, lest it go crazy, then it is giving the wrong

> answer- and your faith in computers is diminished.- Hide quoted text -
>
> - Show quoted text -

[Bart Goddard]

"DonH" <donlhu...@bigpond.com> wrote in
news:2KYkr.5997$v14...@viwinnwfe02.internal.bigpond.com:

Are you saying you can't answer my question? If you can't answer
my question, then how can you make the assertion above with
any confidence? You seem, on one hand, to want to not deal with
abstracts, and so you changed numbers to apples. But then
you write 6 x 1 = 6, but get confounded in that 6 meands 6 apples,
but 1 doesn't seem to mean 1 apples. (If it does, then the
answer is 6 apples squared, whatever that means.) If it doesn't
then you've failed in your attempt to "deal with actual entities",
because 1, is again, abstract.

So first decide whether the 1 in 6 x 1 means "one"(abstract)
or "one apple." Then try to answer my first question above.

[DonH]

"Frederick Williams" <freddyw...@btinternet.com> wrote in message

news:4F940CF3...@btinternet.com...

> DonH wrote:
>
>> (Zero, while a mathematical symbol, is not a number, but denotes the
>> absence
>> of number?)
>
> Strictly speaking, if zero is a mathematical symbol, then it is not a
> number. But the man in the street speaks (where numerals and numbers
> are concerned) of symbols and things symbolized as if they were the
> same.
>
> It is understandable that the ancients found zero hard to understand,
> but consider: you have eleven pounds in your bank account and you take
> eleven pounds out. Your statement will read
>
> ... £11 ...
> ... £0 ...
>
> that '0' doesn't indicate absence of number, it indicates absence of
> money.
>

# This illustrates the need for denoting what it is we are enumerating.
Every number should be associated with a physical entity.
" '0' doesn't indicate absence of number" - but it does: absence of a
number of pounds sterling.
Reminds me of an actual case, in which a person was sent a demand for $0
by the IRS. He phone them, and said he didn't owe anything. He got
reminder notices, demanding payment, but such ceased when he mailed them his
cheque for $0.

[Bart Goddard]

"DonH" <donlhu...@bigpond.com> wrote in news:R4Zkr.5998$v14.1462
@viwinnwfe02.internal.bigpond.com:

> Every number should be associated with a physical entity.

Then you're in trouble. What does (6 apples) X (6 apples)
enumerate? Is it 36 apple-squareds ? And what is 6 pears/2 apples?
Or even simpler, what is is (6 apples)/(2 apples)? It should be
3 somethings, but what does apple/apple enumerate?

If you insist on your statement above, then the only math
one can do is count sheep. Multiplication and division are
impossible.

There are two approaches to teaching calculus. The popular
one is to try to cut it up into tiny enough pieces that it
will fit into a freshman's small brain. The better approach
is to endeavor to make the freshman's brain bigger so that
calculus will fit comfortably inside it while remaining
intact.

Your attempt to avoid abstraction reminds me of the popular
method. You can eschew abstraction if you like (lots of
people do) but you can't simultaneously to mathematics.

[Virgil]

To have an unambiguous meaning for 0^0, it should be the case that

lim{x -> 0} 0^x = lim{x -> 0} x^0.

So until 0 = 1, it won't happen.
--

[Peter Webb]

"Tonico" <Toni...@yahoo.com> wrote in message
news:b46e5681-c7ea-4aec...@l18g2000vbx.googlegroups.com...

_______________________________________________
No. Doesn't work. The normal extension of a function f(x) from rationals to
irrationals is to define the value for an irrational x to be the limit of
the rational sequence (eg Cauchy sequence) that defines the irrational, eg
the limit of f(a_1), f(a_2) etc as a_n approaches the irrational number.
This doesn't work for exponentiation of a negative number, as different
rational sequences will produce different limits.

For rational powers, we have many solutions based on the branch/cut that is
chosen. For negative numbers raised to an irrational power, there are no
solutions and the function is undefined.

[Tonico]

On Apr 23, 5:05 am, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
> "Tonico" <Tonic...@yahoo.com> wrote in message

> solutions and the function is undefined.-


I almost beg to difer but the truth is I'm not that sure. I'd wish
David Ullrich or some other expert in complex functions would step
into this and help us (or, at least, me) clear this interesting issue.

Tonio

[Dan Christensen]

On Apr 20, 5:22 pm, "DonH" <donlhumphr...@bigpond.com> wrote:
> Well, according to my computer Basic program it does.
>    So, if you have a lot of nothing and multiply it by itself no times, you
> get "one" - what?
>    At least this is consistent with the zero power of any integer, as all
> give 1.
>    Can any of this be proved?

> (Zero, while a mathematical symbol, is not a number, but denotes the absence
> of number?)

I have always informally justified 0^0 being undefined as follows:

x^0 = x^(1-1) = x^1 / x^1 = x/x is undefined for x=0

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com
Also see demo video

[Jussi Piitulainen]

Dan Christensen writes:

> I have always informally justified 0^0 being undefined as follows:
>
> x^0 = x^(1-1) = x^1 / x^1 = x/x is undefined for x=0

Then x^n is undefined for any n at x = 0. Think x^3 = x^4 / x^1.

Then I suppose also 0 * 0 is undefined since it's 0^2. And 0 itself is
undefined since it's 0^1. And 0 * x is undefined for all x since it's
(0 * x)^1 = 0^1 * x^1.

I suggest you don't want to follow your own argument after all.

[Dan Christensen]

On Apr 23, 12:45 am, Jussi Piitulainen <jpiit...@ling.helsinki.fi>
wrote:

Ooops! Let me rephrase. Only if x =/= 0 can we say that x^0 = x/x = 1.

Does this mean that x^0 is "undefined" for x=0? It depends how you
define exponentiation for the integers. IIRC, the recursive definition
is usually given as: x^1=x and x^(n+1)=x^n * x. How do we attach a
meaning to x^0? Only if x=/=0 can we write x^n = x^(n+1)/x. Then, for
n=0, we have

x^0 = x^1 / x = 1

Thus, using the above definition, it seems we cannot attach any
meaning to 0^0, i.e. it is undefined.

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com

Also see video demo

[Jussi Piitulainen]

Dan Christensen writes:

> On Apr 23, 12:45 am, Jussi Piitulainen wrote:
> > Dan Christensen writes:
> > > I have always informally justified 0^0 being undefined as
> > > follows:
> >
> > > x^0 = x^(1-1) = x^1 / x^1 = x/x  is undefined for x=0
> >
> > Then x^n is undefined for any n at x = 0. Think x^3 = x^4 / x^1.
> >
> > Then I suppose also 0 * 0 is undefined since it's 0^2. And 0
> > itself is undefined since it's 0^1. And 0 * x is undefined for all
> > x since it's (0 * x)^1 = 0^1 * x^1.
> >
> > I suggest you don't want to follow your own argument after all.
>
> Ooops! Let me rephrase. Only if x =/= 0 can we say that x^0 = x/x = 1.
>
> Does this mean that x^0 is "undefined" for x=0? It depends how you
> define exponentiation for the integers. IIRC, the recursive
> definition is usually given as: x^1=x and x^(n+1)=x^n * x. How do we
> attach a meaning to x^0? Only if x=/=0 can we write x^n =
> x^(n+1)/x. Then, for n=0, we have
>
> x^0 = x^1 / x = 1
>
> Thus, using the above definition, it seems we cannot attach any
> meaning to 0^0, i.e. it is undefined.

Yes, if you leave it undefined, then it will be undefined.

If you want it defined, you define it, and you want all other things
to work well when you do that. You could start with x^0 = 1. It works.
Nothing else works.

Some want to define 0^0 because they want the binomial theorem to be
valid without unnecessary restrictions. (This seems related to the
polynomials that somebody already mentioned: they like to think of the
constant term as the coefficient of x^0 even when x = 0.)

Some like to define m^n as the number of functions from an n-element
set to an m-element set. There is one (rather vacuous) function from
the empty set to any set. By this definition m^0 = 1 for all m. (This
definition was already cited in the present thread.)

[David C. Ullrich]

On Sat, 21 Apr 2012 10:38:29 -0600, "G. A. Edgar"
<ed...@math.ohio-state.edu.invalid> wrote:

>> x^0 = 1 for any x other than 0
>> 0^x = 0 for any x other than 0

Since this is appearing in reply to a post of mine, I'd
like to point out that I didn't say the above. For some
reason we're responding to something Virgil said,
in a reply to my post.

>
>Woh! 0^(-1)=0 That will prove a lot of interesting things for me!

[Frederick Williams]

Dan Christensen wrote:

> Ooops! Let me rephrase. Only if x =/= 0 can we say that x^0 = x/x = 1.
>
> Does this mean that x^0 is "undefined" for x=0? It depends how you
> define exponentiation for the integers. IIRC, the recursive definition
> is usually given as: x^1=x and x^(n+1)=x^n * x. How do we attach a
> meaning to x^0? Only if x=/=0 can we write x^n = x^(n+1)/x. Then, for
> n=0, we have

If you are going to define exponentiation for the integers, then you
should recall that integers can be negative. Now consider x^(n+m) where
n and m are the negatives of one another.

[Frederick Williams]

DonH wrote:

>
> [... ] I'd claim x^0 is actually a nonsense, as it tries to use a

> non-number (which is what zero is) in a calculation. Zero merely denotes
> the absence of number.

It makes my draw drop with amazement, the way non-mathematicians tell
mathematicians(*) what they mean and what the facts of their subject
are. Are you aware that you're just making yourself look like a
complete idiot?

Should the commutative law be restated

x + y = y + x, except when either or both of x and y are zero.

And, if so, why is it never stated thus? And what about all the other
laws of arithmetic which pervade not just mathematics but everyday life
too?

(* Of whom I am not one. I am just an interested bystander.)

[Pubkeybreaker]

On Apr 23, 8:49 am, Frederick Williams <freddywilli...@btinternet.com>
wrote:

> DonH wrote:
>
> > [... ] I'd claim x^0 is actually a nonsense, as it tries to use a
> > non-number (which is what zero is) in a calculation.  Zero merely denotes
> > the absence of number.
>
> It makes my draw drop with amazement, the way non-mathematicians tell
> mathematicians(*) what they mean and what the facts of their subject
> are.  Are you aware that you're just making yourself look like a
> complete idiot?

The comparison is unfair. What do you have against idiots?

[Frederick Williams]

Frederick Williams wrote:

> It makes my draw drop with amazement,

Draw? Jaw probably.

[Pubkeybreaker]

On Apr 22, 10:05 pm, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
> "Tonico" <Tonic...@yahoo.com> wrote in message

>
> news:b46e5681-c7ea-4aec...@l18g2000vbx.googlegroups.com...
> On Apr 22, 5:06 pm, Pubkeybreaker <pubkeybrea...@aol.com> wrote:
>
>
>
>
>
> > On Apr 22, 9:34 am, Frederick Williams <freddywilli...@btinternet.com>
> > wrote:
>
> > > Pubkeybreaker wrote:
>
> > > > On Apr 22, 2:22 am, Dave <dave_and_da...@juno.com> wrote:
>
> > > > > Whether x^a is complex or undefined when x < 0 and a is irrational
> > > > > depends on whether you say that sqrt(-1) is undefined or i, doesn't
> > > > > it?
>
> > > > No, it doesn't.
>
> > > Well, I'm not so sure. If "you say" that sqrt(-1) is undefined, then
> > > that implies that reals only are being considered; and in that case, if
> > > x^a were (non-real) complex, then it would be undefined.
>
> > x^a is undefined EVEN WHEN sqrt(-1) IS defined.
>
> > (hint: x^a for a irrational has infinitely many possible values)
>
> Well, I'm not sure: since we can put x^a = e^{a Log(x)} , with Log =
> complex logarithm (Log Z = log|Z| + i*arg(z), with log = the usual
> real logarithm function) it's all a matter of chosing some branch of
> this function

Yes, *we can take the principal branch* of the logarithm function.

x^a in this instance does have infinitely many values, UNLESS one
restricts
to (as you say) the principal branch. But this restriction was NOT
part of
the problem statement!

AS ORIGINALLY STATED, x^a is undefined.

Under the slection of principal branch, x^a becomes defined.

But it is a matter of definition.

[Dan Christensen]

On Apr 23, 3:18 am, Jussi Piitulainen <jpiit...@ling.helsinki.fi>
wrote:

> Dan Christensen writes:
> > On Apr 23, 12:45 am, Jussi Piitulainen wrote:
> > > Dan Christensen writes:
> > > > I have always informally justified 0^0 being undefined as
> > > > follows:
>
> > > > x^0 = x^(1-1) = x^1 / x^1 = x/x  is undefined for x=0
>
> > > Then x^n is undefined for any n at x = 0. Think x^3 = x^4 / x^1.
>
> > > Then I suppose also 0 * 0 is undefined since it's 0^2. And 0
> > > itself is undefined since it's 0^1. And 0 * x is undefined for all
> > > x since it's (0 * x)^1 = 0^1 * x^1.
>
> > > I suggest you don't want to follow your own argument after all.
>
> > Ooops! Let me rephrase. Only if x =/= 0 can we say that x^0 = x/x = 1.
>
> > Does this mean that x^0 is "undefined" for x=0? It depends how you
> > define exponentiation for the integers. IIRC, the recursive
> > definition is usually given as: x^1=x and x^(n+1)=x^n * x. How do we
> > attach a meaning to x^0? Only if x=/=0 can we write x^n =
> > x^(n+1)/x. Then, for n=0, we have
>
> > x^0 = x^1 / x = 1
>
> > Thus, using the above definition, it seems we cannot attach any
> > meaning to 0^0, i.e. it is undefined.
>
> Yes, if you leave it undefined, then it will be undefined.
>
> If you want it defined, you define it, and you want all other things
> to work well when you do that. You could start with x^0 = 1. It works.
> Nothing else works.
>

Nothing? Why couldn't you just add to the above recursive definition,
say, the "fact" that 0^0=19 (or any other integer)? It would "work" in
the sense that it would be consistent with any statement formally
derived from the original definition.

[Frederick Williams]

Dan Christensen wrote:

> Why couldn't you just add to the above recursive definition,
> say, the "fact" that 0^0=19 (or any other integer)? It would "work" in
> the sense that it would be consistent with any statement formally
> derived from the original definition.

Because, instead of writing (for example)

If ...x^y... then ...

one would be forever writing

If ...x^y (except when x = 0 = y) ... then ...
If ...0^0... then ....

Why be wilfully stupid? It would be like using the word 'door' to mean
what it usually means except when talking about No. 42, The High Street,
when it means window. Why the fuck would anybody do that?

[Jussi Piitulainen]

Dan Christensen writes:
> On Apr 23, 3:18 am, Jussi Piitulainen

You mean first define x^0 = 1 for all x and then add 0^0 = 19,
immediately making 1 = 19? I don't think so.

You mean first define x^0 = 1 for all non-0 x and then add 0^0 = 19?
The binomial theorem gives unusual results with 0^0 = 19:

(x + y)^n = sum for k = 0, ..., n of C(n,k) x^k y^(n - k)

(x + 0)^2 = C(2,0) x^0 0^2 + C(2,1) x^1 0^1 + C(2,2) x^2 0^0
= 0 + 0 + x^2 * 19

Assuming x^2 != 0, again 1 = 19. No, I don't think so.

That's what I mean by works.

[Dan Christensen]

On Apr 23, 10:18 am, Jussi Piitulainen <jpiit...@ling.helsinki.fi>

No.

> I don't think so.
>
> You mean first define x^0 = 1 for all non-0 x and then add 0^0 = 19?

Yes.

x^0 = 19 if x=0; 1 otherwise

(Note: I am NOT claiming that 0^0 = 19. I am trying to make the case
that 0^0 should be left undefined.)

> The binomial theorem gives unusual results with 0^0 = 19:
>
> (x + y)^n = sum for k = 0, ..., n of C(n,k) x^k y^(n - k)
>

[snip]

This theorem could not be formally derived from the usual definition
of integer exponentiation (above). You could, however, prove:

(x + y)^n = x^n if y = 0; sum for k = 0, ..., n of C(n,k) x^k y^(n -
k) otherwise

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com

Also see video demo.

[Jussi Piitulainen]

Dan Christensen writes:

> On Apr 23, 10:18 am, Jussi Piitulainen wrote:
> > You mean first define x^0 = 1 for all non-0 x and then add 0^0 =
> > 19?
>
> Yes.
>
> x^0 = 19 if x=0; 1 otherwise
>
> (Note: I am NOT claiming that 0^0 = 19. I am trying to make the case
> that 0^0 should be left undefined.)

Yes. Why?

And why not show something that works _better_ with 0^0 undefined? Now
you are adding unnecessary conditions to a useful general theorem and
gaining nothing, which makes it _worse_.

> > The binomial theorem gives unusual results with 0^0 = 19:
> >
> > (x + y)^n = sum for k = 0, ..., n of C(n,k) x^k y^(n - k)
>
> [snip]
>
> This theorem could not be formally derived from the usual definition
> of integer exponentiation (above). You could, however, prove:
>
> (x + y)^n = x^n if y = 0; sum for k = 0, ..., n of C(n,k) x^k y^(n -
> k) otherwise

You need to handle separately also the cases where x = 0, and leave
undefined the cases where x + y = 0 and n = 0. For no gain.

[Dan Christensen]

On Apr 23, 10:17 am, Frederick Williams

<freddywilli...@btinternet.com> wrote:
> Dan Christensen wrote:
> > Why couldn't you just add to the above recursive definition,
> > say, the "fact" that 0^0=19 (or any other integer)? It would "work" in
> > the sense that it would be consistent with any statement formally
> > derived from the original definition.
>
> Because, instead of writing (for example)
>
>    If ...x^y... then ...
>
> one would be forever writing
>
>    If ...x^y (except when x = 0 = y) ... then ...

Yes.

>    If ...0^0... then ....
>

More like...

Otherwise, if x = 0 = y... then....


> Why be wilfully stupid?  [snipping more of the same]

Do try to be civil, Freddie.

[DonH]

"Bart Goddard" <godd...@netscape.net> wrote in message

news:XnsA03DA8C3E275Ago...@74.209.136.95...

> "DonH" <donlhu...@bigpond.com> wrote in news:R4Zkr.5998$v14.1462
> @viwinnwfe02.internal.bigpond.com:
>
>> Every number should be associated with a physical entity.
>
> Then you're in trouble. What does (6 apples) X (6 apples)
> enumerate? Is it 36 apple-squareds ? And what is 6 pears/2 apples?
> Or even simpler, what is is (6 apples)/(2 apples)? It should be
> 3 somethings, but what does apple/apple enumerate?
>
> If you insist on your statement above, then the only math
> one can do is count sheep. Multiplication and division are
> impossible.
>

# This is where Science differs from Philosophy or Religion - Science DOES
count things.
What is the basis of the Theory of Biological Evolution, or of AGW? Lots
of facts.
Which is why Set Theory is irrelevant to Science, whereas Classification
(a sub-set) is.
As for "(6 apples) X (6 apples)", this is a nonsense, as we've already
established that we are dealing with apples. 6 x 6, of apples, can be
re-stated as addition (multiplication is a shorthand way of adding).
Once you get away from actual things, then there's a tendency to drift
into the stratosphere - as the European financial crisis demonstrates, where
money is now more important than people.

[Dan Christensen]

On Apr 23, 11:47 am, Jussi Piitulainen <jpiit...@ling.helsinki.fi>
wrote:

> Dan Christensen writes:
> > On Apr 23, 10:18 am, Jussi Piitulainen wrote:
> > > You mean first define x^0 = 1 for all non-0 x and then add 0^0 =
> > > 19?
>
> > Yes.
>
> > x^0 = 19 if x=0; 1 otherwise
>
> > (Note: I am NOT claiming that 0^0 = 19. I am trying to make the case
> > that 0^0 should be left undefined.)
>
> Yes. Why?
>

Because it is unnecessary to assume 0^0 = 1.

> And why not show something that works _better_ with 0^0 undefined? Now
> you are adding unnecessary conditions to a useful general theorem and
> gaining nothing, which makes it _worse_.
>

You gain some certainty. You aren't just fudging your axioms to get
the required result.

> > > The binomial theorem gives unusual results with 0^0 = 19:
>
> > > (x + y)^n = sum for k = 0, ..., n of C(n,k) x^k y^(n - k)
>
> > [snip]
>
> > This theorem could not be formally derived from the usual definition
> > of integer exponentiation (above). You could, however, prove:
>
> > (x + y)^n = x^n if y = 0; sum for k = 0, ..., n of C(n,k) x^k y^(n -
> > k) otherwise
>
> You need to handle separately also the cases where x = 0,
> and leave
> undefined the cases where x + y = 0 and n = 0.

Yes. Just as you cannot divide by zero, so you can cannot raise 0 to
the power of 0.

In the case of the binomial theorem, I think you can prove...

(x+y)^n = x^n if y=0 and not x = n = 0
= y^n if x=0 and not y = n = 0
= 0 if x+y = 0 and n > 0
= sum for k = 0, ..., n of C(n,k) x^k y^(n - k) if x =/= 0 and
y =/= 0 and not x+y = n = 0

(Have I covered all the cases?)

> For no gain.

You introduce what amounts to an unnecessary axiom (0^0 = 1) to deal
with a few pathological cases in number theory that can be easily
handled in other ways. A questionable "gain" in my view.

[Frederick Williams]

Dan Christensen wrote:

> You gain some certainty. You aren't just fudging your axioms to get
> the required result.

[...]

>
> In the case of the binomial theorem, I think you can prove...
>
> (x+y)^n = x^n if y=0 and not x = n = 0
> = y^n if x=0 and not y = n = 0
> = 0 if x+y = 0 and n > 0
> = sum for k = 0, ..., n of C(n,k) x^k y^(n - k) if x =/= 0 and
> y =/= 0 and not x+y = n = 0
>
> (Have I covered all the cases?)

That parenthetical remark is very revealing. By having to deal with
particular cases separately, the possibility of overlooking one arises.
Were is this certainty that has been gained?

> > For no gain.
>
> You introduce what amounts to an unnecessary axiom (0^0 = 1) to deal
> with a few pathological cases in number theory that can be easily
> handled in other ways.

How do you know it's a few? We're talking about _every_ time
expressions of the form x^y appear.

Also, 0^0 = 1 is not an axioms. Either it is a consequence of axioms
and definitions "upstream", or it is a definition itself.

[Dan Christensen]

On Apr 23, 5:58 pm, Frederick Williams <freddywilli...@btinternet.com>
wrote:

> Dan Christensen wrote:
> > You gain some certainty. You aren't just fudging your axioms to get
> > the required result.
>
> [...]
>
>
>
> > In the case of the binomial theorem, I think you can prove...
>
> > (x+y)^n = x^n if y=0 and not x = n = 0
> >         = y^n if x=0 and not y = n = 0
> >         = 0 if x+y = 0 and n > 0
> >         = sum for k = 0, ..., n of C(n,k) x^k y^(n - k) if x =/= 0 and
> > y =/= 0 and not x+y = n = 0
>
> > (Have I covered all the cases?)
>
> That parenthetical remark is very revealing.  By having to deal with
> particular cases separately, the possibility of overlooking one arises.

A formal proof would no doubt clarify matters.

> Were is this certainty that has been gained?
>

I would be more certain of the result if I did not have to resort to
seemingly ad hoc definitions/axioms like 0^0=1.

> > > For no gain.
>
> > You introduce what amounts to an unnecessary axiom (0^0 = 1) to deal
> > with a few pathological cases in number theory that can be easily
> > handled in other ways.
>
> How do you know it's a few?  We're talking about _every_ time
> expressions of the form x^y appear.

Likewise for division by zero. Every time you have x/y, you must test
for y=0.

>
> Also, 0^0 = 1 is not an axioms.  Either it is a consequence of axioms
> and definitions "upstream", or it is a definition itself.
>

I doubt that it can be derived from the usual definitions and axioms
of set and number theory, but I remain open to the possibility.

[Frederick Williams]

Dan Christensen wrote:
>
> On Apr 23, 5:58 pm, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
> > Dan Christensen wrote:

> > > In the case of the binomial theorem, I think you can prove...
> >
> > > (x+y)^n = x^n if y=0 and not x = n = 0
> > > = y^n if x=0 and not y = n = 0
> > > = 0 if x+y = 0 and n > 0
> > > = sum for k = 0, ..., n of C(n,k) x^k y^(n - k) if x =/= 0 and
> > > y =/= 0 and not x+y = n = 0
> >
> > > (Have I covered all the cases?)
> >
> > That parenthetical remark is very revealing. By having to deal with
> > particular cases separately, the possibility of overlooking one arises.
>
> A formal proof would no doubt clarify matters.
>
> > Were is this certainty that has been gained?
> >
>
> I would be more certain of the result if I did not have to resort to
> seemingly ad hoc definitions/axioms like 0^0=1.

It isn't ad hoc. One indication that it isn't is that the multiplicity
of cases in '(x+y)^n = ...' is unnecessary.

> > > > For no gain.
> >
> > > You introduce what amounts to an unnecessary axiom (0^0 = 1) to deal
> > > with a few pathological cases in number theory that can be easily
> > > handled in other ways.
> >
> > How do you know it's a few? We're talking about _every_ time
> > expressions of the form x^y appear.
>
> Likewise for division by zero. Every time you have x/y, you must test
> for y=0.

The fact is 0^0 can always be sensibly specified, in general x/0
cannot. So the two cases aren't as similar as you may think.

> > Also, 0^0 = 1 is not an axioms. Either it is a consequence of axioms
> > and definitions "upstream", or it is a definition itself.
>
> I doubt that it can be derived from the usual definitions and axioms
> of set and number theory, but I remain open to the possibility.

That's very gracious of you.

[Bart Goddard]

"DonH" <donlhu...@bigpond.com> wrote in
news:nAglr.6117$%E2....@viwinnwfe01.internal.bigpond.com:

I believe, and this is saying a lot, that you are the dumbest
person I've ever run across. Each "sentence" you post is
dumber than the previous one. I have to killfile you now
because I don't want to know how much dumber things will get.

[Frederick Williams]

DonH wrote:

> > "DonH" <donlhu...@bigpond.com> wrote in news:R4Zkr.5998$v14.1462
> > @viwinnwfe02.internal.bigpond.com:
> >
> >> Every number should be associated with a physical entity.

> As for "(6 apples) X (6 apples)", this is a nonsense, as we've already
> established that we are dealing with apples. 6 x 6, of apples, can be
> re-stated as addition (multiplication is a shorthand way of adding).

To all of us bar you, 6 x 6 apples is just 36 apples. But you want
every number to be associated with a physical entity, so, for you it
must mean (6 ...) x (6 apples). What physical entity is that ...?

[Pubkeybreaker]

On Apr 24, 11:15 am, Frederick Williams
<freddywilli...@btinternet.com> wrote:
> DonH wrote:
> > > "DonH" <donlhumphr...@bigpond.com> wrote in news:R4Zkr.5998$v14.1462

> > > @viwinnwfe02.internal.bigpond.com:
>
> > >>    Every number should be associated with a physical entity.
> >     As for "(6 apples) X (6 apples)", this is a nonsense, as we've already
> > established that we are dealing with apples.  6 x 6, of apples, can be
> > re-stated as addition (multiplication is a shorthand way of adding).
>
> To all of us bar you, 6 x 6 apples is just 36 apples.  But you want
> every number to be associated with a physical entity, so, for you it
> must mean (6 ...) x (6 apples).  What physical entity is that ...?
>

Isn't it obvious? 6 apples x 6 apples = 36 apples^2.

And you thought that apples were round.........

[Helmut Richter]

On Fri, 20 Apr 2012, Virgil wrote:

> x^0 = 1 for any x other than 0

x^0 is 1 because if a product has no factors at all, it is reasonable to
take the neutral element of multiplication, so that the multiplication
with no factor does not change the result. This argument is independent
of x.

> 0^x = 0 for any x other than 0

0^x is 0 because a product is zero if it has at least one factor which is
zero. This argument is meaningless for x=0 because then there is not "at
least one factor" which is 0.

> Virtually the only reason to define ANY value for 0^0 is so that
> calculators and computers do not throw a fit when they try to evaluate
> it.

The are many other reasons. For instance, it is useful when the formula

exp(x) = Sum (i=0...oo) x^i/i!

is also defined for x=0 and yields the correct value which would not be
the case when 0^0 were undefined or different from 1.

--
Helmut Richter

[Dan Christensen]

On Apr 24, 10:01 am, Frederick Williams

<freddywilli...@btinternet.com> wrote:
> Dan Christensen wrote:
>
> > On Apr 23, 5:58 pm, Frederick Williams <freddywilli...@btinternet.com>
> > wrote:
> > > Dan Christensen wrote:
> > > > In the case of the binomial theorem, I think you can prove...
>
> > > > (x+y)^n = x^n if y=0 and not x = n = 0
> > > >         = y^n if x=0 and not y = n = 0
> > > >         = 0 if x+y = 0 and n > 0
> > > >         = sum for k = 0, ..., n of C(n,k) x^k y^(n - k) if x =/= 0 and
> > > > y =/= 0 and not x+y = n = 0
>
> > > > (Have I covered all the cases?)
>
> > > That parenthetical remark is very revealing.  By having to deal with
> > > particular cases separately, the possibility of overlooking one arises.
>
> > A formal proof would no doubt clarify matters.
>
> > > Were is this certainty that has been gained?
>
> > I would be more certain of the result if I did not have to resort to
> > seemingly ad hoc definitions/axioms like 0^0=1.
>
> It isn't ad hoc.

It isn't included in the usual recursive definition: x^1 = x and x^(n
+1) = x^n * x. And I don't think it can be derived from this
definition. This makes me think it is ad hoc.

> One indication that it isn't is that the multiplicity
> of cases in '(x+y)^n = ...' is unnecessary.
>

It must be possible to prove a workable version of BT without
resorting to 0^0=1. As we see here, the different cases are not
onerous. Just avoid 0^0. Just like you normally avoid x/0. It looks
easy enough to work around.

> > > > > For no gain.
>
> > > > You introduce what amounts to an unnecessary axiom (0^0 = 1) to deal
> > > > with a few pathological cases in number theory that can be easily
> > > > handled in other ways.
>
> > > How do you know it's a few?  We're talking about _every_ time
> > > expressions of the form x^y appear.
>
> > Likewise for division by zero. Every time you have x/y, you must test
> > for y=0.
>
> The fact is 0^0 can always be sensibly specified, in general x/0
> cannot.  So the two cases aren't as similar as you may think.
>

Really, it needs a stronger basis than it can be used to generate cool
results. Especially if you can obtain equivalent results without it.

[Dan Christensen]

On Apr 24, 1:18 pm, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote:

Have a look at the following informal proof of BT:
http://www.mcs.sdsmt.edu/~ecorwin/cs251/bin_thm/bin_thm.html

The author doesn't take into consideration that 0^0 may be undefined,
but the only bases used there are x, y and x+y. Assume neither are
zero and the proof would work, and the formula would apply. The other
cases can be handled trivially (see my previous postings here).

[Frederick Williams]

Dan Christensen wrote:

>
> It isn't included in the usual recursive definition: x^1 = x and x^(n
> +1) = x^n * x. And I don't think it can be derived from this
> definition. This makes me think it is ad hoc.

It's not surprising if a definition of x^n for n = 1, 2, 3, ... cannot
reveal what 0^0 is.

[Frederick Williams]

Dan Christensen wrote:

>
> Have a look at the following informal proof of BT:
> http://www.mcs.sdsmt.edu/~ecorwin/cs251/bin_thm/bin_thm.html

It doesn't help your case. The author writes "n = 0 works in the sense
that both sides are 1,...". The formula he is talking about is

(x + y)^n = sum_{j=0}^n (n choose j) x^{n-j} y^j

>
> The author doesn't take into consideration that 0^0 may be undefined,
> but the only bases used there are x, y and x+y. Assume neither are
> zero and the proof would work,

Assume x and y are zero, or x + y is zero, and it still works.

[DonH]

"Bart Goddard" <godd...@netscape.net> wrote in message

news:XnsA03F5EB8E48B3go...@74.209.136.99...

# Aw, what have you got against dumb people? Why not deaf or blind?
All I can say is, that if 0^0=1, then this post has taken a long time to
conclude, one way or other.

[Tonico]

On Apr 25, 3:35 am, "DonH" <donlhumphr...@bigpond.com> wrote:
> "Bart Goddard" <goddar...@netscape.net> wrote in message
>
> news:XnsA03F5EB8E48B3go...@74.209.136.99...
>
>
>
>
>
> > "DonH" <donlhumphr...@bigpond.com> wrote in
> >news:nAglr.6117$%E2....@viwinnwfe01.internal.bigpond.com:
>
> >> "Bart Goddard" <goddar...@netscape.net> wrote in message
> >>news:XnsA03DA8C3E275Ago...@74.209.136.95...
> >>> "DonH" <donlhumphr...@bigpond.com> wrote in news:R4Zkr.5998$v14.1462

> conclude, one way or other.-


Perhaps you concluded 0^0 = 1, but nobody else, as far as I can see,
did. You got lot of opinions, ideas, definitions and stuff. That's
all.

Tonio

[Dan Christensen]

On Apr 24, 4:54 pm, Frederick Williams <freddywilli...@btinternet.com>
wrote:

> Dan Christensen wrote:
>
> > Have a look at the following informal proof of BT:
> >http://www.mcs.sdsmt.edu/~ecorwin/cs251/bin_thm/bin_thm.html
>
> It doesn't help your case.  The author writes "n = 0 works in the sense
> that both sides are 1,...".  The formula he is talking about is
>
>   (x + y)^n = sum_{j=0}^n (n choose j) x^{n-j} y^j
>

Look harder....

There, the author is talking about the base case (proof by induction).
You have a zero exponent on the y in the 1st term, and on the x in
the 2nd term. When x or y is 0, the result will be undefined.

>
>
> > The author doesn't take into consideration that 0^0 may be undefined,
> > but the only bases used there are x, y and x+y. Assume neither are
> > zero and the proof would work,
>
> Assume x and y are zero, or x + y is zero, and it still works.
>

Each has an exponent that can be 0. In that case, they would be
undefined. So, you establish the formula for non-zero x, y and x+y,
and then establish the required result for each zero case.

[Dan Christensen]

On Apr 24, 4:39 pm, Frederick Williams <freddywilli...@btinternet.com>
wrote:

> Dan Christensen wrote:
>
> > It isn't included in the usual recursive definition:  x^1 = x and x^(n
> > +1) = x^n * x. And I don't think it can be derived from this
> > definition. This makes me think it is ad hoc.
>
> It's not surprising if a definition of x^n for n = 1, 2, 3, ... cannot
> reveal what 0^0 is.
>

Also for n = 0, -1, -2, -3, ...

Again, for x=/=0, we have x^n = (x^n+1)/x.

So, for x=/=0, we have x^0 = 1, x^(-1) = 1/x, x^(-2) = 1/(x^2), etc.

[Virgil]

Those who argue that 0^0 must have some unique value should note that
0^0 cannot be required to have a unique value until

lim_{x ->0} x^0 = lim_{y -> 0} 0^y
--

[Jussi Piitulainen]

I don't see any problem. The definition is not motivated by limits.
Limits remain the same as before, and you will keep warning people
that 0^0 is an indeterminate limiting form, or whatever the term is.

[Dan Christensen]

At a more fundamental level -- in the very definition of integer (or
natural number) exponentiation -- 0^0 is left undefined. But your
argument for the reals puts the final nail in coffin! Thanks.

[Jussi Piitulainen]

Dan Christensen writes:

> Because it is unnecessary to assume 0^0 = 1.

You are of course free to work in a restricted system where you leave
it out. Other people need it defined. I think my main theme below is
that 0^0 is _naturally_ 1, and it is artificial to exclude it. So: we
disagree.

0^0 is the number of functions from the empty set to the empty set. In
general, m^n is the number of functions from an n-element set to an
m-element set. The number of ways to distribute n marbles in m boxes.

0^0 is the empty product. In general, x^n is the product of n x's, and
then x^0 is the empty product.

Similarly, a factorial n! is the product of the n numbers from 1 to n,
and 0! = 1 is the product of no numbers. It also has a natural
combinatorial interpretation.

It is usual to define powers 0, 1, 2, ... so that x^0 is the neutral
element.

> Yes. Just as you cannot divide by zero, so you can cannot raise 0 to
> the power of 0.

Division by zero is quite different. Zero cannot have a multiplicative
inverse, call it w, because then we would have both 0*w = 0 (zero does
that to any number) and 0*w = 1 (the multiplicative inverse does that
to the number), which is a contradiction. I'm taking for granted that
(a = b and a = c) implies a = c, and not 0 = 1.

I think division by zero would continue to cause all manner of
problems even if it were arbitrarily defined. In contrast, 0^0 does
not cause any problems at all.

> In the case of the binomial theorem, I think you can prove...
>
> (x+y)^n = x^n if y=0 and not x = n = 0
> = y^n if x=0 and not y = n = 0
> = 0 if x+y = 0 and n > 0
> = sum for k = 0, ..., n of C(n,k) x^k y^(n - k) if x =/= 0 and
> y =/= 0 and not x+y = n = 0
>
> (Have I covered all the cases?)

I'm told it's also valid for negative and non-integer exponents when x
and y are such that the infinite series converges. One actually leaves
k unbounded, making the expression simpler. In the usual case, C(n, k)
becomes zero outside the usual bounds, and then the sum is finite.

My source says the theorem is "too important to be arbitrarily
restricted", which is what happens when one leaves 0^0 undefined.

> You introduce what amounts to an unnecessary axiom (0^0 = 1) to deal
> with a few pathological cases in number theory that can be easily
> handled in other ways. A questionable "gain" in my view.

Not few and not pathological. It turns up and behaves naturally. Its
exclusion is artificial.

And not handled easily. It's a lot of unnecessary special cases that
turn out to be not special after all. You demonstrate this above.

I'm not sure what axiom system you are referring to, or what purpose
you have in mind when you imply that 0^0 = 1 is unnecessary.

[Jesse F. Hughes]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> On Apr 24, 4:39 pm, Frederick Williams <freddywilli...@btinternet.com>

> wrote:
>> Dan Christensen wrote:
>>
>> > It isn't included in the usual recursive definition:  x^1 = x and x^(n
>> > +1) = x^n * x. And I don't think it can be derived from this
>> > definition. This makes me think it is ad hoc.
>>
>> It's not surprising if a definition of x^n for n = 1, 2, 3, ... cannot
>> reveal what 0^0 is.
>>
>

> Yes, 0^0 is UNDEFINED!

Er, Dan? 1^0 is also left undefined by that definition. As are
2^{1/2}, e^{i*pi} and so on.

Duh.

--
Jesse F. Hughes

"The Hammer has arrived."
-- James S. Harris, Feb. 14 2006

[Frederick Williams]

Dan Christensen wrote:
>
> On Apr 24, 4:39 pm, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
> > Dan Christensen wrote:
> >
> > > It isn't included in the usual recursive definition: x^1 = x and x^(n
> > > +1) = x^n * x. And I don't think it can be derived from this
> > > definition. This makes me think it is ad hoc.
> >
> > It's not surprising if a definition of x^n for n = 1, 2, 3, ... cannot
> > reveal what 0^0 is.
> >
>
> Also for n = 0, -1, -2, -3, ...

Ok, I misunderstood your use of the phrase 'recursive definition'.
Let's see how you define x^-3:

x^-3 = x^-4 * x = (x^-5 * x) * x = ...

Compare x^3:

x^3 = x^2 * x = (x^1 * x) * x = (x * x) * x.

The point of recursive definitions, as I understand the phrase, is that
such a sequence comes to an end. In the x^3 case it comes to an end
because the natural numbers are well ordered. In the x^-3 it doesn't
because the integers aren't.

[Frederick Williams]

Dan Christensen wrote:
>
> On Apr 24, 4:39 pm, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
> > Dan Christensen wrote:
> >
> > > It isn't included in the usual recursive definition: x^1 = x and x^(n
> > > +1) = x^n * x. And I don't think it can be derived from this
> > > definition. This makes me think it is ad hoc.
> >
> > It's not surprising if a definition of x^n for n = 1, 2, 3, ... cannot
> > reveal what 0^0 is.
> >
>
> Also for n = 0, -1, -2, -3, ...
>
> Again, for x=/=0, we have x^n = (x^n+1)/x.
>
> So, for x=/=0, we have x^0 = 1, x^(-1) = 1/x, x^(-2) = 1/(x^2), etc.

Let's see how 'the usual recursive definition: x^1 = x and x^(n +1) =
x^n * x' (your words) works in the case n =0:

x^0 = x^-1 * x = (x^-2 * x) * x = ((x^-3 * x) * x) * x = ...

[Frederick Williams]

Dan Christensen wrote:
>
> On Apr 24, 4:39 pm, Frederick Williams <freddywilli...@btinternet.com>

> wrote:
> > Dan Christensen wrote:
> >
> > > It isn't included in the usual recursive definition: x^1 = x and x^(n
> > > +1) = x^n * x. And I don't think it can be derived from this
> > > definition. This makes me think it is ad hoc.
> >
> > It's not surprising if a definition of x^n for n = 1, 2, 3, ... cannot
> > reveal what 0^0 is.
> >
>

> Yes, 0^0 is UNDEFINED!

If I look in the dictionary for the definition of 'xylophone', I will
find that 'trombone' is undefined. The answer is to look elsewhere in
the dictionary for the definition of 'trombone'.

[Dan Christensen]

On Apr 25, 2:22 am, Jussi Piitulainen <jpiit...@ling.helsinki.fi>
wrote:

> Dan Christensen writes:
> > Because it is unnecessary to assume 0^0 = 1.
>
> You are of course free to work in a restricted system where you leave
> it out. Other people need it defined. I think my main theme below is
> that 0^0 is _naturally_ 1, and it is artificial to exclude it. So: we
> disagree.
>
> 0^0 is the number of functions from the empty set to the empty set. In
> general, m^n is the number of functions from an n-element set to an
> m-element set. The number of ways to distribute n marbles in m boxes.
>
> 0^0 is the empty product. In general, x^n is the product of n x's, and
> then x^0 is the empty product.
>

Seems like a lot of hand waving to me. I have always seen
exponentiation defined recursively for integer or natural number
exponents n as:

x^1 = x
x^(n+1)= x^n * x.

For integer n, and non-zero x, we have x^n=x^(n+1)/x. Thus x^0 = 1 for
non-zero x and x^0 is left undefined for x=0.

> Similarly, a factorial n! is the product of the n numbers from 1 to n,
> and 0! = 1 is the product of no numbers. It also has a natural
> combinatorial interpretation.
>

The factorial function is usually defined recursively on the natural
numbers as follows:

1! = 1
(n+1)! = n! * (n+1)

Since n+1 is non-zero, we also have n! = (n+1)!/(n+1) for ALL natural
numbers n. Thus 0!=1 (if you include 0 in N).

> It is usual to define powers 0, 1, 2, ... so that x^0 is the neutral
> element.
>
> > Yes. Just as you cannot divide by zero, so you can cannot raise 0 to
> > the power of 0.
>
> Division by zero is quite different. Zero cannot have a multiplicative
> inverse, call it w, because then we would have both 0*w = 0 (zero does
> that to any number) and 0*w = 1 (the multiplicative inverse does that
> to the number), which is a contradiction. I'm taking for granted that
> (a = b and a = c) implies a = c, and not 0 = 1.

Good point. x/y is is undefined for y=0. As a direct result, x^0 is
undefined for x=0 (see above).

>
> I think division by zero would continue to cause all manner of
> problems even if it were arbitrarily defined. In contrast, 0^0 does
> not cause any problems at all.
>
> > In the case of the binomial theorem, I think you can prove...
>
> > (x+y)^n = x^n if y=0 and not x = n = 0
> >         = y^n if x=0 and not y = n = 0
> >         = 0 if x+y = 0 and n > 0
> >         = sum for k = 0, ..., n of C(n,k) x^k y^(n - k) if x =/= 0 and
> > y =/= 0 and not x+y = n = 0
>
> > (Have I covered all the cases?)
>
> I'm told it's also valid for negative and non-integer exponents when x
> and y are such that the infinite series converges. One actually leaves
> k unbounded, making the expression simpler. In the usual case, C(n, k)
> becomes zero outside the usual bounds, and then the sum is finite.
>
> My source says the theorem is "too important to be arbitrarily
> restricted", which is what happens when one leaves 0^0 undefined.
>

Too big to fail? I have argued here that BT does not require 0^0 to be

defined.


> > You introduce what amounts to an unnecessary axiom (0^0 = 1) to deal
> > with a few pathological cases in number theory that can be easily
> > handled in other ways. A questionable "gain" in my view.
>
> Not few and not pathological. It turns up and behaves naturally. Its
> exclusion is artificial.
>

I disagree.


> And not handled easily.

A matter of opinion, and not a decisive argument.

> It's a lot of unnecessary special cases that
> turn out to be not special after all. You demonstrate this above.
>
> I'm not sure what axiom system you are referring to, or what purpose
> you have in mind when you imply that 0^0 = 1 is unnecessary.

See my reply to Fred on this.

[Dan Christensen]

On Apr 25, 7:13 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > On Apr 24, 4:39 pm, Frederick Williams <freddywilli...@btinternet.com>
> > wrote:
> >> Dan Christensen wrote:
>
> >> > It isn't included in the usual recursive definition:  x^1 = x and x^(n
> >> > +1) = x^n * x. And I don't think it can be derived from this
> >> > definition. This makes me think it is ad hoc.
>
> >> It's not surprising if a definition of x^n for n = 1, 2, 3, ... cannot
> >> reveal what 0^0 is.
>
> > Yes, 0^0 is UNDEFINED!
>
> Er, Dan?  1^0 is also left undefined by that definition.

No, it isn't. (See my reply to Fred.)

> As are
> 2^{1/2}, e^{i*pi} and so on.
>

We are talking about integer exponents. 0 is an integer and x^0 is
left undefined for x=0 because you cannot divide by 0. (Also in my
reply to Fred.)

[Dan Christensen]

On Apr 25, 9:08 am, Frederick Williams <freddywilli...@btinternet.com>

wrote:
> Dan Christensen wrote:
>
> > On Apr 24, 4:39 pm, Frederick Williams <freddywilli...@btinternet.com>
> > wrote:
> > > Dan Christensen wrote:
>
> > > > It isn't included in the usual recursive definition:  x^1 = x and x^(n
> > > > +1) = x^n * x. And I don't think it can be derived from this
> > > > definition. This makes me think it is ad hoc.
>
> > > It's not surprising if a definition of x^n for n = 1, 2, 3, ... cannot
> > > reveal what 0^0 is.
>
> > Also for n = 0, -1, -2, -3, ...
>
> > Again, for x=/=0, we have x^n = (x^n+1)/x.
>
> > So, for x=/=0, we have x^0 = 1, x^(-1) = 1/x, x^(-2) = 1/(x^2), etc.
>
> Let's see how 'the usual recursive definition:  x^1 = x and x^(n +1) =
> x^n * x' (your words) works in the case n =0:
>
> x^0 = x^-1 * x = (x^-2 * x) * x = ((x^-3 * x) * x) * x = ...
>

Recall, for non-zero x, we have x^n = x^(n+1)/x.

Thus, for non-zero x, x^0 = x^(0+1)/x = x^1/x = x/x = 1

[Dan Christensen]

On Apr 25, 9:05 am, Frederick Williams <freddywilli...@btinternet.com>

wrote:
> Dan Christensen wrote:
>
> > On Apr 24, 4:39 pm, Frederick Williams <freddywilli...@btinternet.com>
> > wrote:
> > > Dan Christensen wrote:
>
> > > > It isn't included in the usual recursive definition:  x^1 = x and x^(n
> > > > +1) = x^n * x. And I don't think it can be derived from this
> > > > definition. This makes me think it is ad hoc.
>
> > > It's not surprising if a definition of x^n for n = 1, 2, 3, ... cannot
> > > reveal what 0^0 is.
>
> > Also for n = 0, -1, -2, -3, ...
>

You snipped this part:

Again, for x=/=0, we have x^n = (x^(n+1))/x.

So, for x=/=0, we have x^0 = 1, x^(-1) = 1/x, x^(-2) = 1/(x^2), etc.

> Ok, I misunderstood your use of the phrase 'recursive definition'.
> Let's see how you define x^-3:
>
> x^-3 = x^-4 * x = (x^-5 * x) * x = ...
>

Actually, continuing from the part you snipped, x^(-3) = x^(-2) / x =
(1/(x^2)) / x = 1/(x^3)

[Frederick Williams]

Dan Christensen wrote:
>
> On Apr 25, 9:08 am, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
> > Dan Christensen wrote:
> >
> > > On Apr 24, 4:39 pm, Frederick Williams <freddywilli...@btinternet.com>
> > > wrote:
> > > > Dan Christensen wrote:
> >
> > > > > It isn't included in the usual recursive definition: x^1 = x and x^(n
> > > > > +1) = x^n * x. And I don't think it can be derived from this
> > > > > definition. This makes me think it is ad hoc.
> >
> > > > It's not surprising if a definition of x^n for n = 1, 2, 3, ... cannot
> > > > reveal what 0^0 is.
> >
> > > Also for n = 0, -1, -2, -3, ...
> >
> > > Again, for x=/=0, we have x^n = (x^n+1)/x.
> >
> > > So, for x=/=0, we have x^0 = 1, x^(-1) = 1/x, x^(-2) = 1/(x^2), etc.
> >
> > Let's see how 'the usual recursive definition: x^1 = x and x^(n +1) =
> > x^n * x' (your words) works in the case n =0:
> >
> > x^0 = x^-1 * x = (x^-2 * x) * x = ((x^-3 * x) * x) * x = ...
> >
>
> Recall, for non-zero x, we have x^n = x^(n+1)/x.

"Recall" that here:
news:7cccf62a-3273-45db...@q13g2000vbd.googlegroups.com,

you wrote:

"It isn't included in the usual recursive definition: x^1 = x
and x^(n+1) = x^n * x."

The "it" being 0^0 = 1. So are you now giving up on "the usual
recursive definition"? I have no objection to "the usual recursive
definition" for defining x^n for n = 1, 2, ..., but it has no relevance
to your claim about 0^0. Are you admitting that now?

> Thus, for non-zero x, x^0 = x^(0+1)/x = x^1/x = x/x = 1

[Frederick Williams]

Dan Christensen wrote:
>
> [...] I have always seen

> exponentiation defined recursively for integer or natural number
> exponents n as:
>
> x^1 = x
> x^(n+1)= x^n * x.
>
> For integer n, and non-zero x, we have x^n=x^(n+1)/x. Thus x^0 = 1 for
> non-zero x and x^0 is left undefined for x=0.

If one sought to define 0^n by the equation x^n=x^(n+1)/x then it won't
work because you can divide by 0. The moral of that story is, one
doesn't seek to define 0^n by the equation x^n=x^(n+1)/x.

>
> The factorial function is usually defined recursively on the natural
> numbers as follows:
>
> 1! = 1
> (n+1)! = n! * (n+1)

If 0 is a natural number, that doesn't defined ! recursively on the
natural numbers at all.

> Since n+1 is non-zero, we also have n! = (n+1)!/(n+1) for ALL natural
> numbers n.

Not from your recursion above we don't, because you didn't allow n = 0.

> Thus 0!=1 (if you include 0 in N).
>
> > It is usual to define powers 0, 1, 2, ... so that x^0 is the neutral
> > element.
> >
> > > Yes. Just as you cannot divide by zero, so you can cannot raise 0 to
> > > the power of 0.
> >
> > Division by zero is quite different. Zero cannot have a multiplicative
> > inverse, call it w, because then we would have both 0*w = 0 (zero does
> > that to any number) and 0*w = 1 (the multiplicative inverse does that
> > to the number), which is a contradiction. I'm taking for granted that
> > (a = b and a = c) implies a = c, and not 0 = 1.
>
> Good point. x/y is is undefined for y=0. As a direct result, x^0 is
> undefined for x=0 (see above).

That x/0 is undefined has got nothing to do with the definition of x^0.
Oh, to be sure, _you_ can define x^0 so that it makes use of x/0, but so
what? I could define 0^0 to be 6, but it would carry no conviction.

[Rotwang]

On 25/04/2012 15:52, Dan Christensen wrote:
> On Apr 25, 2:22 am, Jussi Piitulainen<jpiit...@ling.helsinki.fi>
> wrote:
>> Dan Christensen writes:
>>> Because it is unnecessary to assume 0^0 = 1.
>>
>> You are of course free to work in a restricted system where you leave
>> it out. Other people need it defined. I think my main theme below is
>> that 0^0 is _naturally_ 1, and it is artificial to exclude it. So: we
>> disagree.
>>
>> 0^0 is the number of functions from the empty set to the empty set. In
>> general, m^n is the number of functions from an n-element set to an
>> m-element set. The number of ways to distribute n marbles in m boxes.
>>
>> 0^0 is the empty product. In general, x^n is the product of n x's, and
>> then x^0 is the empty product.
>>
>
> Seems like a lot of hand waving to me. I have always seen
> exponentiation defined recursively for integer or natural number
> exponents n as:
>
> x^1 = x
> x^(n+1)= x^n * x.

The first book I checked, namely Goldrei's /Classic Set Theory/, defines
natural number exponentiation by the following recursion:

x^0 = 1
x^{n + 1} = x^n*x

This is what I would expect.


--
Hate music? Then you'll hate this:

http://tinyurl.com/psymix

[Frederick Williams]

Dan Christensen wrote:
>
> [...] I have always seen

> exponentiation defined recursively for integer or natural number
> exponents n as:
>
> x^1 = x
> x^(n+1)= x^n * x.

Since Rotwang has give a reference (news:jn99nu$f4b$1...@dont-email.me) to
a book in which x^0 = 1, I think the onus is on you to give a reference
to a book where you have seen it defined as above. Actually, your use
of the word 'always' obliges you to give a number of such references.

[Jesse F. Hughes]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> On Apr 25, 7:13 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
>> > On Apr 24, 4:39 pm, Frederick Williams <freddywilli...@btinternet.com>
>> > wrote:
>> >> Dan Christensen wrote:
>>
>> >> > It isn't included in the usual recursive definition:  x^1 = x and x^(n
>> >> > +1) = x^n * x. And I don't think it can be derived from this
>> >> > definition. This makes me think it is ad hoc.
>>
>> >> It's not surprising if a definition of x^n for n = 1, 2, 3, ... cannot
>> >> reveal what 0^0 is.
>>
>> > Yes, 0^0 is UNDEFINED!
>>
>> Er, Dan?  1^0 is also left undefined by that definition.
>
> No, it isn't. (See my reply to Fred.)

Yes, it is.

You extended the definition using the rule that x^n = x^{n+1}/x, but
that rule does not follow for n = 0 using the above definition. You've
amended the definition thus.

>> As are
>> 2^{1/2}, e^{i*pi} and so on.
>>
>
> We are talking about integer exponents. 0 is an integer and x^0 is
> left undefined for x=0 because you cannot divide by 0. (Also in my
> reply to Fred.)

Say it as often as you like, but the more or less universal convention
is that 0^0 is defined to be 1. And the sky has not fallen because of
that convention.

But, if you prefer not to use that convention, then don't. Just take
the time to remind others of your personal idiosyncrasies when it
matters.

--
"It seems to me that in wartime Americans shouldn't be attacking each
other in this way on a *worldwide* forum. Then again, I know I'm an
American, but I have no way of knowing that you are, which would
explain a lot." --James Harris, on why Yanks should accept his proof

[Dan Christensen]

Odd. In Wiki, they have, as I would expect:

b^1 = b
b^(n+1) = b^n * b

http://en.wikipedia.org/wiki/Exponentiation

> x^{n + 1} = x^n*x
>

[Dan Christensen]

On Apr 25, 12:35 pm, Frederick Williams

<freddywilli...@btinternet.com> wrote:
> Dan Christensen wrote:
>
> > [...] I have always seen
> > exponentiation defined recursively for integer or natural number
> > exponents n as:
>
> > x^1 = x
> > x^(n+1)= x^n * x.
>
> > For integer n, and non-zero x, we have x^n=x^(n+1)/x. Thus x^0 = 1 for
> > non-zero x and x^0 is left undefined for x=0.
>
> If one sought to define 0^n by the equation x^n=x^(n+1)/x then it won't
> work because you can divide by 0.

My point exactly.

> The moral of that story is, one
> doesn't seek to define 0^n by the equation x^n=x^(n+1)/x.
>

Yes.

>
>
> > The factorial function is usually defined recursively on the natural
> > numbers as follows:
>
> > 1! = 1
> > (n+1)! = n! * (n+1)
>
> If 0 is a natural number, that doesn't defined ! recursively on the
> natural numbers at all.
>

Yes, it does. See the following.

> > Since n+1 is non-zero, we also have n! = (n+1)!/(n+1) for ALL natural
> > numbers n.
>
> Not from your recursion above we don't, because you didn't allow n = 0.
>

False. The definition works even if you allow n = 0.

> > Thus 0!=1 (if you include 0 in N).
>
> > > It is usual to define powers 0, 1, 2, ... so that x^0 is the neutral
> > > element.
>
> > > > Yes. Just as you cannot divide by zero, so you can cannot raise 0 to
> > > > the power of 0.
>
> > > Division by zero is quite different. Zero cannot have a multiplicative
> > > inverse, call it w, because then we would have both 0*w = 0 (zero does
> > > that to any number) and 0*w = 1 (the multiplicative inverse does that
> > > to the number), which is a contradiction. I'm taking for granted that
> > > (a = b and a = c) implies a = c, and not 0 = 1.
>
> > Good point. x/y is is undefined for y=0. As a direct result, x^0 is
> > undefined for x=0 (see above).
>
> That x/0 is undefined has got nothing to do with the definition of x^0.

It has everything to with 0^0 being undefined.

[Dan Christensen]

On Apr 25, 12:23 pm, Frederick Williams

My argument is actually based on the usual recursive definition for
the case of non-zero x.

> I have no objection to "the usual recursive
> definition" for defining x^n for n = 1, 2, ..., but it has no relevance
> to your claim about 0^0.  Are you admitting that now?
>

Of course not.

> > Thus, for non-zero x,  x^0 = x^(0+1)/x = x^1/x = x/x = 1
>

Like I said.

[Dan Christensen]

On Apr 25, 2:14 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > On Apr 25, 7:13 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> >> > On Apr 24, 4:39 pm, Frederick Williams <freddywilli...@btinternet.com>
> >> > wrote:
> >> >> Dan Christensen wrote:
>
> >> >> > It isn't included in the usual recursive definition:  x^1 = x and x^(n
> >> >> > +1) = x^n * x. And I don't think it can be derived from this
> >> >> > definition. This makes me think it is ad hoc.
>
> >> >> It's not surprising if a definition of x^n for n = 1, 2, 3, ... cannot
> >> >> reveal what 0^0 is.
>
> >> > Yes, 0^0 is UNDEFINED!
>
> >> Er, Dan?  1^0 is also left undefined by that definition.
>
> > No, it isn't. (See my reply to Fred.)
>
> Yes, it is.
>
> You extended the definition using the rule that x^n = x^{n+1}/x,

That is a consequence of the definition for non-zero x.

> but
> that rule does not follow for n = 0 using the above definition.  You've
> amended the definition thus.
>

Actually, it is a theorem.

> >> As are
> >> 2^{1/2}, e^{i*pi} and so on.
>
> > We are talking about integer exponents. 0 is an integer and x^0 is
> > left undefined for x=0 because you cannot divide by 0. (Also in my
> > reply to Fred.)
>
> Say it as often as you like, but the more or less universal convention
> is that 0^0 is defined to be 1.

A needless convention as it turns out, but suit yourself.

> And the sky has not fallen because of
> that convention.
>

If you want to fudge your axioms/definitions to make your results come
out nicely, go right ahead. Engineers and architects have been doing
so for millennia.

[Frederick Williams]

Dan Christensen wrote:

>
> Odd. In Wiki, they have, as I would expect:
>
> b^1 = b
> b^(n+1) = b^n * b
>
> http://en.wikipedia.org/wiki/Exponentiation

You, being either a liar or an idiot, forgot to put in the text:

"Formally, powers with positive integer exponents may be defined by the
initial condition

b^1 = b

and the recurrence relation

b^(n+1) = b^n * b"

Since that is about "positive integer exponents" it has no bearing on
b^0.

You, being either a liar or an idiot, forgot to read on to
http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero_power.

[Jussi Piitulainen]

Dan Christensen writes:

> I have always seen exponentiation defined recursively for integer or
> natural number exponents n as:
>
> x^1 = x
> x^(n+1)= x^n * x.
>
> For integer n, and non-zero x, we have x^n=x^(n+1)/x. Thus x^0 = 1
> for non-zero x and x^0 is left undefined for x=0.

...

> > Division by zero is quite different. Zero cannot have a
> > multiplicative inverse, call it w, because then we would have both
> > 0*w = 0 (zero does that to any number) and 0*w = 1 (the
> > multiplicative inverse does that to the number), which is a
> > contradiction. I'm taking for granted that (a = b and a = c)
> > implies a = c, and not 0 = 1.
>
> Good point. x/y is is undefined for y=0. As a direct result, x^0 is
> undefined for x=0 (see above).

I make no such point. The failure to define x^0 by a rule that is not
valid when x=0 is entirely yours, and only you are impressed by it.

> > > (x+y)^n = x^n if y=0 and not x = n = 0
> > >         = y^n if x=0 and not y = n = 0
> > >         = 0 if x+y = 0 and n > 0
> > >         = sum for k = 0, ..., n of C(n,k) x^k y^(n - k) if x =/= 0 and
> > > y =/= 0 and not x+y = n = 0
> >
> > > (Have I covered all the cases?)

...

> > My source says the theorem is "too important to be arbitrarily
> > restricted", which is what happens when one leaves 0^0 undefined.
>
> Too big to fail? I have argued here that BT does not require 0^0 to
> be defined.

No, "too important to be arbitrarily restricted". They want to use it
without the unnecessary system of cases and conditions that you
introduce above.

I take it you do not expect definitions to be motivated by theorems.

> > I'm not sure what axiom system you are referring to, or what
> > purpose you have in mind when you imply that 0^0 = 1 is
> > unnecessary.
>
> See my reply to Fred on this.

I saw nothing about your motivation. I guess that you find recursive
definitions pleasing, fine, but the one Rotwang cited is simpler and
it defines 0^0 = 1 without fuss. You seem afraid of 0^0 when you
really should avoid x/0. And you haven't seen any book where 0^0 = 1
is taken for granted or explicitly defined. These are my guesses. I
would like to understand but I remain puzzled.

People seem to leave 0^0 undefined in order to leave 0^0 undefined.
Is it a zen thing?

[Jussi Piitulainen]

Dan Christensen writes:

> On Apr 25, 12:45 pm, Rotwang wrote:
> > The first book I checked, namely Goldrei's /Classic Set Theory/,
> > defines natural number exponentiation by the following recursion:
> >
> > x^0 = 1
>
> Odd. In Wiki, they have, as I would expect:
>
> b^1 = b
> b^(n+1) = b^n * b
>
> http://en.wikipedia.org/wiki/Exponentiation
>
> > x^{n + 1} = x^n*x

That Wikipedia article has a lot more specifically on 0^0. Could you
live with this: "If the exponent is zero, some authors define 0^0=1,
whereas others leave it undefined"?

Or even this: "In most settings not involving continuity in the
exponent, interpreting 0^0 as 1 simplifies formulas and eliminates the
need for special cases in theorems"?

See also the section titled "In abstract algebra" and what is done
with x^0 when multiplication has a two-sided identity element 1.

[Virgil]

In article
<c465f603-2567-4a40...@dc2g2000vbb.googlegroups.com>,

But we can never have an unambiguous definition for 0^0 until

Lim_{x -> 0} 0^x = Lim_{y -> 0} y^0
--

[Jesse F. Hughes]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> On Apr 25, 2:14 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
>> > On Apr 25, 7:13 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> >> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
>> >> > On Apr 24, 4:39 pm, Frederick Williams <freddywilli...@btinternet.com>
>> >> > wrote:
>> >> >> Dan Christensen wrote:
>>
>> >> >> > It isn't included in the usual recursive definition:  x^1 = x and x^(n
>> >> >> > +1) = x^n * x. And I don't think it can be derived from this
>> >> >> > definition. This makes me think it is ad hoc.
>>
>> >> >> It's not surprising if a definition of x^n for n = 1, 2, 3, ... cannot
>> >> >> reveal what 0^0 is.
>>
>> >> > Yes, 0^0 is UNDEFINED!
>>
>> >> Er, Dan?  1^0 is also left undefined by that definition.
>>
>> > No, it isn't. (See my reply to Fred.)
>>
>> Yes, it is.
>>
>> You extended the definition using the rule that x^n = x^{n+1}/x,
>
> That is a consequence of the definition for non-zero x.

No, the consequence of the definition for non-zero x is:

For all n >= 1, x^n = x^{n+1}/x.

If you extend it to n < 1, then you are *extending the definition*.

>
>> but
>> that rule does not follow for n = 0 using the above definition.  You've
>> amended the definition thus.
>>
>
> Actually, it is a theorem.

The theorem is stated above.

>> >> As are
>> >> 2^{1/2}, e^{i*pi} and so on.
>>
>> > We are talking about integer exponents. 0 is an integer and x^0 is
>> > left undefined for x=0 because you cannot divide by 0. (Also in my
>> > reply to Fred.)
>>
>> Say it as often as you like, but the more or less universal convention
>> is that 0^0 is defined to be 1.
>
> A needless convention as it turns out, but suit yourself.

A useful convention, pretty much everyone else has agreed.

>
>> And the sky has not fallen because of
>> that convention.
>>
>
> If you want to fudge your axioms/definitions to make your results come
> out nicely, go right ahead. Engineers and architects have been doing
> so for millennia.

You really are a twit.

The theorem:

For all x, y and z,

z^x * z^y = z^{x+y}

is clean and easy to understand. Compare to your analogous theorem:

For all x, y and z, if z != 0 or (x != 0 and y != 0 and x+y != 0) then

z^x * z^y = z^{x+y}.

You say the definition 0^0 is a fudge. Well, I'll take that "fudge"
over such needlessly complicated theorems as the latter. And the
"fudge" is no more of a fudge than extending the definition to x^0 for
non-zero x. It's a sensible extension of a definition.

--
Jesse F. Hughes

"Sir, if you won't say sorry for killing father and mother, then I'll
shoot you." -- Quincy P. Hughes, screenwriter

[David W. Cantrell]

Jussi Piitulainen <jpii...@ling.helsinki.fi> wrote:
> Dan Christensen writes:
> > On Apr 25, 12:45 pm, Rotwang wrote:
> > > The first book I checked, namely Goldrei's /Classic Set Theory/,
> > > defines natural number exponentiation by the following recursion:
> > >
> > > x^0 = 1
> >
> > Odd. In Wiki, they have, as I would expect:
> >
> > b^1 = b
> > b^(n+1) = b^n * b
> >
> > http://en.wikipedia.org/wiki/Exponentiation
> >
> > > x^{n + 1} = x^n*x
>
> That Wikipedia article has a lot more specifically on 0^0. Could you
> live with this: "If the exponent is zero, some authors define 0^0=1,
> whereas others leave it undefined"?
>
> Or even this: "In most settings not involving continuity in the
> exponent, interpreting 0^0 as 1 simplifies formulas and eliminates the
> need for special cases in theorems"?

Related to the above quotation: Proponents of 0^0 = 1 in this thread have
mentioned instances in which the exponent is understood to be an integer.

But it should be mentioned that there are also settings _involving
continuity_ in the exponent in which "interpreting 0^0 as 1 simplifies
formulas and eliminates the need for special cases in theorems". Here's one
of my favorites:

Consider the function x^r, where r is real. Calculus texts often state its
derivative as

r * x^(r - 1)

without any explicit restrictions. Now consider the case when r = 1. The
graph of the function x^1 is of course a line through the origin with
slope = 1. We should be able to use the derivative to obtain 1 for the
slope of the tangent line at _any_ x. Replacing r by 1 in the general
formula, we get

1 * x^(1 - 1)

which simplifies to x^0. But if 0^0 were undefined, then we could not
conclude from this result that the tangent to the graph of x^1 has
slope = 1 at x = 0, as would obviously be desired.

David W. Cantrell