PROVE sqrt(2)+sqrt(3) is a irrational number!!!
ChineseBoy who loves maths
Then how to prove sqrt(2)+sqrt(3) is a irrational number?
William Hughes
wrote:
> The proof of sqrt(2) is a irrational number is well-known.
>
> Then how to prove sqrt(2)+sqrt(3) is a irrational number?
look at (sqrt(2) + sqrt(3))*(sqrt(2) * sqrt(3)) = 5 + 2*sqrt(6)
Now show 5 + 2*sqrt(6) is irrational (hint the proof that sqrt(6) is
irrational is almost the same as for sqrt(2))
conclude that a*a is irrational. What can you say about a?
- William Hughes
ChineseBoy who loves maths
> On Sep 30, 8:33 pm, ChineseBoy who loves maths <h2...@yahoo.com.cn>
> wrote:
>
> > The proof of sqrt(2) is a irrational number is well-known.
>
> > Then how to prove sqrt(2)+sqrt(3) is a irrational number?
>
>
> Now show 5 + 2*sqrt(6) is irrational (hint the proof that sqrt(6) is
> irrational is almost the same as for sqrt(2))
>
> conclude that a*a is irrational. What can you say about a?
>
> - William Hughes
You still need to prove:
1)a rational number times a irrational number is a irrational number.
2)a rational number plus a irrational number is a irrational number.
3)If a*a is irrational,then a is irrational.
G. Frege
<h2...@yahoo.com.cn> wrote:
>
>You still need to prove:
>
> 1) a rational number times a irrational number is a irrational number.
>
Lets write <i> to denote "an irrational number" and <r> to denote "a
rational number".
Claim: <r> * <i> = <i>.
Proof: Assume <r> * <i> = <r>, then <i> = <r> / <r> = <r>.
Contradiction! []
>
> 2) a rational number plus a irrational number is a irrational number.
>
Claim: r + i = i.
Proof: Assume <r> + <i> = <r>, then <i> = <r> - <r> = <r>.
Contradiction! []
>
> 3) If a*a is irrational, then a is irrational.
>
Claim: a*a is <i> => a is <i>
Proof: Suppose a*a is <i>. Now assume a is <r>. Then a*a is <r> too.
Contradiction! []
F.
--
E-mail: info<at>simple-line<dot>de
G. Frege
<h2...@yahoo.com.cn> wrote:
>
> How to prove sqrt(2) + sqrt(3) is a irrational number?
>
__ __
Assume v(2) + v(3) is rational. I.e. there are some integers n, m
__ __
(m =/= 0) such that v(2) + v(3) = n/m. Squaring both sides of the
__ __
equation gives: 2 + 2 * v(2) * v(3) + 3 = n^2/m^2. This would lead
__ __ __ __
to v(2) * v(3) = (n^2/m^2 - 5) / 2. With other words, v(2) * v(3)
__ __
is rational. But v(2) * v(3) is an irrational number. (see below)
Contradiction! qed.
__ __
Claim: v(2) * v(3) is an irrational number.
__ __
Proof: Assume v(2) * v(3) is rational. I.e. there are some integers
__ __
n, m (m =/= 0) such that v(2) * v(3) = n/m. Squaring both sides of
the equation gives: 2 * 3 = n^2/m^2. This would lead to: 2 * 3 * m^2
= n^2. Now both sides of the equation should have the same prime
factorization (by the fundamental theorem of arithmetic). But this
is impossible here. (Justification left as an exercise to the reader.)
Contradiction! qed.
ChineseBoy who loves maths
G. Frege
<wpih...@hotmail.com> wrote:
>>
>> How to prove sqrt(2)+sqrt(3) is a irrational number?
>>
> Look at (sqrt(2) + sqrt(3))*(sqrt(2) * sqrt(3)) = 5 + 2*sqrt(6)
> ^
?
I guess you meant "+" here.
>
> Now show 5 + 2*sqrt(6) is irrational. (Hint the proof that sqrt(6) is
> irrational is almost the same as for sqrt(2).)
>
> Conclude that a*a [with a := sqrt(2) + sqrt(3)] is irrational.
> What can you say about a [knowing that a is a real number]?
G. Frege
>>
>> 3) If a*a is irrational, then a is irrational.
>>
> Claim: a*a is <i> => a is <i>
>
> Proof: Suppose a*a is <i>. Now assume a is <r>. Then a*a is <r> too.
> Contradiction! []
>
Actually, 3) does not hold, and hence the alleged proof is not (quite)
correct.
4__ __
Counterexample: Let a = -i * v(2). Then a*a = -v(2). With other words,
a*a is irrational. But a is NOT irrational (i.e. not a real number
which is not rational), since it is a (purely) imaginary number.
Actually, the given proof only proves that a is not <r> (i.e. not
rational):
Proof: Suppose a*a is <i>. Now assume a is <r>.
Then a*a is <r> too. Contradiction!
Hence the assumption that a is <r> is false,
hence a is not <r>.
Now if we suppose that a is a real number, then we can proceed:
Hence a is <i> (since a is a real number, but not <r>).
qed.
This means that we actually proved the theorem:
3') If a*a is irrational, and a e IR, then a is irrational.
(Which is sufficient for our purpose.)
G. Frege
<h2...@yahoo.com.cn> wrote:
>
> You still need to prove:
>
> 1) a rational number times a irrational number is a irrational number.
> 2) a rational number plus a irrational number is a irrational number.
> 3) if a*a is irrational, and a is a real number, then a is irrational.
>
Theorems:
1') For every x and every y: if x is a rational number and y is an
irrational number, then x*y is an irrational number.
2') For every x and every y: if x is a rational number and y is an
irrational number, then x+y is an irrational number.
3') For every x: if x*x is an irrational number, and x is a real
number, then x is an irrational number.
First let's define:
x is a real number =df x e R
x is a rational number =df x e Q
x is an irrational number =df x e R\Q
where x e R\Q means x e R & x !e Q. (Reminder: Q c R.)
Now we may reformulate the theorems from above:
1'') AxAy(x e Q & y e R\Q -> x*y e R\Q).
2'') AxAy(x e Q & y e R\Q -> x+y e R\Q).
3'') Ax(x*x e R\Q & x e R -> x e R\Q).
Proofs:
1. Suppose a e Q & b e R\Q. Hence a e Q, and hence a e R (since Q c
R), as well as b e R and b !e Q. Hence a*b e R (since R is closed
under multiplication). Assume a*b e Q. Let c = a*b. Hence c e Q, and
b = c/a e Q (since Q is closed under division). Contradiction! Hence
a*b !e Q. Hence a*b e R\Q. qed.
2. Suppose a e Q & b e R\Q. Hence a e Q, and hence a e R (since Q c
R), as well as b e R and b !e Q. Hence a+b e R (since R is closed
under addition). Assume a+b e Q. Let c = a+b. Hence c e Q, and
b = c-a e Q (since Q is closed under subtraction). Contradiction!
Hence a+b !e Q. Hence a+b e R\Q. qed.
3. Suppose a*a e R\Q & a e R. Hence a*a e R and a*a !e Q, as well
as a e R. Assume a e Q. Then a*a e Q (since Q is closed under
multiplication). Contradiction! Hence a !e Q. Hence a e R\Q. qed.
G. Frege
Typo corrected:
> 4__ __
> Counterexample: Let a = i * v(2). Then a*a = -v(2). With other words,
> ^^^
> a*a is irrational. But a is NOT irrational (i.e. not a real number
> which is not rational), since it is a (purely) imaginary number.
>
Pubkeybreaker
ChineseBoy who loves maths wrote:
> The proof of sqrt(2) is a irrational number is well-known.
>
> Then how to prove sqrt(2)+sqrt(3) is a irrational number?
Find the minimal polynomial, then apply the rational roots theorem.
Tonico
wrote:
> The proof of sqrt(2) is a irrational number is well-known.
>
> Then how to prove sqrt(2)+sqrt(3) is a irrational number?
************************************************
Assume r = sqrt(2) + sqrt(3) is rational ==> (r - sqrt(2))^2 = 3
==>
r^2 - 2r*sqrt(2) + 2 = 3 ==> sqrt(2) = [r^2 - 1]/2r , and since
you're assuming r is ration, the number [r^2 - 1]/2r is rational ==>
Srqt(2) is rational and there you've your contradiction.
Regards
Tonio
Dave L. Renfro
>> Then how to prove sqrt(2)+sqrt(3) is a irrational number?
Pubkeybreaker wrote:
> Find the minimal polynomial, then apply the rational
> roots theorem.
Or any polynomail with sqrt(2) + sqrt(3) as a root,
although you'll want one in which you can prove
all the rational root possibilities are not equal
to sqrt(2) + sqrt(3). Also, if you get the expected
4'th degree polynomial [x = sqrt(2) + sqrt(3). Subtract
sqrt(2), square both sides, isolate sqrt(2) term, square
again, set equal to zero to get (x^2 - 1)^2 - 8x], then
you only need to prove the rational roots theorem for
quartics, which is less abstract than doing so in
general. Incidentally, a proof of the rational roots
theorem (for quartics, or for n'th degree polynomials)
can be structured so that it's very similar to the
proof that sqrt(2) is irrational.
Dave L. Renfro
A N Niel
ChineseBoy who loves maths <h2...@yahoo.com.cn> wrote:
> On Oct 1, 8:44 am, William Hughes <wpihug...@hotmail.com> wrote:
> > On Sep 30, 8:33 pm, ChineseBoy who loves maths <h2...@yahoo.com.cn>
> > wrote:
> >
> > > The proof of sqrt(2) is a irrational number is well-known.
> >
> > > Then how to prove sqrt(2)+sqrt(3) is a irrational number?
> >
> > look at (sqrt(2) + sqrt(3))*(sqrt(2) + sqrt(3)) = 5 + 2*sqrt(6)
> >
> > Now show 5 + 2*sqrt(6) is irrational (hint the proof that sqrt(6) is
> > irrational is almost the same as for sqrt(2))
> >
> > conclude that a*a is irrational. What can you say about a?
> >
> > - William Hughes
>
> You still need to prove:
> 1)a rational number times a irrational number is a irrational number.
false, despite the "proof" given
> 2)a rational number plus a irrational number is a irrational number.
true
> 3)If a*a is irrational,then a is irrational.
true
>
Ray Vickson
wrote:
> The proof of sqrt(2) is a irrational number is well-known.
>
> Then how to prove sqrt(2)+sqrt(3) is a irrational number?
Let a = sqrt(3) + sqrt(2) and b = sqrt(3) - sqrt(2). Since a*b = 1,
either both a and b are rational or both are irrational. They cannot
be rational, since their sum (or difference) is irrational.
R.G. Vickson
G. Frege
<ann...@nym.alias.net.invalid> wrote:
>>
>> You still need to prove:
>> 1)a rational number times a irrational number is a irrational number.
>>
> false, despite the "proof" given
>
You are right, I missed the case that the rational number is 0.
>>
>> 2)a rational number plus a irrational number is a irrational number.
>
> true
>
>> 3)If a*a is irrational,then a is irrational.
>>
> true
>
False.
4__ __
Counterexample: Let a = i * v(2). Then a*a = -v(2). With other words,
G. Frege
>>
>> You still need to prove:
>>
>> 1) a rational number times a irrational number is a irrational number.
>>
> Lets write <i> to denote "an irrational number" and <r> to denote "a
> rational number".
>
> Claim: <r> * <i> = <i>.
>
> Proof: Assume <r> * <i> = <r>, then <i> = <r> / <r> = <r>.
> Contradiction! []
>
Actually, the claim is wrong, and "proof" is not correct. :-(
I missed the case, that <r> = 0.
So let's correct that slip:
1) a rational number (which is not 0) times a irrational number is a
irrational number.
Claim: <r> =/= 0 => <r> * <i> = <i>.
Proof: Suppose <r> =/= 0. Assume <r> * <i> = <r>, then <i> = <r> / <r>
= <r>. Contradiction! []
G. Frege
>>
>> You still need to prove:
>>
>> 1) a rational number times a irrational number is a irrational number.
>>
A. N. Niel observed that this doesn't hold in general.
Of course this means that my alleged proof is erroneous, let's see:
>
> 1. Suppose a e Q & b e R\Q. Hence a e Q, and hence a e R (since Q c
> R), as well as b e R and b !e Q. Hence a*b e R (since R is closed
> under multiplication). Assume a*b e Q. Let c = a*b. Hence c e Q, and
> b = c/a e Q (since Q is closed under division).
~~~~~~~~~~~
Nope! This step is only allowed if a =/= 0. :-(
So let's consider (instead) the following claim:
1) a rational number which is not 0 times a irrational number is a
irrational number.
1') For every x and every y: if x is a rational number =/= 0 and y is
an irrational number, then x*y is an irrational number.
1'') AxAy(x e Q & x =/= 0 & y e R\Q -> x*y e R\Q).
Proof:
Suppose a e Q & a =/= 0 & b e R\Q. Hence a e Q, and hence a e R (since
Q c R), as well as a =/= 0, b e R and b !e Q. Hence a*b e R (since R
is closed under multiplication). Assume a*b e Q. Let c = a*b. Hence
c e Q, and b = c/a e Q (since Q is closed under division; and a =/=0).
Contradiction! Hence a*b !e Q. Hence a*b e R\Q. qed.
Dave L. Renfro
> So let's consider (instead) the following claim:
>
> 1) a rational number which is not 0 times a irrational
> number is a irrational number.
Here's something to play with.
We know an irrational number to an irrational power can be
rational. For instance, a^b is an example for a = b = sqrt(2)
or for a = sqrt(2)^sqrt(2) and b = sqrt(2).
QUESTION: Can a^(b^c) be rational with a, b, c all irrational?
Dave L. Renfro
Dave Seaman
> G. Frege wrote (in part):
>> So let's consider (instead) the following claim:
>>
>> 1) a rational number which is not 0 times a irrational
>> number is a irrational number.
> Here's something to play with.
> We know an irrational number to an irrational power can be
> rational. For instance, a^b is an example for a = b = sqrt(2)
> or for a = sqrt(2)^sqrt(2) and b = sqrt(2).
False in the case of a = b = sqrt(2), by Gelfond-Schneider.
> QUESTION: Can a^(b^c) be rational with a, b, c all irrational?
a = c = sqrt(2) and b = sqrt(2)^sqrt(2).
--
Dave Seaman
Oral Arguments in Mumia Abu-Jamal Case heard May 17
U.S. Court of Appeals, Third Circuit
<http://www.abu-jamal-news.com/>
quasi
Very nice.
quasi
tommy1729
no he does not !!!
william hughes proof is the simplest yet correct.
if you prove a*a is irrational then a is also irrational because a sqrt cannot make an irrational number rational.
HIS PROOF IS CORRECT !!!
ha !
tommy1729
Gerry Myerson
G. Frege <nomail@invalid> wrote:
> On Mon, 01 Oct 2007 03:28:59 +0200, G. Frege <nomail@invalid> wrote:
>
> >>
> >> 3) If a*a is irrational, then a is irrational.
> >>
> > Claim: a*a is <i> => a is <i>
> >
> > Proof: Suppose a*a is <i>. Now assume a is <r>. Then a*a is <r> too.
> > Contradiction! []
> >
> Actually, 3) does not hold, and hence the alleged proof is not (quite)
> correct.
> 4__ __
> Counterexample: Let a = -i * v(2). Then a*a = -v(2). With other words,
> a*a is irrational. But a is NOT irrational (i.e. not a real number
> which is not rational), since it is a (purely) imaginary number.
I don't know where people get this idea.
An irrational number is a number that is not rational.
In particular, every complex number with non-zero
imaginary part is irrational.
This is the standard definition in Number Theory.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Fuckwit
wrote:
>
> if you prove a*a is irrational then a is also irrational [...]
>
No.
4___ ___
Counterexample: Let a = i * V 2 . Then a*a = -V 2 . This means:
a*a is irrational, but a is NOT irrational (i.e. not a real number
which is not rational), since it is a (purely) imaginary number.
F.
G. Frege
<ge...@maths.mq.edi.ai.i2u4email> wrote:
>>
>> a is NOT irrational (i.e. not a real number which is not rational),
>> since it is a (purely) imaginary number.
>>
> I don't know where people get this idea.
>
"In mathematics, an irrational number is any real number that is not a
rational number [...]."
Source:
http://en.wikipedia.org/wiki/Irrational_number
>
> An irrational number is a number that is not rational.
> In particular, every complex number with non-zero
> imaginary part is irrational.
>
> This is the standard definition in Number Theory.
>
So Wikipedia got it wrong?
Arturo Magidin
Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote:
[...]
>I don't know where people get this idea.
>An irrational number is a number that is not rational.
Usually, I see it defined as a ->real<- number that is not
rational.
>In particular, every complex number with non-zero
>imaginary part is irrational.
Never seen i (the complex square root of -1) called "irrational",
though it would of course go with the definition you provide.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
G. Frege
>>>
>>> a is NOT irrational (i.e. not a real number which is not rational),
>>> since it is a (purely) imaginary number.
>>>
>> I don't know where people get this idea.
>>
>
> "In mathematics, an irrational number is any real number that is not a
> rational number [...]."
>
> Source:
> http://en.wikipedia.org/wiki/Irrational_number
>
Here's another source:
"An irrational number is a real number which cannot be represented as
a ratio of two integers."
(http://planetmath.org/encyclopedia/IrrationalNumber.html)
or
"An irrational number is any real number that is not rational."
(http://mathforum.org/dr.math/faq/faq.integers.html)
.
:
So does this answer your quesion, "where people get this idea" [from]?
G. Frege
(Arturo Magidin) wrote:
>>
>> I don't know where people get this idea.
>> An irrational number is a number that is not rational.
>>
> Usually, I see it defined as a ->real<- number that is not
> rational.
>
Which is in agreement with what Wikipedia says:
"In mathematics, an irrational number is any real number that is not a
rational number [...]."
Source:
http://en.wikipedia.org/wiki/Irrational_number
G. Frege
>>>
>>> The proof of sqrt(2) is a irrational number is well-known.
>>> Then how to prove sqrt(2)+sqrt(3) is a irrational number?
>>>
>> Let a = sqrt(3) + sqrt(2) and b = sqrt(3) - sqrt(2). Since a*b = 1,
>> either both a and b are rational or both are irrational. They cannot
>> be rational, since their sum (or difference) is irrational.
>
> Very nice.
>
Indeed!
G. Frege
>
> "In mathematics, an irrational number is any real number that is not a
> rational number [...]."
>
> (http://en.wikipedia.org/wiki/Irrational_number)
>
"Usually, I see it defined as a real number that is not
rational." (Arturo Magidin)
So I hope you don't mind that I'm using the mentioned definition for
the notion /irrational number/. After all it doesn't seem to be that
uncommon.
Gerry Myerson
G. Frege <nomail@invalid> wrote:
No, it just pushes the question back a step - where do wikipedia,
planetmath, and dr.math get this idea?
You can't do much mathematics without i,
and you can't call i rational,
so you either throw your hands up in the air,
or you call i irrational.
Wikipedia, planetmath, and dr.math to the contrary notwithstanding,
this is the absolutely standard use in (sufficiently) advanced math.
Jorn Steuding, Diophantine Analysis, page 6: "... a number is said
to be rational if it can be represented as a quotient of two integers,
the denominator being nonzero; all other numbers (in the set R of
real and the set C of complex numbers, respectively) are called
irrational."
Miller and Takloo-Bighash, An Invitation to Modern Number Theory,
page 113: "If alpha is not in Q, we say alpha is irrational. Clearly,
not all numbers are rational (for example, sqrt(-1))."
Rainer Rosenthal
> On Tue, 02 Oct 2007 04:41:28 +0200, G. Frege <nomail@invalid> wrote:
>
>> "In mathematics, an irrational number is any real number that is not a
>> rational number [...]."
>>
A stubborn person is a poster in sci.math who sticks to
his wrong beliefs contrary to what mathematicians say.
This does not exclude other people from being called
stubborn as well (imagine!).
Or to put it less aggressive (sorry): a number is called
rational if it's the quotient a/b of integers a and b <> 0.
As there are Gaussian integers you are free to define
Gaussian rationals as a/b with a and b Gaussian integers,
b <> 0. And (in the context!) you are allowed to call them
rational. It's all about context and you (back to aggression,
sorry) stubbornly deny this dependency. Gerry Myerson had
said everything that was to be said; my posting is complete-
ly obsolete and you can forget it immediately, thanks!
Best regards,
Rainer Rosenthal
r.ros...@web.de
Tonico
wrote:
> In article <n4c3g3t5mqd5fk7npbic4qvg9a1d0fb...@4ax.com>,
************************************************************8
As other guys around, I have never seen a definition of irrational
number that didn't imply the idea that we're talking about real
numbers.
According to the above quoted authors, and taking into account that
"number" can be applied to a rather wide range of objects, why not
call "irrational" number to the elements of any field of positive
characteristic?
John Steuding, according to the above above, decides that "...all
other numbers (in the set R of real and the set C of complex numbers,
respectively) are called irrational."
Why? Why to stop in C? Why not continue into C(x) = the field of
rational functions on a symbol x?
Miller and Takloo are even more spectacular: they want to give an
example of a non-rational number and thus they choose, according to
the above quote,...the square root of -1! Why? Misteries...
It may be a split decision, but I have this feeling that the huge
majority of mathematicians consider rational and irrational numbers as
being part of the real ones, and complex numbers don't take part in
this.
Am I right?
Regards
Tonio
G. Frege
<r.ros...@web.de> wrote:
>
> [laber laber]
>
> Rainer Rosenthal
>
Tipp: Wenn man nichts zu sagen hat, einfach mal die Klappe halten.
G. Frege
<ge...@maths.mq.edi.ai.i2u4email> wrote:
>>
>> So does this answer your quesion, "where people get this idea" [from]?
>>
> No, it just pushes the question back a step - where do wikipedia,
> planetmath, [, A. Magidin, ..., ] and dr.math get this idea?
>
I guess they got it from some other sources (books, working
mathematicians, etc.)
>
> You can't do much mathematics without i,
> and you can't call i rational,
>
Given the usual definition of rational, I'd agree.
>
> so you either throw your hands up in the air,
> or you call i irrational.
>
No. You might call i not rational.
There's no need to call i /irrational/. Especially _if_ you define an
/irrational number/ as a real number which is not rational, i is not
an irrational number.
>
> Wikipedia, planetmath, and dr.math to the contrary notwithstanding,
> this is the absolutely standard use in (sufficiently) advanced math.
>
This may very well be the case. You know, it might be possible that
there are different definitions of /irrational number/ out there.
Actually, both sources you quote are from
the field of number theory, right?
I'd guess that the definition mentioned by Wikipedia, planetmath,
dr.math, and A. Magidin, ... is rather common in (Real) Analysis, etc.
>
> Jorn Steuding, Diophantine Analysis, page 6: "... a number is said
> to be rational if it can be represented as a quotient of two integers,
> the denominator being nonzero; all other numbers (in the set R of
> real and the set C of complex numbers, respectively) are called
> irrational."
>
No one can stop Mr. Steuding to use this definition of /irrational/.
>
> Miller and Takloo-Bighash, An Invitation to Modern Number Theory,
> page 113: "If alpha is not in Q, we say alpha is irrational. Clearly,
> not all numbers are rational (for example, sqrt(-1))."
>
Same, same.
Phil Carmody
> In article <n4c3g3t5mqd5fk7np...@4ax.com>,
> G. Frege <nomail@invalid> wrote:
>
> > On Tue, 02 Oct 2007 04:41:28 +0200, G. Frege <nomail@invalid> wrote:
> > >>> a is NOT irrational (i.e. not a real number which is not rational),
> > >>> since it is a (purely) imaginary number.
> > >>>
> > >> I don't know where people get this idea.
> > >
> > > "In mathematics, an irrational number is any real number that is not a
> > > rational number [...]."
> >
> > "An irrational number is a real number which cannot be represented as
> > a ratio of two integers."
> > (http://planetmath.org/encyclopedia/IrrationalNumber.html)
> >
> > (http://mathforum.org/dr.math/faq/faq.integers.html)
> >
>
> No, it just pushes the question back a step - where do wikipedia,
> planetmath, and dr.math get this idea?
>
> You can't do much mathematics without i,
> and you can't call i rational,
> so you either throw your hands up in the air,
> or you call i irrational.
Or you can both not call irrational and not call it rational.
Let's see how your style of argument works in other contexts:
You can't do much mathematics without 0,
and you can't call 0 positive,
so you either throw your hands up in the air,
or you call 0 negative.
If you don't think the above is close enough to yours, feel free
to, replace 0 with i.
> Wikipedia, planetmath, and dr.math to the contrary notwithstanding,
> this is the absolutely standard use in (sufficiently) advanced math.
Let's play 'spot the "no true Scotsman" fallacy', shall we?
Phil
--
Dear aunt, let's set so double the killer delete select all.
-- Microsoft voice recognition live demonstration
G. Frege
>>
>> and you can't call i rational,
>>
> Given the usual definition of rational, I'd agree.
>
>> so you either throw your hands up in the air,
>> or you call i irrational.
>>
> No. You might call i not rational.
>
> There's no need to call i /irrational/. Especially _if_ you define an
> /irrational number/ as _a real number which is not rational_, i is not
> an irrational number.
>
Remember the theory of relations: there are /reflexive/ relations, and
/irreflexive/ relations (and non-reflexive relations). But an
irreflexive relation is not just (=is not defined as) a relation which
is not reflexive.
Dave L. Renfro
> You can't do much mathematics without 0,
> and you can't call 0 positive,
> so you either throw your hands up in the air,
> or you call 0 negative.
For Bourbakians, 0 is both positive and negative,
although it's neither strictly positive nor strictly
negative.
Dave L. Renfro
hagman
I first wanted to agre that "ir-" = "not", hence i is irrational
(as it is not rational).
But then you aould have to say that "non-" = "not", hence i is non-
negative
as it is not negative, although "non-negative" is generally
used to stand for "positiev or zero".
Thus one might wish agree that the terms irrational and non-negative
should only be applied to real numbers, whereas all non-real complex
numbers are neither rational nor irrational (and neither positive nor
nonnegative).
But in that case the proof in the earlier post becomes correct again
if one adds the premise that all numbers are considered real:
If x is rational, then x^2 is ratuional.
Thus:
If x is a real number such that x^2 is irrational, then x is
irrational.
This is unproblematic as the OP investigated real numbers
(namely sqrt(2)+sqrt(3)) anyway.
G. Frege
wrote:
>
> If x is a real number such that x^2 is irrational, then x is
> irrational.
>
Exactly. :-)
I discovered the missing "premise" ("x is a real number") when I tried
to prove the formal version of the original "theorem", namely
Ax(x*x e R\Q -> x e R\Q).
Now if we add "x e R" we get
Ax(x*x e R\Q & x e R -> x e R\Q) ,
which can easily be proved:
Suppose a*a e R\Q & a e R. Hence a*a e R and a*a !e Q, as well
as a e R. Assume a e Q. Then a*a e Q (since Q is closed under
multiplication). Contradiction! Hence a !e Q. Hence a e R\Q. qed.
On the other hand...
>
> This is unproblematic as the OP investigated real numbers
> (namely sqrt(2)+sqrt(3)) anyway.
>
Right.
G. Frege
>>>>
>>>> /a/ is NOT irrational (i.e. not a real number that is not rational)
>>>>
>>> I don't know where people get this idea.
>>>
Well, here are some sources:
"In mathematics, an irrational number is any real number that is not a
rational number [...]."
(http://en.wikipedia.org/wiki/Irrational_number)
"An irrational number is a real number which cannot be represented as
a ratio of two integers."
(http://planetmath.org/encyclopedia/IrrationalNumber.html)
"An irrational number is any real number that is not rational."
(http://mathforum.org/dr.math/faq/faq.integers.html)
"Irrational Numbers: Real numbers that are not rational."
(http://www.mathwords.com/i/irrational_numbers.htm)
~~~~~~~~~~~~~~
"ir·ra·tion·al num·ber (plural ir·ra·tion·al num·bers)
noun
Definition:
real number that is not rational: any real number that cannot be
expressed as the exact ratio of two integers, e.g. V2 and pi.
(http://encarta.msn.com/dictionary_1861692625/irrational_number.html)
Though this definition might not be the one you prefer it seems to be
rather common.
Arturo Magidin
G. Frege <nomail@invalid> wrote:
[...]
>This may very well be the case. You know, it might be possible that
>there are different definitions of /irrational number/ out there.
>
> Actually, both sources you quote are from
> the field of number theory, right?
>
>I'd guess that the definition mentioned by Wikipedia, planetmath,
>dr.math, and A. Magidin, ... is rather common in (Real) Analysis, etc.
My experience is that dichotomy rational/irrational is usually
reserved for contexts in which we are working with real numbers; in
algebraic number theory, when complex numbers are at work, the
dichotomy runs first and foremost along algebraic/transcendental, then
along algebraic integer/algebraic number, then by degree. Sometimes
along the real/nonreal line as well.
Ivan Niven, in his book "Irrational Numbers" (The Carus Mathematical
Monographs vol. 11, published by the MAA, first published in 1956,
fifth printing in 2005) opens by saying (after some other words; this
is the end of the very first paragraph) "A rational number is one that
can be put in the form h/k where h and k are integers with k =/=
0. Real numbers like sqrt(2) which are not rational are said to be
irrational."
Likewise in chapter 4 he says: "Given an irrational number a, it is
clear that there are rational numbers h/k as close to a, so that
|a - h/k| is small." This would not hold if we extend the notion of
"irrational" to include any complex number that is not rational.
Many of my Number Theory books (Lang's "Algebraic Number Theory",
Niven/Zuckerman/Montgomery "The Theory of Numbers", Marcus's "Number
Fields", Burton's "Elementary Number Theory") do not seem to define
"rational number", taken it for granted. However:
* Burton discusses continued fractions. He defines an "infinite
continued fraction" as an expression of the form
a + (b_1/(a_1 + (b_2/( a_2 + (b_3/( a_3 + ... ))))))
where a_0, a_1, ...., and b_1, b_2, ... are real numbers. Then Theorem
15.7 states "Every irrational number has a unique representation as an
infinite continued fraction[.]" This would be false if we allow
non-real numbers to be called irrational.
* William LeVeque's "Fundamentals of Number Theory" (Dover, 1977) in
the chapter on Diophantine approximations, writes:
"... if x belongs to R\Q (the set of real-but-not-rational (=
irrational) numbers.)"
thus restricting the definition of "irrational" to real numbers.
* On the other hand, Serge Lang's "Introduction to Diophantine
Approximations" New Expanded Edition, Springer-Verlag 1995, writes
in p 7: "Let a be a real irrational number", suggesting there are
other kinds of irrationals.
G. Frege
Thanx, Arturo for sharing your knowledge and thoughts. Your
observations seem (by and large) to justify Wikipedia's claim:
"In mathematics, an irrational number is any real number
that is not a rational number [...]."
(http://en.wikipedia.org/wiki/Irrational_number)
But it seems that especially in _number theory_ /irrational/ is
(sometimes?) defined in a way to embrace the field of complex numbers.
Actually that's exactly what Gerry Myerson claimed in his _first_ post
concerning this question:
"An irrational number is a number that is not rational.
In particular, every complex number with non-zero
imaginary part is irrational.
This is the standard definition in Number Theory."
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
>
> * On the other hand, Serge Lang's "Introduction to Diophantine
> Approximations" New Expanded Edition, Springer-Verlag 1995, writes
> in p 7: "Let a be a real irrational number", suggesting there are
> other kinds of irrationals.
>
Indeed, some quotes provided by Gerry Myerson:
Jorn Steuding, Diophantine Analysis, page 6: "... a number is said
to be rational if it can be represented as a quotient of two integers,
the denominator being nonzero; all other numbers (in the set R of
real and the set C of complex numbers, respectively) are called
irrational."
Miller and Takloo-Bighash, An Invitation to Modern Number Theory,
page 113: "If alpha is not in Q, we say alpha is irrational. Clearly,
not all numbers are rational (for example, sqrt(-1))."
Arturo Magidin
G. Frege <nomail@invalid> wrote:
>On Tue, 2 Oct 2007 17:19:02 +0000 (UTC), mag...@math.berkeley.edu
>(Arturo Magidin) wrote:
[...]
>Thanx, Arturo for sharing your knowledge and thoughts. Your
>observations seem (by and large) to justify Wikipedia's claim:
>
> "In mathematics, an irrational number is any real number
> that is not a rational number [...]."
>
> (http://en.wikipedia.org/wiki/Irrational_number)
>
>But it seems that especially in _number theory_ /irrational/ is
>(sometimes?) defined in a way to embrace the field of complex numbers.
>
>Actually that's exactly what Gerry Myerson claimed in his _first_ post
>concerning this question:
Not really...
> "An irrational number is a number that is not rational.
> In particular, every complex number with non-zero
> imaginary part is irrational.
>
> This is the standard definition in Number Theory."
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
You'll note that both LeVeque's book and Burton's book are Number
Theory books whose definition disagrees with the above. Lang seems to
agree, but only by implication (and for all we know, he may restrict
"irrational" in the complex setting to numbers of the form a + bi with
a and b real, not both rational). Dedekind's "Theory of Algebraic
Integers" states irrationals must be real numbers, though of course
that is a very dated reference. Erdos and Suranyi's "Topics in the
Theory of Numbers" likewise restrict irrationals to being real numbers
by implication of their theorems on continued fractions.
I disagree that the quoted definition is standard in Number Theory. In
my experience, very few advanced number theory texts even use the
notion of "irrational" (most using rational, and speaking about
elements of R\Q), or by implication restrict them to real
numbers. Most Number Theory books routinely talk about approximating
irrational numbers by sequences of rational numbers, a statement that
would be trivially false using the definition that is above asserted
to be "standard".
G. Frege
(Arturo Magidin) wrote:
>
> [...]
>
Thanx again, Arturo. I think this should settle the question.
Moreover it might help Gerry Myerson to understand "where people get
this idea."
Dave Seaman
> My experience is that dichotomy rational/irrational is usually
> reserved for contexts in which we are working with real numbers; in
> algebraic number theory, when complex numbers are at work, the
> dichotomy runs first and foremost along algebraic/transcendental, then
> along algebraic integer/algebraic number, then by degree. Sometimes
> along the real/nonreal line as well.
What about the Gelfond-Schneider theorem? It states that a^b is
transcendental if:
(1) a is algebraic, not equal to 0 or 1, and
(2) b is algebraic and irrational.
An example given at <http://mathworld.wolfram.com/GelfondsTheorem.html>
states that Gelfond's constant e^pi is therefore transcendental, because
e^pi = (e^i*pi)^(-i) = (-1)^(-i). Notice that the exponent -i is
algebraic and irrational (because it isn't rational), and therefore the
theorem applies.
--
Dave Seaman
Oral Arguments in Mumia Abu-Jamal Case heard May 17
U.S. Court of Appeals, Third Circuit
<http://www.abu-jamal-news.com/>
Arturo Magidin
Dave Seaman <dse...@no.such.host> wrote:
>On Tue, 2 Oct 2007 17:19:02 +0000 (UTC), Arturo Magidin wrote:
>
>> My experience is that dichotomy rational/irrational is usually
>> reserved for contexts in which we are working with real numbers; in
>> algebraic number theory, when complex numbers are at work, the
>> dichotomy runs first and foremost along algebraic/transcendental, then
>> along algebraic integer/algebraic number, then by degree. Sometimes
>> along the real/nonreal line as well.
>
>What about the Gelfond-Schneider theorem? It states that a^b is
>transcendental if:
>
> (1) a is algebraic, not equal to 0 or 1, and
> (2) b is algebraic and irrational.
Well, what it states depends on your definitions. Here is how Niven
quotes the Gelfond-Schneider theorem (p 134, Chapter 10, of
"Irrational Numbers"):
Theorem 10.1 If a and ba re algebraic numbers, with a not equal to
0 or 1, and if b is not a real rational number, then any value of
a^b is transcendental.
Remark. The hypothesis that "b is not a real rational number" is
usually stated in the form "b is irrational." Our wording is an
attempt to avoid the suggestions that b must be a real number. Such
a number as b=2+3i, sometimes called a "complex rational number",
satisfies the hypothesis of the theorem. [...]
LeVeque ("Topics in Number Theory" volume II, p 188) writes:
"[Gelfond's 1929] result was that a^b is transcendental for
algebraic a different from 0 or 1, if beta is an imaginary
quadratic irrationality."
The old saw about standards and there being so many to choose from
comes to mind.
quasi
(Arturo Magidin) wrote (in part):
>My experience is that dichotomy rational/irrational is usually
>reserved for contexts in which we are working with real numbers; in
>algebraic number theory, when complex numbers are at work, the
>dichotomy runs first and foremost along algebraic/transcendental, then
>along algebraic integer/algebraic number, then by degree. Sometimes
>along the real/nonreal line as well.
I agree with the above.
But I think it's ok for an author to use a non-standard (or less
standard) interpretation, as long as it's declared. If something is
almost universally standard, choosing a nonstandard interpretation is
probably unwise, but still legal, as long as they declare their
intended meaning. However, if they redefine too many standard terms,
the text becomes unreadable.
>Ivan Niven, in his book "Irrational Numbers" (The Carus Mathematical
>Monographs vol. 11, published by the MAA, first published in 1956,
>fifth printing in 2005) opens by saying (after some other words; this
>is the end of the very first paragraph) "A rational number is one that
>can be put in the form h/k where h and k are integers with k =/=
>0. Real numbers like sqrt(2) which are not rational are said to be
>irrational."
>
>Likewise in chapter 4 he says: "Given an irrational number a, it is
>clear that there are rational numbers h/k as close to a, so that
>|a - h/k| is small." This would not hold if we extend the notion of
>"irrational" to include any complex number that is not rational.
>
>Many of my Number Theory books (Lang's "Algebraic Number Theory",
>Niven/Zuckerman/Montgomery "The Theory of Numbers", Marcus's "Number
>Fields", Burton's "Elementary Number Theory") do not seem to define
>"rational number", taken it for granted. However:
Niven & Zuckerman (& Montgomery) do provide an intended meaning (for
their text) of the term "irrational number" ...
"A rational number is one that is expressible as the quotient of two
integers. Real numbers that are not rational are said to be
irrational."
quoted from page 297 of:
An Introduction to the Theory of Numbers, 5th Ed (1991)
Niven / Zuckerman / Montgomery
Of course, you could argue that they are only defining "real
irrationals", but I think the more natural interpretation of the above
quote is that the authors are defining "irrational" to mean real but
not rational.
I think the rationale (no pun intended) for restricting the term
"irrational" to only real numbers is that, in practice, the descriptor
"irrational" is primarily used to make statements about real numbers.
Thus, if "real" is implicitly assumed, you don't have to constantly
qualify theorems by saying, for example ...
"If x is irrational and real then (some conclusion)."
or
"If (some hypothesis) then x is irrational and real."
quasi
Arturo Magidin
G. Frege <nomail@invalid> wrote:
>On Tue, 2 Oct 2007 02:45:08 +0000 (UTC), mag...@math.berkeley.edu
>(Arturo Magidin) wrote:
>
>>>
>>> I don't know where people get this idea.
>>> An irrational number is a number that is not rational.
>>>
>> Usually, I see it defined as a ->real<- number that is not
>> rational.
>>
>
>Which is in agreement with what Wikipedia says:
I would be loath to call upon Wikipedia as an authority here. We are
talking about a source that is at best a tertiary source (using
Wikipedia's own source classification), and as such will reflect a
distillation or summary by editors of definitions they have seen.
I would not use the Britannica or other "encyclopedic" sources either
(even if they are purely mathematical, like Mathworld, or even books
like Schwartzman's "The Words of Mathematics: An etymological
dictionary of mathematical terms used in English").
In this respect, looking at textbooks is really the best way to try to
figure out (i) whether a particular definition actually occurs in the
literature; and to a lesser extent (ii) which definition seems to be
"most common".
For the record, here's another one in which "irrational" is restricted
to real numbers: "The Geometry of Numbers" by C.D. Olds, Anneli Lax,
and Guiliana Davidoff (Anneli Lax New Mathematical Library Volume 41,
MAA, 2000). In Chapter 10, they state the following theorem of
Tchebychev:
THEOREM 10.1. Let theta be irrational and suppose that a is any
real number for which the equation x - theta*y - a = 0 has no
solutions in integers p, q. Then for any given positive e, there
are infinitely many pairs of integers p,q such that
| q(p - theta*q - a) | < 2,
and, at the same time, such that | p - theta*q - a | < e.
If we allow theta=i, then x - iy - a has no integer solutions for x
and y, but clearly |p - i*q - a| < e would have only finitely many
solutions for e<1, since it would require q=0.
Phil Carmody
Then insert "strictly" twice. 't ain't hard.
Michael Press
<gerry-A6B842....@sunb.ocs.mq.edu.au>,
Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email>
wrote:
> In article <u3q0g31se5bi3la4f...@4ax.com>,
> G. Frege <nomail@invalid> wrote:
>
> > On Mon, 01 Oct 2007 03:28:59 +0200, G. Frege <nomail@invalid> wrote:
> >
> > >>
> > >> 3) If a*a is irrational, then a is irrational.
> > >>
> > > Claim: a*a is <i> => a is <i>
> > >
> > > Proof: Suppose a*a is <i>. Now assume a is <r>. Then a*a is <r> too.
> > > Contradiction! []
> > >
> > Actually, 3) does not hold, and hence the alleged proof is not (quite)
> > correct.
> > 4__ __
> > Counterexample: Let a = -i * v(2). Then a*a = -v(2). With other words,
> > a*a is irrational. But a is NOT irrational (i.e. not a real number
> > which is not rational), since it is a (purely) imaginary number.
>
> I don't know where people get this idea.
> An irrational number is a number that is not rational.
> In particular, every complex number with non-zero
> imaginary part is irrational.
>
> This is the standard definition in Number Theory.
If a number field contains sqrt{-1},
then sqrt{-1} is also an integer,
since it is a root of the monic polynomial xx + 1.
How do you manage that?
--
Michael Press
Gerry Myerson
mag...@math.berkeley.edu (Arturo Magidin) wrote:
Gelfond's 1929 result is not the Gelfond-Schneider Theorem,
just an important step on the way.
Here's Hardy & Wright, The Theory of Numbers, 4th edition, p. 176:
"It has been proved more recently that alpha^beta is transcendental
if alpha and beta are algebraic, alpha is not 0 or 1, and beta is
irrational. This shows in particular that e^(-pi), which is one of the
values of i^(2i), is transcendental."
So evidently Hardy & Wright are happy to call 2i irrational.
Here's Alan Baker, A Concise Introduction to the Theory of Numbers,
p. 54:
"In 1900, Hilbert raised as the seventh of his famous list of 23
problems, the question of proving the transcendence of 2^sqrt2 and,
more generally, that of alpha^beta for algebraic alpha not 0 or 1
and algebraic irrational beta. ...in 1929 ... Gelfond succeeded in
verifying the special case that e^pi = (-1)^(-i) is transcendental...."
So Baker (who won a Fields medal for his work on transcendence) is
happy to call -i irrational.
Burger and Tubbs, Making Transcendence Transparent, p. 113:
"Theorem 5.1 (The Gelfond-Schneider Theorem) Suppose alpha
and beta are algebraic numbers with alpha not 0, 1, and beta
irrational. Then alpha^beta is transcendental.
".... The transcendence of e^pi also follows immediately if we view
e^pi = i^(-2i)...."
I'm beginning to think that the discussion of whether i is irrational
is similar to the discussion that occasionally breaks out here of
whether 0 is a natural number. Both sides of the discussion can cite
supporting sources ad nauseam, and neither point of view is going
to cause anyone to prove false results, so long as people make clear
just which definition is in use at any given time.
In a context where only real numbers are under discussion (e.g.,
continued fractions, approximation by rationals) no harm is done by
saying "irrational" where I would prefer to say "real irrational." In a
context where nonreal numbers take part, I think one has to interpret
"irrational" to include all those a + b i with b not zero.
Gerry Myerson
Phil Carmody <thefatphi...@yahoo.co.uk> wrote:
> Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> writes:
> >
> > You can't do much mathematics without i,
> > and you can't call i rational,
> > so you either throw your hands up in the air,
> > or you call i irrational.
>
> Or you can both not call irrational and not call it rational.
>
> Let's see how your style of argument works in other contexts:
>
> You can't do much mathematics without 0,
> and you can't call 0 positive,
> so you either throw your hands up in the air,
> or you call 0 negative.
No, that's why we have the word "nonpositive."
Just as the positive integers come in three flavors (composite,
prime, and unit), so do the reals: positive, negative, and zero.
But the complex numbers come in only two flavors: rational,
and irrational.
lwa...@lausd.net
wrote:
> I don't know where people get this idea.
> An irrational number is a number that is not rational.
> In particular, every complex number with non-zero
> imaginary part is irrational.
>
> This is the standard definition in Number Theory.
And so the debate continues over whether the set of
irrationals is R\Q or C\Q. This came up earlier in
one of the so-called "crank" threads, in which
finite guy was trying to prove that the irrationals
don't belong on the real line. I pointed out that
ironically, the "crank" was right, if one uses
Myerson's definition of irrationals (since C\Q is
not a subset of R).
So apparently, there is no general consensus as to
whether R\Q or C\Q are the irrationals. I must
admit that the only context in which I've ever
seen C\Q is in the aforementioned Gelfond-Schneider
theorem, otherwise it usually refers to R\Q.
William Hughes
wrote:
> In article <fdu2p4$ub...@agate.berkeley.edu>,
> magi...@math.berkeley.edu (Arturo Magidin) wrote:
>
>
>
> > In article <fdu26g$89...@mailhub227.itcs.purdue.edu>,
About time too.
> that the discussion of whether i is irrational
> is similar to the discussion that occasionally breaks out here of
> whether 0 is a natural number.
No? Ya think?
> Both sides of the discussion can cite
> supporting sources ad nauseam, and neither point of view is going
> to cause anyone to prove false results, so long as people make clear
> just which definition is in use at any given time.
>
> In a context where only real numbers are under discussion (e.g.,
> continued fractions, approximation by rationals) no harm is done by
> saying "irrational" where I would prefer to say "real irrational." In a
> context where nonreal numbers take part, I think one has to interpret
> "irrational" to include all those a + b i with b not zero.
Piffle. It is obvious to anyone with two brain cells to rub
together that the only reasonable extention of rational/irrational
to complex values, is to say that a complex value is rational
if and only if it is the ratio of Gaussian integers.
Calling any imaginary value irrational, just to make the
statement of Gelfond-Schneider a bit simpler is dumb.
Your citations above only prove that even the best mathematicians
have lapses.
- William Hughes
P.S. You started out inflamatory, then wrote a semi-conciliatory
reponse,
ending with "I was really right anyway". I am not impressed.
Arturo Magidin
Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote:
>In article <fdu2p4$uba$1...@agate.berkeley.edu>,
> mag...@math.berkeley.edu (Arturo Magidin) wrote:
[...]
>> The old saw about standards and there being so many to choose from
>> comes to mind.
>
>Gelfond's 1929 result is not the Gelfond-Schneider Theorem,
Thank you, I know. LeVeque's statement of the Gelfond-Schneider
theorem happened to agree with the use in Niven (that is, using words
other than simply "irrational" so as not to restrict itself to real
numbers), but its statement of Gelfond's 1929 Theorem, with "imaginary
quadratic irrationalities", by contrast, seemed to lean the other
way.
[...]
>I'm beginning to think that the discussion of whether i is irrational
>is similar to the discussion that occasionally breaks out here of
>whether 0 is a natural number. Both sides of the discussion can cite
>supporting sources ad nauseam, and neither point of view is going
>to cause anyone to prove false results, so long as people make clear
>just which definition is in use at any given time.
I wonder what I was refering to above when alluding to the old saw
about standards...
Gerry Myerson
William Hughes <wpih...@hotmail.com> wrote:
> On Oct 2, 8:02 pm, Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email>
> wrote:
> >
> > I'm beginning to think
>
> About time too.
I've been posting here for over 15 years. I don't recall having ever
posted a personal insult aimed at you. I don't know why you choose
to aim one at me, but may I say I don't think it does anything for
your image here.
> > In a context where only real numbers are under discussion (e.g.,
> > continued fractions, approximation by rationals) no harm is done by
> > saying "irrational" where I would prefer to say "real irrational." In a
> > context where nonreal numbers take part, I think one has to interpret
> > "irrational" to include all those a + b i with b not zero.
>
> Piffle. It is obvious to anyone with two brain cells to rub
> together that the only reasonable extention of rational/irrational
> to complex values, is to say that a complex value is rational
> if and only if it is the ratio of Gaussian integers.
> Calling any imaginary value irrational, just to make the
> statement of Gelfond-Schneider a bit simpler is dumb.
You're saying G H Hardy and Alan Baker are dumb
and don't have two brain cells each. Given a choice between
your intellect (or mine) and theirs, who do you more people
would cite as dumb?
There is an interesting mathematical question here. Why, among all
subrings of the ring of algebraic integers, might one single out the
ring of Gaussian integers as special? or, why, among all subfields
of the complex numbers, single out Q(i)? Sure, i generates the complex
numbers, in the sense that C = R(i), but if alpha is any nonreal
complex number, then C = R(alpha), so there's nothing special about
i there. Z is special because it is generated by 1, and 1 is the
multiplicative identity; but i is no identity. It's a root of unity, but
there are lots of those around. So what's so special about the Gaussian
integers?
Gerry Myerson
Michael Press <rub...@pacbell.net> wrote:
How do I manage what?
sqrt(-1) is not an integer; it is an algebraic integer.
Given enough context, it is permissible to abuse the terminology
and refer to sqrt(-1) as an integer, but whenever people do that
they know they really mean it's an algebraic integer. Hey, if you're
working in the field Q(sqrt2), you might refer to sqrt2 as an integer.
Are you suggesting then that it's wrong to call sqrt2 an irrational?
quasi
<wpih...@hotmail.com> wrote:
>Piffle. It is obvious to anyone with two brain cells to rub
>together that the only reasonable extention of rational/irrational
>to complex values, is to say that a complex value is rational
>if and only if it is the ratio of Gaussian integers.
That doesn't make sense to me.
If you declare the quotient of 2 Gaussian integers rational, then why
wouldn't you do the same for the elements of Z[sqrt(2)].
If a,b are ordinary integers, then
a + b*sqrt(2) is an algebraic integer, right?
Note that it's not much different in form than
a + b*sqrt(-1)
Thus, if you are going to regard
(a + b*i) / (c + d*i)
as rational, then why wouldn't
(a + b*sqrt(2)) / (c + d*sqrt(2))
also be considered rational?
Of course that would mean sqrt(2) is rational.
This argues against regarding the quotient of 2 Gaussian integers as
rational.
The fact is, as far as I can see, most sources define irrational as
reals which are not rational.
However, there's nothing really wrong with calling i irrational. The
justification is simple -- i is not rational.
Also, just look at it ...
sqrt(-1)
It's a square root that doesn't simplify.
In that regard, it's conceptually just as irrational as
sqrt(2)
My point is this ...
Defining irrationals as reals which are not rational, while perhaps
too limiting in some respects, appears to be the most standard
interpretation.
Defining irrationals as complexes which are not rational is used by
some authors, but not most. However it certainly seems to be a
reasonable definition, and the greater generality makes it useful for
some theorems.
Defining irrationals as complex numbers which are not the quotient of
2 gaussian integers gives too much priority to a particular quadratic
extension of Z, namely Z[i]. Such a definition seems a little strange
to me, and I doubt it would be useful as a general concept of
irrationality.
quasi
Arturo Magidin
quasi <qu...@null.set> wrote:
>On Tue, 02 Oct 2007 17:48:03 -0700, William Hughes
><wpih...@hotmail.com> wrote:
>
>>Piffle. It is obvious to anyone with two brain cells to rub
>>together that the only reasonable extention of rational/irrational
>>to complex values, is to say that a complex value is rational
>>if and only if it is the ratio of Gaussian integers.
>
>That doesn't make sense to me.
>
>If you declare the quotient of 2 Gaussian integers rational, then why
>wouldn't you do the same for the elements of Z[sqrt(2)].
I suspect the reason is quite simple: under that definition, the
"rational" complex numbers would be exactly those that can be
expressed as a + bi with a and b real rationals, and the "irrationals"
would be everything else. In other words, if you identify the complex
numbers with the plane, then the "rationals" would be Q x Q, the
points with rational coordinates. This would allow all the
approximation theorems to go through.
That said, I also suspect that William is being more than a bit
sarcastic in his preamble, judging from his postscript.
>The fact is, as far as I can see, most sources define irrational as
>reals which are not rational.
>
>However, there's nothing really wrong with calling i irrational. The
>justification is simple -- i is not rational.
Neither is the number 2... in GF(3) (there it corresponds to an
equivalence class of integers, hence not an integer or a ratio of
integers).
The definition of rational as "a number that can be expresssed as
quotient of two integers" assumes a scope for "number". It is not
always immediately clear what that scope is supposed to be.
[...]
>Defining irrationals as reals which are not rational, while perhaps
>too limiting in some respects, appears to be the most standard
>interpretation.
>
>Defining irrationals as complexes which are not rational is used by
>some authors, but not most. However it certainly seems to be a
>reasonable definition, and the greater generality makes it useful for
>some theorems.
Certainly, as already noted ad nauseam, the two most common uses give
either R\Q or C\Q as the meaning for "irrational". Both are "valid",
in the sense that both are reasonable meanings for the term and so
long as they are used consistently, are unlikely to cause any
problems. Just like any other number of multiple conventions (is the
empty set a subsemigroup of a semigroup [by 'subsemigroup' I mean a
set with a binary associative operation]? Yes, according to some
authors, no, according to others; each has advantages; e.g., the
intersection of two subsemigroups is not necessarily a subsemigroup if
you do not allow the empty semigroup... Hell, for that matter,
"semigroup" doesn't always means the same thing, nor does "monoid")
I certainly did not dispute the ->validity<- of saying "irrationals"
means C\Q. Rather, I disputed the assertion that it was the
->standard<- meaning "in Number Theory", and presented evidence that
explained where people might get the idea that "irrational" may mean
R\Q instead. But at the same time, I also provided evidence that it
->sometimes<- means C\Q. (For what it is worth, since I am summarizing
what I said, I also disputed the use of Wikipedia or PlanetMath as a
good place to look for what may or may not be the "standard meaning",
assuming such a thing existed).
Michael Press
<gerry-D04D24....@sunb.ocs.mq.edu.au>,
Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email>
wrote:
> In article <rubrum-0FC979....@newsclstr03.news.prodigy.net>,
> Michael Press <rub...@pacbell.net> wrote:
>
> > In article
> > <gerry-A6B842....@sunb.ocs.mq.edu.au>,
> > Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email>
> > wrote:
> >
> > > I don't know where people get this idea.
> > > An irrational number is a number that is not rational.
> > > In particular, every complex number with non-zero
> > > imaginary part is irrational.
> > >
> > > This is the standard definition in Number Theory.
> >
> > If a number field contains sqrt{-1},
> > then sqrt{-1} is also an integer,
> > since it is a root of the monic polynomial xx + 1.
> > How do you manage that?
>
> How do I manage what?
Well, you do answer this anyway.
> sqrt(-1) is not an integer; it is an algebraic integer.
> Given enough context, it is permissible to abuse the terminology
> and refer to sqrt(-1) as an integer, but whenever people do that
> they know they really mean it's an algebraic integer. Hey, if you're
> working in the field Q(sqrt2), you might refer to sqrt2 as an integer.
> Are you suggesting then that it's wrong to call sqrt2 an irrational?
In Q(sqrt2) what is an irrational number?
In Q(sqrt2), sqrt2 is the ratio of two integers.
Since they are not rational integers,
then we can call sqrt2 an irrational integer.
Still, what is the utility of the definition?
--
Michael Press
quasi
(Arturo Magidin) wrote:
>In article <ra36g3lktcj1hijaf...@4ax.com>,
>quasi <qu...@null.set> wrote:
>>On Tue, 02 Oct 2007 17:48:03 -0700, William Hughes
>><wpih...@hotmail.com> wrote:
>>
>>>Piffle. It is obvious to anyone with two brain cells to rub
>>>together that the only reasonable extention of rational/irrational
>>>to complex values, is to say that a complex value is rational
>>>if and only if it is the ratio of Gaussian integers.
>>
>>That doesn't make sense to me.
>>
>>If you declare the quotient of 2 Gaussian integers rational, then why
>>wouldn't you do the same for the elements of Z[sqrt(2)].
>
>I suspect the reason is quite simple: under that definition, the
>"rational" complex numbers would be exactly those that can be
>expressed as a + bi with a and b real rationals, and the "irrationals"
>would be everything else. In other words, if you identify the complex
>numbers with the plane, then the "rationals" would be Q x Q, the
>points with rational coordinates. This would allow all the
>approximation theorems to go through.
Ok, that's a good defense.
Thus, it seems now that the 3 definitions previously discussed all
have some validity.
>That said, I also suspect that William is being more than a bit
>sarcastic in his preamble, judging from his postscript.
Well, the 2 brain cells comment was simply wrong.
quasi
Gerry Myerson
Michael Press <rub...@pacbell.net> wrote:
I don't think there is any.
I'm not sure I know what we're arguing about.
I maintain that sqrt(-1) is irrational.
You note that (in some contexts) sqrt(-1) is an integer.
In the context of this thread, I took it that you were putting
this forth as evidence against my position that sqrt(-1) is
irrational. If that's not what you were trying to do, then I'm
afraid I don't know wnat your point was in bringing up
sqrt(-1). If that was what you were trying to do, then my
point was that the same argument that suggests sqrt(-1)
isn't irrational also suggests sqrt2 isn't irrational, which
doesn't look too good. Or, the same argument you use to
call sqrt2 irrational, shows that sqrt(-1) is irrational, as
it isn't the ratio of two rational integers.
I hope I've made myself less unclear than usual.
Gerry Myerson
quasi <qu...@null.set> wrote:
> On Wed, 3 Oct 2007 04:35:14 +0000 (UTC), mag...@math.berkeley.edu
> (Arturo Magidin) wrote:
>
> >In article <ra36g3lktcj1hijaf...@4ax.com>,
> >quasi <qu...@null.set> wrote:
> >>On Tue, 02 Oct 2007 17:48:03 -0700, William Hughes
> >><wpih...@hotmail.com> wrote:
> >>
> >>>Piffle. It is obvious to anyone with two brain cells to rub
> >>>together that the only reasonable extention of rational/irrational
> >>>to complex values, is to say that a complex value is rational
> >>>if and only if it is the ratio of Gaussian integers.
> >>
> >>That doesn't make sense to me.
> >>
> >>If you declare the quotient of 2 Gaussian integers rational, then why
> >>wouldn't you do the same for the elements of Z[sqrt(2)].
> >
> >I suspect the reason is quite simple: under that definition, the
> >"rational" complex numbers would be exactly those that can be
> >expressed as a + bi with a and b real rationals, and the "irrationals"
> >would be everything else. In other words, if you identify the complex
> >numbers with the plane, then the "rationals" would be Q x Q, the
> >points with rational coordinates. This would allow all the
> >approximation theorems to go through.
>
> Ok, that's a good defense.
It still privileges i in a way I find unconvincing.
Let w be a complex cube root of 1.
Why not let the "rational complex numbers" be
those that can be expressed as a + b w with a, b
real rationals? Q(w) is still a 2-dimensional vector
space over Q, thus it is Q x Q with respect to the basis
{1, w}. The approximation theorems go through, no?
Robert Israel
> In article <ra36g3lktcj1hijaf...@4ax.com>,
> quasi <qu...@null.set> wrote:
> >On Tue, 02 Oct 2007 17:48:03 -0700, William Hughes
> ><wpih...@hotmail.com> wrote:
> >
> >>Piffle. It is obvious to anyone with two brain cells to rub
> >>together that the only reasonable extention of rational/irrational
> >>to complex values, is to say that a complex value is rational
> >>if and only if it is the ratio of Gaussian integers.
> >
> >That doesn't make sense to me.
> >
> >If you declare the quotient of 2 Gaussian integers rational, then why
> >wouldn't you do the same for the elements of Z[sqrt(2)].
>
> I suspect the reason is quite simple: under that definition, the
> "rational" complex numbers would be exactly those that can be
> expressed as a + bi with a and b real rationals, and the "irrationals"
> would be everything else. In other words, if you identify the complex
> numbers with the plane, then the "rationals" would be Q x Q, the
> points with rational coordinates. This would allow all the
> approximation theorems to go through.
How do you mean? In what way does Hurwitz's theorem on Diophantine
approximation, say, go through?
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Phil Carmody
> But the complex numbers come in only two flavors: rational,
> and irrational.
You are here simply repeating a prior assertion that
you've seen several disagree with. Assertions do not
get more convincing the more they are repeated.
Arturo Magidin
Robert Israel <isr...@math.MyUniversitysInitials.ca> wrote:
>mag...@math.berkeley.edu (Arturo Magidin) writes:
>
>> In article <ra36g3lktcj1hijaf...@4ax.com>,
>> quasi <qu...@null.set> wrote:
>> >On Tue, 02 Oct 2007 17:48:03 -0700, William Hughes
>> ><wpih...@hotmail.com> wrote:
>> >
>> >>Piffle. It is obvious to anyone with two brain cells to rub
>> >>together that the only reasonable extention of rational/irrational
>> >>to complex values, is to say that a complex value is rational
>> >>if and only if it is the ratio of Gaussian integers.
>> >
>> >That doesn't make sense to me.
>> >
>> >If you declare the quotient of 2 Gaussian integers rational, then why
>> >wouldn't you do the same for the elements of Z[sqrt(2)].
>>
>> I suspect the reason is quite simple: under that definition, the
>> "rational" complex numbers would be exactly those that can be
>> expressed as a + bi with a and b real rationals, and the "irrationals"
>> would be everything else. In other words, if you identify the complex
>> numbers with the plane, then the "rationals" would be Q x Q, the
>> points with rational coordinates. This would allow all the
>> approximation theorems to go through.
>
>How do you mean? In what way does Hurwitz's theorem on Diophantine
>approximation, say, go through?
Okay, first, I am not advocating such a definition; in the part that
quasi snipped, I think I made that somewhat clear.
Second, I clearly misspoke and should have said "most", rather than
"all". Certainly things like "every irrational can be approximated by
a sequence of rationals", and the like.
That said, I'm getting a little tired of pummeling this defunct
equine...
William Hughes
> On Wed, 3 Oct 2007 04:35:14 +0000 (UTC), magi...@math.berkeley.edu
> >That said, I also suspect that William is being more than a bit
> >sarcastic in his preamble, judging from his postscript.
>
> Well, the 2 brain cells comment was simply wrong.
>
Well, I suppose I could have been even more over the
top, but I thought it was pretty good.
(I even managed a cliche!)
- William Hughes
Jesse F. Hughes
> Assertions do not get more convincing the more they are repeated.
They do so. Do so, do so, do so, do so! DO SO!!!!
--
"People make mistakes. Better to live today and learn the truth, than
to be one of those poor saps who died deluded, thinking they knew
certain things that they just didn't. Thinking they had proofs that
they didn't." --James S. Harris, almost too sad for a .sig
Phil Carmody
> Phil Carmody <thefatphi...@yahoo.co.uk> writes:
>
> > Assertions do not get more convincing the more they are repeated.
>
>
> They do so. Do so, do so, do so, do so! DO SO!!!!
Well now you put it /that/ way ;-)
Michael Press
<gerry-07169A....@sunb.ocs.mq.edu.au>,
Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email>
wrote:
Thanks. It is not all clear to me, but I am
successfully communicating my confusion.
H.E. Rose in _A_Course_in_Number_Theory_ states the
Gelfond-Schneider theorem as
If a and b are algebraic numbers, a =/= 0 or 1, and b
is irrational then w = a^b is transcendental.
With a bit of thinking the meaning is clear,
but I prefer that statements of theorems to be clear
_without_ thinking. :)
I prefer
If a and b are algebraic numbers, a =/= 0 or 1,
and b not in Q, then w = a^b is transcendental.
--
Michael Press
A little learning is a dangerous thing.
G. Frege
wrote:
>
> H.E. Rose in _A_Course_in_Number_Theory_ states the
> Gelfond-Schneider theorem as
>
> If a and b are algebraic numbers, a =/= 0 or 1, and b
> is irrational then w = a^b is transcendental.
>
> With a bit of thinking the meaning is clear,
> but I prefer that statements of theorems to be clear
> _without_ thinking. :)
>
> I prefer
>
> If a and b are algebraic numbers, a =/= 0 or 1,
> and b not in Q, then w = a^b is transcendental.
>
How about
If a and b are algebraic numbers, a =/= 0 or 1,
and b not rational, then w = a^b is transcendental.
?
(I guess, the definition of /rational/ is less "controversial" than
the definition of /irrational/, no?)
F.
--
E-mail: info<at>simple-line<dot>de
Gerry Myerson
Phil Carmody <thefatphi...@yahoo.co.uk> wrote:
> Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> writes:
> > But the complex numbers come in only two flavors: rational,
> > and irrational.
>
> You are here simply repeating a prior assertion that
> you've seen several disagree with. Assertions do not
> get more convincing the more they are repeated.
Let's see if this is more convincing.
The concepts of rational & irrational were originally developed
by Pythagoras & Co. in the context of the real numbers. Some
millennia later, the complex numbers were discovered/invented/
whatever. Mathematicians being prone to generalization, it
occurred to someone to ask whether the rational/irrational
distinction could be extended to this new domain. Now, you're
certainly within your rights to say, no, it can't; irrational is a
subconcept of real, and nonreal complex numbers are neither
rational nor irrational. But the fact is that mathematicians have
found it useful to extend the concept to the complex numbers,
in a way that is consistent with the pre-existing concept, and in
a way that makes it easier to state some theorems.
An individual can always reject a generalization, but mathematics
will always accept a generalization, if it is useful and is not seen to
lead to contradictions. There's a good reason to call i irrational,
and no good reason not to, so, I suggest, that's the way to go.
G. Frege
>
> So apparently, there is no general consensus as to
> whether R\Q or C\Q are the irrationals. I must
> admit that the only context in which I've ever
> seen C\Q is in the aforementioned Gelfond-Schneider
> theorem, otherwise it usually refers to R\Q.
>
While on the other hand I've several times read that the rational
numbers + the irrational numbers form the real numbers.
Just tried to find a quote:
"The field of all rational and irrational numbers is called the real
numbers, or simply the "reals," and denoted IR."
(http://mathworld.wolfram.com/RealNumber.html)
Michael Press
<0488g3hfkudcbktca...@4ax.com>,
G. Frege <nomail@invalid> wrote:
Yes. It works for me.
--
Michael Press
David Bernier
To get the gcd of 7 and 100 or -24 and 81, all one really needs to do
is perform
repeated additions and subtractions. Eventually, one gets a pair
consisting
of +/- the gcd, and 0; the process stops there. If a, b are non-zero
reals, and
a/b is not in Q, the euclidean algorithm goes on forever (the ancient
Greeks).
For complex numbers, if u and v are both non-zero, and u/v is not a real
number, then arg(u/v) is neither 0 nor pi radians. If the euclidean
algorithm ever
comes to a halt, then by reversing the procedure, u = m*a, v = n*a, with
m, n in Z and some non-zero complex number a. But then
u/v = m/n is in Q. Since u/v is non-real, it is not in Q so the euclidean
algorithm for u and v goes on forever. In that sense, u/v is irrational
as u and v are incommensurable. If we let v = 1, and u is not in R,
so u in C\R, then u and 1 are incommensurable and in that sense
u/1 is irrational, i.e. u is irrational.
Since we're only using addition and subtraction, I wonder if the euclidean
algorithm can be generalized to some abelian groups. But it seems
some sort of norm or metric on the group would be needed ...
David Bernier
G. Frege
Agree with what you said.
Now I claimed (in my original post):
a is NOT irrational (i.e. not a real number which is not rational),
since it is a (purely) imaginary number.
stating in parenthesis the definition if /irrational/ I'm using here.
Strangely, Gerry Myerson questioned that rather common definition of
/irrational/ (as if there were "true" and "false" definitions):
"I don't know where people get this idea. [...]"
Well...
Gerry Myerson
David Bernier <davi...@videotron.ca> wrote:
> To get the gcd of 7 and 100 or -24 and 81, all one really needs to
> do is perform repeated additions and subtractions.
Although one has to know which additions and subtrtactions
to perform. The useful additiions and subtractions are the ones
that decrease the absolute values. There is always a useful
addition or subtraction because you can prove that if a and b
are nonzero then one of the numbers a + b, a - b has absolute
value less than the larger of |a| and |b|.
> Eventually, one gets a pair consisting of +/- the gcd, and 0; the
> process stops there. If a, b are non-zero reals, and a/b is not in
> Q, the euclidean algorithm goes on forever (the ancient Greeks).
>
> For complex numbers, if u and v are both non-zero, and u/v is not a
> real number, then arg(u/v) is neither 0 nor pi radians. If the
> euclidean algorithm ever comes to a halt,
The problem isn't halting, it's starting. If you try to apply your
procedure to, say, 1 and i, then neither 1 + i nor 1 - i has absolute
value smaller than |1| = |i| = 1, so you can't even begin to apply
repeated addition and/or subtraction.
> then by reversing the procedure, u = m*a, v = n*a, with m, n in Z
> and some non-zero complex number a. But then u/v = m/n is in Q.
> Since u/v is non-real, it is not in Q so the euclidean algorithm for
> u and v goes on forever. In that sense, u/v is irrational as u and v
> are incommensurable. If we let v = 1, and u is not in R, so u in
> C\R, then u and 1 are incommensurable and in that sense u/1 is
> irrational, i.e. u is irrational.
>
> Since we're only using addition and subtraction, I wonder if the
> euclidean algorithm can be generalized to some abelian groups. But
> it seems some sort of norm or metric on the group would be needed ...
There is a Euclidean algorithm in, for example, the Gaussian integers,
Z[i], but it uses the multiplication in Z[i], which is not just repeated
addition. The Division Theorem for Z[i] says that given any a, b
in Z[i], b not zero, there exist q, r in Z[i] with a = b q + r and
with |r| < |b|. From this, you can get a Euclidean algorithm, unique
factorization, etc., in Z[i].
Keith Ramsay
On Oct 2, 4:15 am, Phil Carmody <thefatphil_demun...@yahoo.co.uk>
wrote:
|Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> writes:
|> Wikipedia, planetmath, and dr.math to the contrary notwithstanding,
|> this is the absolutely standard use in (sufficiently) advanced
math.
|
|Let's play 'spot the "no true Scotsman" fallacy', shall we?
My impression is that the "true Scotsman" fallacy is much
overdiagnosed. It's only really fair to call him on it if
there's some evidence of abusing the concept of "degree of
advancedness" in mathematics to fit his claim. As far as I
can see, we have a pretty good idea of what counts as more
or less advanced mathematics.
I don't know for sure whether Gerry Myerson is right, but
his claim fits my own experience, at least if we say
"sufficiently advanced number theory". At a certain level
of advancement, as a graduate student, the instances of
number theorists referring to things other than real
numbers as "irrational" started appearing, and I don't
remember anybody insisting upon the opposite until now.
Yes, I too did a double-take when I first heard it, but
just once.
I suspect the case that tipped the balance in favor of the
usage Gerry Myerson and I are accustomed to, is discussions
of fields not "automatically" equipped with an embedding
into the complex numbers. The field
Q[x]/(x^3-2)
has three embeddings into the complex numbers, which take
x to each of the three cube roots of 2, respectively. One
could say, *if* we embed it with x going to the usual (real)
cube root, then some of the elements are rational and the
rest are irrational, and *if* we embed it with x going to
one of the other cube roots, then some of the elements are
rational while the rest are "neither rational nor
irrational, because they're non-real complex numbers". But
there doesn't seem much point in describing the situation
in such terms. Whether you've embedded some elements of the
field into the real line or not is just beside the point.
Some of the elements of the field are rational; the rest
are irrational (in our terms) however you choose to embed
the field into the complex numbers. (And likewise for
"transcendental" elements of the field.) It's a purely
algebraic property of the element, instead of being partly
algebraic and partly how it's approximated by rationals.
It seems to me to be another case where a concept C is taken
to presume some condition P which is needed in some sense
to make it a relevant concept, until at some point enough
time is spent considering cases where it's unknown,
meaningless, or irrelevant whether P holds, but one wants to
talk about the concept C anyway, and it's simpler to make
a minor generalization to C which covers the case where P
doesn't hold. Then some minority of authors will use this
broader definition of the concept. Then there'll be some
kind of argument. Then the argument will get tiresome. But
then the number of people using it will increase. And
finally, everybody will use it, and they'll forget that it
was ever an issue.
For example, if you have some set X which is known to be
empty, whether "every element of X is an element of Y" for
a second set Y is not an interesting question. In common
parlance, "every X is a Y" presumes that there are some Xs.
If there aren't, then it's thought of as meaningless or
false. At some point, in mathematics, it must have gotten
to be enough of a pain to have to treat the case of X being
empty separately, and the convention that when there aren't
any Xs, "every X is a Y" is true (vacuously) became
standard.
So perhaps the convention of calling complex numbers that
aren't real irrational hasn't appeared in your grocer's
freezer section yet. :-) Maybe your friends and family
haven't developed a taste for it yet. Get ahead of the
curve! Jump on the bandwagon today! Make your terminology
more like the hip fashions of the elite. You can't get
the U.S. to convert to metric, but you can be the first on
your block to use the mathematical terminology of tomorrow.
Keith Ramsay
G. Frege
<renf...@cmich.edu> wrote:
>
> Here's something to play with.
>
> We know an irrational number to an irrational power can be
> rational. For instance, a^b is an example for a = b = sqrt(2)
> or for a = sqrt(2)^sqrt(2) and b = sqrt(2).
>
> QUESTION: Can a^(b^c) be rational with a, b, c all irrational?
>
Yes.
Proof: sqrt(2)^sqrt(2) either is rational or not.
Assume it's rational. Then let b = c = sqrt(2). Hence b^c e Q.
This means that there are integers x, y (y =/= 0) such that b^c = x/y.
Since b^c > 0 there are _pos._ integers n, m such that b^c = n/m with
gcd(n,m) = 1. n > 1 and n > m (since b^c = sqrt(2)^sqrt(2) > 1.6).
Now let a = n-th root of 2^m: a = 2^(m/n). a is irrational [easy!],
and a^(b^c) = (2^(m/n))^(n/m) = 2^(m/n * n/m) = 2 e Q.
Now assume sqrt(2)^sqrt(2) is /not/ rational. Then let a = c = sqrt(2)
and b = sqrt(2)^sqrt(2). Then b^c = 2 and a^(b^c) = sqrt(2)^2 = 2 e Q.
Hence in either case there are irrational numbers a, b, c such that
a^(b^c) is rational. qed.
Bill Dubuque
>
> [...] At a certain level of advancement, as a graduate student,
> the instances of number theorists referring to things
As I mentioned the last time this query arose here [1]
(in a thread titled "Is i irrational?" on Mar 7, 2006)
a Google Books search quickly confirms that this modern
usage is widespread among eminent mathematicians, e.g.
Conway, Gelfond, Manin, Ribenboim, Shafarevich, Waldschmidt
(esp. in diophantine approximation, e.g. Thue-Siegel-Roth
theorem, Gelfond-Schneider theorem, etc).
--Bill Dubuque
[1] http://google.com/groups?threadm=y8zbqwgfv1r.fsf%40nestle.csail.mit.edu
Bill Dubuque
>David Bernier <davi...@videotron.ca> wrote:
>>
>> Since we're only using addition and subtraction, I wonder if the
>> euclidean algorithm can be generalized to some abelian groups. But
>> it seems some sort of norm or metric on the group would be needed ...
>
> There is a Euclidean algorithm in, for example, the Gaussian integers,
> Z[i], but it uses the multiplication in Z[i], which is not just repeated
> addition. The Division Theorem for Z[i] says that given any a, b
> in Z[i], b not zero, there exist q, r in Z[i] with a = b q + r and
> with |r| < |b|. From this, you can get a Euclidean algorithm, unique
> factorization, etc., in Z[i].
In fact there is a pretty generalization to arbitrary PIDs.
Namely the so-called Dedekind-Hasse criterion states that
a domain is a PID iff given any two nonzero elts a, b in D,
either a|b or some D-linear combination is "smaller" than a.
It's clear that such a domain must be principal (since then
the "smallest" elt in an ideal must divide all other elts).
Conversely, since a PID is UFD, an adequate metric is
the number of prime factors (if not a|b then their gcd c
must have fewer prime factors; for if (a,b) = (c) then
c|a _properly_, else a|c|b contra hypothesis). Clearly
the Euclidean descent via the Division Algorithm is just a
special case, so Euclidean -> PID (-> {UFD,Bezout} -> GCD)
--Bill Dubuque