Ten equals 9.9999....
Richard Tjoa
9.9999... (nine point nine repeating) = 10 (ten)
_
9.9 = x (yes, I mean "repeating" with the bar over it.)
_
99.9 = 10x (multiply by 10)
_
- 9.9 = - x (subtract that first thing)
--------------
90 = 9x (oooh, no more decimal)
10 = x (divide by 9)
_
9.9 = 10 (voila! [QED])
-Richard
(My personal favorite)
How about 1 = 2 later?
Michael Neylon
>Ok, here's something interesting...
>9.9999... (nine point nine repeating) = 10 (ten)
> _
> 9.9 = x (yes, I mean "repeating" with the bar over it.)
> _
> 99.9 = 10x (multiply by 10)
> _
> - 9.9 = - x (subtract that first thing)
> --------------
> 90 = 9x (oooh, no more decimal)
> 10 = x (divide by 9)
> _
> 9.9 = 10 (voila! [QED])
>
>
If I remember correctly, there was a discussion last year that
9999999999(repeating) === 1
so this is no fallacy (9.99999... = 9 + .999999... = 9 + 1 = 10)
Mike Neylon aka Masem the Great and Almighty Thermo GOD
// Only the | Univ. of Toledo - Chemical Engineering - Senior
\X/ AMIGA! | NASA Lewis Research Center - Summer Intern
\_____________|_____s...@ariel.lerc.nasa.gov________________/
-+ How do YOU spell 'potato'? How about 'lousy'? +-
"Me and Spike are big Malcolm 10 supporters."-J.S.,PLCL
David Karr
>9.9999... (nine point nine repeating) = 10 (ten)
Ah, but the first equality is a mathematical truth (provided you
define decimal representation so that you are permitted to write an
infinite string of 9's in the first place (*)), while the second
equality is a falsehood.
So these really are two different topics altogether.
(*) You're depending on a view of decimal fractions which says to get
a decimal number you first throw together any sequence of digits, then
determine what real number that corresponds to. An alternative view
is that a decimal fraction is defined only as the limit of a
deterministic series of approximations to a certain real number. Thus
you get .3333... for 1/3, but you always get 1.0000... for 1, never
.9999...
-- David Karr (ka...@cs.cornell.edu)
Robert Cutler
Actually (repeating)9999999999 === -1
It follows from the fact that .999999999 === 1
9999999999.999999999999 = x
9999999999.999999999999 = 10 x
multiply by ten and they are still equal. hence x = 0 and since .99999 = 1
(repeating)999999999. must equal -1 to add up to zero.
If you'll accept that I have a really neat trick with infinity
mod and infinitely large prime number. (Pascals Triangle)
t911...@phillip.edu.au
> Ok, here's something interesting...
> 9.9999... (nine point nine repeating) = 10 (ten)
There's the error. 9.9999... does not equal 10, it approaches 10.
> _
> 9.9 = x (yes, I mean "repeating" with the bar over it.)
> _
> 99.9 = 10x (multiply by 10)
> _
> - 9.9 = - x (subtract that first thing)
> --------------
> 90 = 9x (oooh, no more decimal)
> 10 = x (divide by 9)
> _
> 9.9 = 10 (voila! [QED])
>
>
> -Richard
> (My personal favorite)
> How about 1 = 2 later?
--
Adam Eberbach t911...@phillip.edu.au fnord
Computer Science student, Phillip Institute of Technology.
Dave Hare
>In article <1992Jul24.1...@pasteur.Berkeley.EDU>, tj...@cory.Berkeley.EDU (Richard Tjoa) writes:
>
>> Ok, here's something interesting...
>> 9.9999... (nine point nine repeating) = 10 (ten)
>
>There's the error. 9.9999... does not equal 10, it approaches 10.
No, it equals 10. This type of question has subtleties that too often
are too easily dismissed. Consider:
- What exactly is a non-terminating decimal?
- How will you determine that two numbers are *not* equal
- If x is a non-terminating decimal, how do you define 10*x? (Recall
that for ordinary multiplication, the computation proceeds from
least significant digit to most significant digit, but x has no
least significant digit.) And how do you define the other arithmetic
operations?
The problem is that we are cavalierly taught in grade school that if a
non-terminating decimal is periodic, then we just multiply it by 10^{the
period}, subtract the original, divide both sides by 10^{the period}-1
and reduce to lowest terms, and we have that number as an exact rational.
These steps require justification. HOWEVER, that justification is possible,
and once it is done, then statements such as 9.99999... = 10 make perfect
sense.
Dave Hare
t911...@phillip.edu.au
> In article <1992Jul25.1...@phillip.edu.au> t911...@phillip.edu.au writes:
>>In article <1992Jul24.1...@pasteur.Berkeley.EDU>, tj...@cory.Berkeley.EDU (Richard Tjoa) writes:
>>
>>> Ok, here's something interesting...
>>> 9.9999... (nine point nine repeating) = 10 (ten)
>>
>>There's the error. 9.9999... does not equal 10, it approaches 10.
>
> No, it equals 10. This type of question has subtleties that too often
> are too easily dismissed. Consider:
Alright, if 9.999... = 10, then why isn't .33333... exactly 1/3?
I don't see how they are equal. I know that 9.999... with infinite 9's
is very close to 10, but that .0000...0001 is what you need to make
exactly 10.
>
> - What exactly is a non-terminating decimal?
Let's keep it simple - we're talking about recurring 9's.
>
> - How will you determine that two numbers are *not* equal
>
Why does it need a formal mathematical proof?
> - If x is a non-terminating decimal, how do you define 10*x? (Recall
> that for ordinary multiplication, the computation proceeds from
> least significant digit to most significant digit, but x has no
> least significant digit.) And how do you define the other arithmetic
> operations?
10*x is easy. Shift the decimal place one to the right, giving
99.9999.... which is still not equal to 10. (or 100 either.)
>
> The problem is that we are cavalierly taught in grade school that if a
> non-terminating decimal is periodic, then we just multiply it by 10^{the
> period}, subtract the original, divide both sides by 10^{the period}-1
> and reduce to lowest terms, and we have that number as an exact rational.
>
> These steps require justification. HOWEVER, that justification is possible,
> and once it is done, then statements such as 9.99999... = 10 make perfect
> sense.
>
> Dave Hare
Adam Eberbach t911...@phillip.edu.au fnord
t911...@phillip.edu.au
> In article <Bry20...@watdragon.waterloo.edu>, deg...@daisy.waterloo.edu (Dave Hare) writes:
>> In article <1992Jul25.1...@phillip.edu.au> t911...@phillip.edu.au writes:
>>>In article <1992Jul24.1...@pasteur.Berkeley.EDU>, tj...@cory.Berkeley.EDU (Richard Tjoa) writes:
>>>
>>>> Ok, here's something interesting...
>>>> 9.9999... (nine point nine repeating) = 10 (ten)
>>>
>>>There's the error. 9.9999... does not equal 10, it approaches 10.
>>
>> No, it equals 10. This type of question has subtleties that too often
>> are too easily dismissed. Consider:
>
> Alright, if 9.999... = 10, then why isn't .33333... exactly 1/3?
After consulting a maths lecturer I'll admit I'm wrong. It still seems
right though, after working with devices accurate to a finite number
of decimal places.
>>
>> The problem is that we are cavalierly taught in grade school that if a
>> non-terminating decimal is periodic, then we just multiply it by 10^{the
>> period}, subtract the original, divide both sides by 10^{the period}-1
>> and reduce to lowest terms, and we have that number as an exact rational.
>>
My teachers used to love saying "don't believe calculators ever! They will
tell you that things like 9.999... = 10!" :-)
>> These steps require justification. HOWEVER, that justification is possible,
>> and once it is done, then statements such as 9.99999... = 10 make perfect
>> sense.
>>
>> Dave Hare
>
--
Adam Eberbach | The clever men at Oxford
t911...@phillip.edu.au | Know all that there is to be knowed.
Computer Science student, | But they none of them know one half as much
Phillip Institute of Technology.| As intelligent Mr. Toad!
Dave Hare
>In article <Bry20...@watdragon.waterloo.edu>, deg...@daisy.waterloo.edu (Dave Hare) writes:
>> In article <1992Jul25.1...@phillip.edu.au> t911...@phillip.edu.au writes:
>>>In article <1992Jul24.1...@pasteur.Berkeley.EDU>, tj...@cory.Berkeley.EDU (Richard Tjoa) writes:
>>>
>>>> Ok, here's something interesting...
>>>> 9.9999... (nine point nine repeating) = 10 (ten)
>>>
>>>There's the error. 9.9999... does not equal 10, it approaches 10.
>>
>> No, it equals 10. This type of question has subtleties that too often
>> are too easily dismissed. Consider:
>
>Alright, if 9.999... = 10, then why isn't .33333... exactly 1/3?
>
>I don't see how they are equal. I know that 9.999... with infinite 9's
>is very close to 10, but that .0000...0001 is what you need to make
>exactly 10.
Um, how many 0's are there before that 1?
>> - What exactly is a non-terminating decimal?
>
>Let's keep it simple - we're talking about recurring 9's.
Even so. What do you think 9.999999.... means?
>> - How will you determine that two numbers are *not* equal
>
>Why does it need a formal mathematical proof?
Everything in mathematics needs a formal mathematical proof. Otherwise
it's not not mathematics.
>> - If x is a non-terminating decimal, how do you define 10*x? (Recall
>> that for ordinary multiplication, the computation proceeds from
>> least significant digit to most significant digit, but x has no
>> least significant digit.) And how do you define the other arithmetic
>> operations?
>
>10*x is easy. Shift the decimal place one to the right, giving
>99.9999.... which is still not equal to 10. (or 100 either.)
Prove both of those statements (that the result of multiplying by 10 is
99.999.... and that the result is not equal to 100).
>> The problem is that we are cavalierly taught in grade school that if a
>> non-terminating decimal is periodic, then we just multiply it by 10^{the
>> period}, subtract the original, divide both sides by 10^{the period}-1
>> and reduce to lowest terms, and we have that number as an exact rational.
>>
>> These steps require justification. HOWEVER, that justification is possible,
>> and once it is done, then statements such as 9.99999... = 10 make perfect
>> sense.
All this poster has done is to establish my point that while it is true that
9.9999.... = 10, it is not trivial to write down a careful and complete
proof of this fact.
Dave hare
Dave Hare
>In article <1992Jul27.1...@phillip.edu.au>, t911...@phillip.edu.au writes:
>> In article <Bry20...@watdragon.waterloo.edu>, deg...@daisy.waterloo.edu (Dave Hare) writes:
>>> In article <1992Jul25.1...@phillip.edu.au> t911...@phillip.edu.au writes:
>>>>In article <1992Jul24.1...@pasteur.Berkeley.EDU>, tj...@cory.Berkeley.EDU (Richard Tjoa) writes:
>>>>
>>>>> Ok, here's something interesting...
>>>>> 9.9999... (nine point nine repeating) = 10 (ten)
>>>>
>>>>There's the error. 9.9999... does not equal 10, it approaches 10.
>>>
>>> No, it equals 10. This type of question has subtleties that too often
>>> are too easily dismissed. Consider:
>>
>> Alright, if 9.999... = 10, then why isn't .33333... exactly 1/3?
>
>After consulting a maths lecturer I'll admit I'm wrong. It still seems
>right though, after working with devices accurate to a finite number
>of decimal places.
To any finite number of places, 9.9999...999 is not equal to 10. But
9.9999.... is equal to 10. "Devices accurate to a finite number of
decimal places" will not help to resolve this problem.
>My teachers used to love saying "don't believe calculators ever! They will
>tell you that things like 9.999... = 10!" :-)
Then you should have taken your teachers to task on this. It is certainly
true that you should always question the results of finite precision machines,
because, as I have frequently told my students in numerical analysis, when
it comes to floating point arithmetic, computers are always *wrong*.
However, a calculator can never tell you that 9.999... = 10 . The most
it can tell you is that the result of a computation is 10 plus or minus
some error, whose magnitude depends on the details of the computation.
Dave Hare
Mazen Mokhtar
Who said it is not? 0.333... is exactly 1/3.
>I don't see how they are equal. I know that 9.999... with infinite 9's is
>very close to 10, but that .0000...0001 is what you need to make exactly 10.
There is no number such as 0.00...0001, is this a terminating decimal? You
are refering to is zero.
1
0.00...001 is lim -----, which is easily found to be zero.
x->oo x
10
Two real numbers are equal or one is greater than the other.
Consider: 0.00...001,
is it less than 0.00...002 ?
How about 0.00...010 ?
but this is the same as 0.00...001 .
Now, 10 * 0.00...001 = 0.00...010 = 0.00...001, but solve:
10 * x = x ==> x = 0 This is the unique solution.
Therefore, 0.00...001 = 0.
Regards,
Mazen Mokhtar
David Karr
>> In article <1992Jul25.1...@phillip.edu.au> t911...@phillip.edu.au writes:
>>>In article <1992Jul24.1...@pasteur.Berkeley.EDU>, tj...@cory.Berkeley.EDU (Richard Tjoa) writes:
>>>
>>>> Ok, here's something interesting...
>>>> 9.9999... (nine point nine repeating) = 10 (ten)
>
>Alright, if 9.999... = 10, then why isn't .33333... exactly 1/3?
.33333... IS exactly 1/3.
>I don't see how they are equal. I know that 9.999... with infinite 9's
>is very close to 10, but that .0000...0001 is what you need to make
>exactly 10.
>
>>
>> - What exactly is a non-terminating decimal?
>
>Let's keep it simple - we're talking about recurring 9's. [...]
>
>Adam Eberbach t911...@phillip.edu.au fnord
>Computer Science student, Phillip Institute of Technology.
You really haven't faced up to the problem. How many recurring 9's?
An infinite number, right? How many zeros do you have to write in
.0000...0001 in order to make it exactly equal to the difference
between 9.9999... and 10? An infinite number? How do you propose to
write a 1 digit AFTER the infinite number of zeros? There is no
"infinity-th" decimal place!
It is true that 9.9999...9 != 10 for any FINITE number of 9's after
the decimal point. But we are talking about an INFINITE number of
9's, which is a completely different kettle of fish. (Plenty of facts
that are true for all finite numbers are false for an infinite
number.)
The standard--and only sensible--way to interpret a non-terminating
decimal is to consider the decimal digits to be the numerators of the
terms in an infinite series, where the denominators are the successive
powers of ten. For 9.999... we interpret the dots to mean that all
the digits are 9, in which case the series is
9.999... = 9 + 9/10 + 9/100 + 9/1000 + 9/10000 + ...
= sum_n=0^infinity (9/10^n)
Any book on infinite series will tell you that the sum of the above
series is 10. Not "close to 10," but EXACTLY 10. The sum of the
infinite series is defined as the limit of the sum of the first k
terms, as k goes to infinity.
But enough pedantry. I'm a bit surprised about your math instructor,
but maybe you didn't ask the right question. (I could also understand
if your instructor refused to recognize 9.9999... as ANY number on the
grounds that it is improperly constructed, but that's another story.)
-- David Karr (ka...@cs.cornell.edu)
Carl Mueller
>In article <1992Jul27.1...@phillip.edu.au> t911...@phillip.edu.au writes:
>>In article <Bry20...@watdragon.waterloo.edu>, deg...@daisy.waterloo.edu (Dave Hare) writes:
>>> In article <1992Jul25.1...@phillip.edu.au> t911...@phillip.edu.au writes:
>>>>In article <1992Jul24.1...@pasteur.Berkeley.EDU>, tj...@cory.Berkeley.EDU (Richard Tjoa) writes:
>>>>
>>>>> Ok, here's something interesting...
>>>>> 9.9999... (nine point nine repeating) = 10 (ten)
>>[...]
>
>but maybe you didn't ask the right question. (I could also understand
>if your instructor refused to recognize 9.9999... as ANY number on the
>grounds that it is improperly constructed, but that's another story.)
>
>-- David Karr (ka...@cs.cornell.edu)
I'm guessing that the math instructor was tired of seeing things like
.7 for sqrt(2)/2 and 9.99999 when the exact answer is 10. It is
frustrating to have students so reliant on calculators that they
seem not to understand that sqrt(2) is a better answer than 1.414... .
(It is at least more exact -- I personally feel it is better.) In any
case, students should be taught the dangers of roundoff error.
Carl Mueller (mue...@math.wisc.edu)
Joe Keane
> 9.9 = x (yes, I mean "repeating" with the bar over it.)
> _
> 99.9 = 10x (multiply by 10)
> _
> - 9.9 = - x (subtract that first thing)
> --------------
> 90 = 9x (oooh, no more decimal)
I'm afraid you made a mistake in your subtraction. The problem is that
99.999... minus 9.9999... is 89.999..., not 90 like you have.
> 10 = x (divide by 9)
> _
> 9.9 = 10 (voila! [QED])
And then when you divide by 9, you get x = 9.9999... back, so you've only
proven that 9.9999... is equal to itself.
:-)
--
Joe Keane, amateur mathematician
j...@osc.com (uunet!amdcad!osc!jgk)
Kenneth Tolman
>In article <1992Jul24.1...@pasteur.Berkeley.EDU>
>tj...@cory.Berkeley.EDU (Richard Tjoa) writes:
>> _
>> 9.9 = x (yes, I mean "repeating" with the bar over it.)
>> _
>> 99.9 = 10x (multiply by 10)
>> _
>> - 9.9 = - x (subtract that first thing)
>> --------------
>> 90 = 9x (oooh, no more decimal)
>
>I'm afraid you made a mistake in your subtraction. The problem is that
>99.999... minus 9.9999... is 89.999..., not 90 like you have.
Now, who are we to believe? What is the TRUE definition of subtraction?
I am not sure how you get 99.99... - 9.99.. = 89.99..
In any event, let us suppose that for the moment, 10 does equal 9.999...
Then by induction, 10 also equals 9.999.....(infinity)...8
Then by induction, 10 also equals 9.999.....(infinity)...7
Then by induction, 10 also equals 9.999.....(infinity)...6
If we follow this through, than 10 = 9. Then 10 = whatever we want.
That is clearly silly, so I am inclined to agree that 10 != 9.99999....
Seth Breidbart
them. (Actually, a lot of such numbers.)
If you can't give an example of a number between 9.99999... and 10,
please don't post anymore to this thread.
Seth se...@fid.morgan.com
David Karr
>In any event, let us suppose that for the moment, 10 does equal 9.999...
>Then by induction, 10 also equals 9.999.....(infinity)...8
I can deal with the notation 9.999... but what in tarnation is
9.999...(infinity)...8 ? How can a sequence of infinitely many
digits have a LAST digit?
>Then by induction, 10 also equals 9.999.....(infinity)...7
This is not induction. It's not even induction's second cousin.
>Then by induction, 10 also equals 9.999.....(infinity)...6
>
>If we follow this through,
then (swallowing the absurd notion of an infinite terminating decimal)
every number that we generate contains an infinite number of nines
followed by one of the nine non-zero digits:
9.999...992 (just showing some more of the 9's explicitly)
9.999...991
9.999...990
But the 0 in the last number doesn't matter; we can rewrite that number as
9.999...99
i.e. an "infinite" string of 9's ending in a 9, but that's just where you
started! The next number is 9.999...998 and so forth, repeating the first
nine numbers forever.
>than [sic] 10 = 9.
Not so, see above.
Here's an exercise: prove that 9.999... is a real number. Once you have
done that, we can argue about which one it is.
-- David Karr (ka...@cs.cornell.edu)
David Karr
>99.999... minus 9.9999... is 89.999..., not 90 like you have.
Er--excuse me? 89.999... = 90 (assuming it makes any sense to write
89.999..., of course), so your result is technically correct :-), but
what's your justification for that result of the subtraction?
>Joe Keane, amateur mathematician
^^^^^^^
>j...@osc.com (uunet!amdcad!osc!jgk)
-- David Karr (ka...@cs.cornell.edu)
Andrew Pierce
>If two numbers are different, then there's another number between
>them. (Actually, a lot of such numbers.)
I agree with you but it's not that simple. Your statement needs
justification. Mathematicians: is this an axiom of number theory for
reals?
-Andy
ajpi...@med.unc.edu
Kenneth Tolman
>them. (Actually, a lot of such numbers.)
>If you can't give an example of a number between 9.99999... and 10,
>please don't post anymore to this thread.
SAY WHAT? This is the silliest bit of nonsense ever. "If two numbers are
different, then there's another number between them". Let's consider the
integers, and I will give you 2 and 3. Now, if they are different, there
must be an integer between them. Well, what is it?
Now, if you assert that the above is true for REAL's then you better have
some reason, for I can just as easily assert the opposite. As a matter of
fact, I can show you two numbers which are different, which do not have any
number between them...
41.999999.... and 42
Michael Neylon
whole thread...
Dave Hare
No, it is a simple consequence of the order relation for reals. If a and
b are real numbers and a <> b, then either a<b or a>b. Without loss of
generality, assume a<b. Then c = b-a > 0, so c/2 > 0, and c/2 < c, so
a < a+c/2 < a+c = b (note a+c/2 = (a+b)/2 ).
Dave Hare
David F. Skoll
tolman%asylum.cs...@cs.utah.edu (Kenneth Tolman) writes:
>In any event, let us suppose that for the moment, 10 does equal 9.999...
>Then by induction, 10 also equals 9.999.....(infinity)...8
>Then by induction, 10 also equals 9.999.....(infinity)...7
>Then by induction, 10 also equals 9.999.....(infinity)...6
>If we follow this through, than 10 = 9. Then 10 = whatever we want.
>That is clearly silly, so I am inclined to agree that 10 != 9.99999....
Well, you're wrong.
9.9999... is by definition equal to 10, because a repeating decimal is
defined as the limit of the sequences of partial sums. If you don't like
the definition, well... complain to your local mathemetician. :-)
(Anyway, your "induction" step is nonsensical, because it is impossible
to write or define 9.99999 ... 8 in any sensible way. Perhaps you'd care to
give us a rigorous mathematical definition of just what is meant by
9.99999 ... 8?)
--
David F. SKoll
Kalle Kivimaa
>If two numbers are different, then there's another number between
>them. (Actually, a lot of such numbers.)
41.999999.... and 42
Now, to all those r.p readers who agree with Kenneth I quote:
"[Theorem:] Between any two real numbers there is a rational number and an
irrational number."
Unfortunately, it is not proven in the text, but if you want to
disagree, please contact the author Stanley I. Grossman, Univ. of
Montana and tell him that he is wrong. If you can convince HIM then
come back and post.
--
************************************************************************
* The night sky over the planet Krikkit is the least interesting sight *
* in the entire Universe. -- Douglas Adams, HHGTG *
************************************************************************
Barry Hayes
discussion continue in sci.math -- the followup-to of this message is
so set.
What it means for a number to be a well-defined real, for one real to
be equal to another, and so on all follow from what we might mean by
"real number" and "equal to". Until there is an agreement as to what
those terms mean, let alone such ill-defined terms as "9.999...", we
cannot expect to come to agreement.
For example, I will define the following sequence:
P[n] is the nth digit in the decimal expansion of Pi: P[1] = 1, P[2] = 4. . .
A[n] is 0 if P[n] through P[100+n] are all ones.
A[n] is 1/10^n if P[n] through P[n+100] are all twos.
A[n] is P[n]/10^n
Note that since any P[n] can be computed, any A[n] can be computed.
Now define A as the infinite sum of all the A[n]'s. If you believe in
Pi, you have to believe in A. Well, what's the order between A and
Pi? Depends a lot on Pi, doesn't it?
Some would deny both A and Pi, by the way. "Since constructing Pi
requires an infinite number of calculations, it is not reasonable to
discuss it," they would say.
In short, tell me what you mean by "A is a real number", "A and B are
equal", "A and B are not equal", "10", and "9.999..." and maybe I can
tell you if 10=9.9999...
John William Chamberlain
.3333... = 3 3 3 3 3
--- + ---- + ----- + ------ + ------- .......
10 100 1000 10000 100000
S1 = 3 , S2 = 3 , Sn = 3
-- --2 --n
10 10^ 10^
Sum from 0 to infinity of 3 is 1/3
--n
10^
From my own ideas
if 1/3 + 1/3 + 1/3 = 1
and .33... + .33... + .33... = .99999
since .33.... = 1/3
1/3 + 1/3 + 1/3 must equal .99999..
therefore
.99999... = 1/3 + 1/3 + 1/3 = 1 = .33... + .33... + .33...
and thus
.999... = 1
Thanks for your time
John
The book is : Calculus and Analytic Geometry
By : Thomas and Finney 7th edition
pp: 647-648
.
David LeCompte
>99.999... minus 9.9999... is 89.999..., not 90 like you have.
How do you come up with this?
_ _
99.9 - 10 = 89.9
_
I'll grant that as obvious. But if you claim that 9.9 != 10, then I
don't see your claim at all.
_
No, really, really. 9.9 = 10. It does.
_
9.9 = 9+.9+.09+.009+... forever.
_
.9=.9+.09+.009+... forever.
you REALLY can term by term subtract ALL the 9s after the decimal
points and get all 0s after the decimal point.
Dave LeCompte
--
---------------------
ObSig
David LeCompte
>Now, who are we to believe? What is the TRUE definition of subtraction?
>
>In any event, let us suppose that for the moment, 10 does equal 9.999...
>Then by induction, 10 also equals 9.999.....(infinity)...8
>Then by induction, 10 also equals 9.999.....(infinity)...7
>Then by induction, 10 also equals 9.999.....(infinity)...6
>
>If we follow this through, than 10 = 9. Then 10 = whatever we want.
>That is clearly silly, so I am inclined to agree that 10 != 9.99999....
>
Yeah, if we follow this through, we get 10 - 9.99999(infinity)...x,
where x is a digit.
Where do you get the idea that you can write x as the last digit in
this series? There is no last digit! They're ALL 9s! All of them!
How about the 100th? Yeah, that's a 9.
The millionth? is that a 9? Yup.
The nth, where n is a cardinal number? Sure.
So where's the last 9? There isn't one. You can't write
9.999...(infinity)...999
^^^^^^
This bit makes no sense. The (infinity) indicates that the nines keep
on going longer than the energizer rabbit.
--
---------------------
ObSig
David LeCompte
>>them. (Actually, a lot of such numbers.)
>
>SAY WHAT? This is the silliest bit of nonsense ever. "If two numbers are
>different, then there's another number between them". Let's consider the
>integers, and I will give you 2 and 3. Now, if they are different, there
Oh, yeah. THAT sounds like a valid simplification of the problem. How
about just the two integers 5 and 12. This is as silly a subset as the
integers. I'll easily grant that there are no numbers in the set
{5,12} between 5 and 12, but I have no problem with the claim that Ken
disputes.
>must be an integer between them. Well, what is it?
^^^^^^^
No, a number. Try 3.5. You're oversimplifying.
>
>Now, if you assert that the above is true for REAL's then you better have
>some reason, for I can just as easily assert the opposite. As a matter of
>fact, I can show you two numbers which are different, which do not have any
>number between them...
> 41.999999.... and 42
_
No, 41.9 = 42.
--
---------------------
ObSig
Trent Tobler
From the 13 axioms of the real number system: (I hope I can remember all
of them, it has been a few years...)
in the following, a, b, and c are all real.
1) a + b = b + a
2) a * b = b * a
3) a + ( b + c ) = ( a + b ) + c
4) a * ( b + c ) = ( a * b ) + ( a * c )
5) a + 0 = a
6) a * 1 = a
7) for every a and b, there exists c s.t. a + c = b :== c = b - a
8) for every a not zero and b, there exists c s.t. a * c = b :== c = b / a
9) for a and b, exactly one of the following holds:
a = b
a < b
b < a
10) a < b and b < c implies a < c
11) a + c < b + c
12) a * c < b * c for 0 < c
13) let S be any set of real numbers, and T be the set of all numbers
x with x <= y for all y in S. If T is non empty, then there exists
one element p in T, such that p <= q for every q in T. (or,
if you prefer, use one of the equivilant theorems)
(If you have not taken a number theory course, or read any number theory
books, the first 12 define two number systems, rationals and reals..
#13 restricts it to the reals. Well, actually, we may have to have one
more, and that is that 0 < 1)
I assume you already can prove the common arithmetic properties...
Suppose we have two numbers, a and b, with a < b. Prove there exists
some number c such that a < c < b.
a < b
a + a < b + a and a + b < b + b
( a + a ) / 2 < ( b + a ) / 2 and ( a + b ) / 2 < ( b + b ) / 2
a < ( a + b ) / 2 and ( a + b ) / 2 < b
let c = ( a + b ) / 2.
a < c and c < b
a < c < b
Therefore, for any two distinct numbers a and b, there exists a number
c such that a < c < b.
Is that formal enough for you? ;)
Note, this applies to both the system of rationals, and reals.. we don't
invoke axiom #13 at all.
--
Trent Tobler - tto...@unislc.slc.unisys.com
Matthew T. Russotto
>Alright, if 9.999... = 10, then why isn't .33333... exactly 1/3?
It is.
--
Matthew T. Russotto russ...@eng.umd.edu russ...@wam.umd.edu
Some news readers expect "Disclaimer:" here.
Just say NO to police searches and seizures. Make them use force.
(not responsible for bodily harm resulting from following above advice)
Kelly Waters
>In any event, let us suppose that for the moment, 10 does equal 9.999...
>Then by induction, 10 also equals 9.999.....(infinity)...8
>Then by induction, 10 also equals 9.999.....(infinity)...7
>Then by induction, 10 also equals 9.999.....(infinity)...6
>
Perhaps you could explain your 'induction.' It does not follow from
10=9.9999... that 10 also equals 9.999.....(oo)...8 . Additionally, what
digits follow the 8 ie. 9.999.....(oo)...8999 or ...8888 or ....87999 ?
The basis of 10=9.999.... is that no 'last' digit exists. With that in mind,
your arguement is
If we change the last digit, it does not alter the value.
Since your premise is false, your statement is true. But the fact is that we
cannot change the 'last' digit, and your 'inductive' step is unsupported.
>If we follow this through, than 10 = 9. Then 10 = whatever we want.
>That is clearly silly, so I am inclined to agree that 10 != 9.99999....
>
10 = 9 is indeed silly, but you have not established the connection between
this clearly wrong statement 10 = 9, and the statement 10 = 9.9999....
Kelly Waters
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
favorite nagging question : If I exist, do you? And How about Fred? Or Herman?
________________________________________________________________________________
Mike Levis
> if 1/3 + 1/3 + 1/3 = 1
>
> and .33... + .33... + .33... = .99999
>
> since .33.... = 1/3
>
> 1/3 + 1/3 + 1/3 must equal .99999..
>
> therefore
>
> .99999... = 1/3 + 1/3 + 1/3 = 1 = .33... + .33... + .33...
>
> and thus
>
> .999... = 1
This seems to be the best proof so far, and it is so simple. I
have been neutral on this so far, but that ``.3333... + .3333...
+ .3333... = .9999...'' part convinced me that .9999... = 1.0
Just my 1.9999... cents.
> John
--
========= Mike Levis mle...@ringer.cs.utsa.edu ========= .--.
ftp syrinx.umd.edu (128.8.2.114) for Rush lyrics and info. (OS/2)
::::: Get the Rush FAQL in directory /rush/special. :::::: ~--~
t911...@phillip.edu.au
> In article <1992Jul28.1...@phillip.edu.au> t911...@phillip.edu.au writes:
>>In article <1992Jul27.1...@phillip.edu.au>, t911...@phillip.edu.au writes:
>>> In article <Bry20...@watdragon.waterloo.edu>, deg...@daisy.waterloo.edu (Dave Hare) writes:
>>>> In article <1992Jul25.1...@phillip.edu.au> t911...@phillip.edu.au writes:
>>>>>In article <1992Jul24.1...@pasteur.Berkeley.EDU>, tj...@cory.Berkeley.EDU (Richard Tjoa) writes:
(stuff missing)
>
>>My teachers used to love saying "don't believe calculators ever! They will
>>tell you that things like 9.999... = 10!" :-)
>
> Then you should have taken your teachers to task on this. It is certainly
> true that you should always question the results of finite precision machines,
> because, as I have frequently told my students in numerical analysis, when
> it comes to floating point arithmetic, computers are always *wrong*.
>
Well, I was about 12 at the time and not that interested, just disappointed
that I still had to use fractions.
> However, a calculator can never tell you that 9.999... = 10 . The most
> it can tell you is that the result of a computation is 10 plus or minus
> some error, whose magnitude depends on the details of the computation.
>
> Dave Hare
--
Adam Eberbach | The clever men at Oxford
t911...@phillip.edu.au | Know all that there is to be knowed.
Computer Science student, | But they none of them know one half as much
Phillip Institute of Technology.| As intelligent Mr. Toad!
Seth Breidbart
tolman%asylum.cs...@cs.utah.edu (Kenneth Tolman) writes:
>>them. (Actually, a lot of such numbers.)
>
>>If you can't give an example of a number between 9.99999... and 10,
>>please don't post anymore to this thread.
>SAY WHAT? This is the silliest bit of nonsense ever.
You must not have been on the net for very long.
> "If two numbers are
>different, then there's another number between them". Let's consider the
>integers, and I will give you 2 and 3. Now, if they are different, there
>must be an integer between them. Well, what is it?
2.3. I was referring to real numbers; I should have been more specific.
>Now, if you assert that the above is true for REAL's then you better have
>some reason, for I can just as easily assert the opposite. As a matter of
>fact, I can show you two numbers which are different, which do not have any
>number between them...
> 41.999999.... and 42
Well, the real numbers are dense and well-ordered. Since you say
those two numbers are different, what is (41.99999... + 42)/2?
Seth se...@fid.morgan.com
Seth Breidbart
bha...@Xenon.Stanford.EDU (Barry Hayes) writes:
>For example, I will define the following sequence:
>P[n] is the nth digit in the decimal expansion of Pi: P[1] = 1, P[2] = 4. . .
>A[n] is 0 if P[n] through P[100+n] are all ones.
>A[n] is 1/10^n if P[n] through P[n+100] are all twos.
>A[n] is P[n]/10^n
>
>Note that since any P[n] can be computed, any A[n] can be computed.
>
>Now define A as the infinite sum of all the A[n]'s. If you believe in
>Pi, you have to believe in A. Well, what's the order between A and
>Pi? Depends a lot on Pi, doesn't it?
A is a computable number. I don't know if A < pi, A = pi, or A > pi.
If A != pi, then we'll eventually find out (when the program runs long
enough). If A = pi, we might or might not eventually find out.
Define B as the 10^10^100'th digit in the decimal expansion of pi.
Is B > 5?
Seth se...@fid.morgan.com
Bryan D. Nehl
x = 9.999...
10x = 99.999...
10x -x = 99.999... - 9.999...
9x = 90.0
x = 10
10.0 = x = 9.999...
Bryan.
/* === Bryan Nehl ========== kod...@Kodiakpc.Manhattan.KS.US =========+
[ USDA-ARS-NPA-WERU ][ b...@chepil.weru.ksu.edu ]
[ 913.532.6233 or 913.532.6495 ][ kod...@matt.ksu.ksu.edu ]
+_______________ ...!rutgers!matt.ksu.ksu.edu!kodiak _______________ */
t911...@phillip.edu.au
Hang on.... wasn't this whole thread about why x.9999... is actually equal
to x + 1? (And how wrong I was when I said it wasn't? :-)
Surely then 41.99999... is not different from 42?
--
Adam Eberbach Computer Science student, RMIT. t911...@phillip.edu.au
"We don' need no stinkin' signatures!"
Virginia A. Reed
is), then it can be assumed as zero.
therefore:
9.9999...- 0.000..01 = 9.999...
It was stated before that
x=9.999...
10x=99.999...
9x=90
x=10=9.999...
If 9.999...=10, then
9.9999...-0.000...01=10
add 0.000...01 to both sides:
9.9999...=10.000...01
-0.000...01=0.000...01
taken to its extreme, you could prove that 7=-7, or that 2341=-2341
Luis E. Amaya
>If 9.9999... + 0.00...01=10, and 0.00...01 is infinitely small(which it
>is), then it can be assumed as zero.
it IS zero
>therefore:
>9.9999...- 0.000..01 = 9.999...
>If 9.999...=10, then
>9.9999...-0.000...01=10
>add 0.000...01 to both sides:
>9.9999...=10.000...01
>-0.000...01=0.000...01
>taken to its extreme, you could prove that 7=-7, or that 2341=-2341
----------
Luis
Robert Morse
depressing that I have been sorely tempted to cancel my subscription
and ask for my money back. Let me talk about this. This IS
rec.psychotherapy, right? :-)
Those many people who have correctly explained how to sum an infinite
geometric series have unfortunately been wasting their effort.
Although such methods are valuable, perhaps essential, in demonstrating
the truth of less obvious equalities like 1/7 = 0.142857142857..., they
are just not needed here, and miss the point of confusion.
There is surely no one who does not see that 9.9999... gets closer and
closer to 10, the more 9's you write down before you get tired and
stop. (Please do not say that I am wrong about this, and destroy all
my joy in living.) And surely, everyone can see that if you decide you
want to be oh so ever CLOSE to 10, all you have to do is to write down
enough 9's after the decimal point so as to satisfy your requirement
for closeness.
That, folks, is ALL that 10 = 9.9999... means! That, and nothing
else. It does NOT mean that there is some huge number of nines that
one can write down, so that 10 = 9.9999...(huge-number-of-9's) is
literally true. No, there IS no such huge number. "Infinity" is Not
some unfathomably huge number.
All that "infinity" or dot-dot-dot represents, is a mathematical limit
process. It says, in this case, that the sequence of numbers 9, 9.9,
9.99, 9.999, etc, can approach 10 as closely as you like, simply by
extending the sequence out far enough.
For me, there was one interesting objection that was raised by some
people. They said, well, what about 9.9999...8 and 9.9999...7? You
might as well say that these also represent 10, and it gets to look
like 10 is equal to just too many things, then. Maybe 10 = 9?
Well, there is really nothing wrong with saying that 10 = 9.9999...7,
for example. What this "says" is that the limit of the sequence 9,
9.7, 9.97, 9.997, 9.9997, etc, is 10. That is really okay; many
different sequences of numbers can all have the same limit value.
But let me make two points about this:
(1) One could never demonstrate this way that 10 = 9, as all those
sequences develop an indefinite number (an unlimited number) of 9's
following the decimal point.
(2) None of these sequences leads to any different decimal
representation for 10. That's because, in writing the representation
for any fixed number of digits, one is required to write the
representation which is most accurate. Therefore, one has to write 9.9
and not 9.7, and 9.99 and not 9.97, and so forth.
This "accuracy" rule could get me into trouble here. We should really
write 10.0 as more accurate than 9.9, and 10.00 as more accurate than
9.99, and indeed, that's what anyone would really do in practice.
However, mathematicians prefer the "unending nines" representations
because they can then regard *all* real numbers, without exception, as
non-terminating decimals. It makes things formally more consistent.
Just as they say pi = 3.14..., so they can say 4 = 3.99...
Live and let live.
Robert Morse
mo...@math.psu.edu
David Karr
>There is surely no one who does not see that 9.9999... gets closer and
>closer to 10, the more 9's you write down before you get tired and
>stop. (Please do not say that I am wrong about this, and destroy all
>my joy in living.) And surely, everyone can see that if you decide you
>want to be oh so ever CLOSE to 10, all you have to do is to write down
>enough 9's after the decimal point so as to satisfy your requirement
>for closeness.
>
>That, folks, is ALL that 10 = 9.9999... means! That, and nothing
>else. It does NOT mean that there is some huge number of nines that
>one can write down, so that 10 = 9.9999...(huge-number-of-9's) is
>literally true.
Correct on the second point, because (for example) if I were to set
a Turing machine to writing 9.999... on its tape, it would never halt.
Thus the task literally cannot be done in that sense.
Wrong on the first point, because, as you know:
>All that "infinity" or dot-dot-dot represents, is a mathematical limit
>process. It says, in this case, that the sequence of numbers 9, 9.9,
>9.99, 9.999, etc, can approach 10 as closely as you like, simply by
>extending the sequence out far enough.
And the number represented by 9.999... IS THE LIMIT of that sequence,
and the limit of the sequence is 10. Therefore 9.999... = 10.
Moreover this is a very precise statement.
If you allow me to state that 9.999... is a non-terminating decimal
but it is NOT a valid representation for 10, then you also must not
believe that 1/3 can be written 0.333... . In fact you will have told
me you do not believe in non-terminating decimals.
This may in fact be completely justified, depending on what kind of
mathematics you like to do. I'm not very conversant with constructive
mathematics, but I could believe that a constructivist would deny the
existence of a mathematical object like a non-terminating decimal that
cannot be written out as a finite string. Most mathematicians, however,
seem to have little trouble in assuming the existence of an infinite
sequence of digits chosen (perhaps arbitrarily) from the set {0..9}.
>Well, there is really nothing wrong with saying that 10 = 9.9999...7,
>for example. What this "says" is that the limit of the sequence 9,
>9.7, 9.97, 9.997, 9.9997, etc, is 10. That is really okay; many
>different sequences of numbers can all have the same limit value.
Actually I do have a problem with people writing 9.9999...7 because
there is no obvious "canonical" sequence of terminating decimals to
which this corresponds. If 9, 9.7, 9.97, then when not 7, 9.7, 9.97,
or 9, 9.9, 9.99, 9.997, 9.9997, or even (to carry this to an absurd
degree) 3, 10, 8, 9.997, 15, 9.9997, 9.999997, ... ?
>However, mathematicians prefer the "unending nines" representations
>because they can then regard *all* real numbers, without exception, as
>non-terminating decimals. It makes things formally more consistent.
>Just as they say pi = 3.14..., so they can say 4 = 3.99...
What about 4 = 4.0000... ? :-)
-- David Karr (ka...@cs.cornell.edu)
David LeCompte
--
---------------------
ObSig
David LeCompte
>Just as they say pi = 3.14..., so they can say 4 = 3.99...
*I* say 4.0000... :)
-Dave
--
---------------------
ObSig
Carl Mueller
I bet that 4 = 3.99... is written or said more than 4 = 4.0000... . :-)
Carl Mueller (mue...@math.wisc.edu)
25bs...@springfield-schools.org
> Ok, here's something interesting...
> 9.9999... (nine point nine repeating) = 10 (ten)
> _
> 9.9 = x (yes, I mean "repeating" with the bar over it.)
> _
> 99.9 = 10x (multiply by 10)
> _
> - 9.9 = - x (subtract that first thing)
> --------------
> 90 = 9x (oooh, no more decimal)
> 10 = x (divide by 9)
> _
> 9.9 = 10 (voila! [QED])
>
>
> -Richard
> (My personal favorite)
> How about 1 = 2 later?
Just because it is an infinitecimal, does not mean that you change its properties. If 99.999...-9.999...equals 90, then something tells me that you did not multiply by 10...You just added 90. Which would be the same if the value was 10, but you cannot just use this proof in this proof that is being used in this proof (and so on). Why don't you try subtracting 9.999... another time? It should equal 8 times 9.999... but you actually get this number: 8.000...1 because these are 9s. But this number does not exist, so both of those inputs also do not exist. But you can get any number and combine it with another number in an operation to get 8 and 9.999..., so all of math is false. That means this proof is false. So you are wrong. And we have a paradox here. Am I right? FITE ME M8
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FromTheRafters
Richard Heathfield
> On Friday, July 24, 1992 at 6:25:53 AM UTC-4, Richard Tjoa wrote:
>> Ok, here's something interesting...
>> 9.9999... (nine point nine repeating) = 10 (ten)
>> _
>> 9.9 = x (yes, I mean "repeating" with the bar over it.)
>> _
>> 99.9 = 10x (multiply by 10)
>> _
>> - 9.9 = - x (subtract that first thing)
>> --------------
>> 90 = 9x (oooh, no more decimal)
>> 10 = x (divide by 9)
>> _
>> 9.9 = 10 (voila! [QED])
>>
>>
>> -Richard
>> (My personal favorite)
>> How about 1 = 2 later?
>
> Just because it is an infinitecimal, does not mean that you change its properties. If 99.999...-9.999...equals 90, then something tells me that you did not multiply by 10...You just added 90. Which would be the same if the value was 10, but you cannot just use this proof in this proof that is being used in this proof (and so on). Why don't you try subtracting 9.999... another time? It should equal 8 times 9.999... but you actually get this number: 8.000...1 because these are 9s. But this number does not exist, so both of those inputs also do not exist. But you can get any number and combine it with another number in an operation to get 8 and 9.999..., so all of math is false. That means this proof is false. So you are wrong. And we have a paradox here. Am I right? FITE ME M8
precisely as you can.
--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within
Richard Heathfield
Such is life.
riverman
Richard Heathfield
> Wait. if 0.999.. = 10. then does 0.888....= 9?
If A then B.
A is false.
What, then, can we say about B?
FromTheRafters
>> Wait. if 0.999.. = 10. then does 0.888....= 9?
>
> This is a simple problem in logic, which can be re-stated thusly:
>
> If A then B.
>
> A is false.
>
> What, then, can we say about B?
It is best not to post typographic errors in puzzles.