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### Fermat's Last Theorem - Disproof?

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Date: 05/09/98 at 11:33:28
Subject: Fermat's last theorm - a small issue.

Here is the question I have.

I would like to ask you whether there is any problem with the
following disproof of Fermat's last theorem.

I read in _More joy of Mathematics: exploring mathematical insights
and concepts_ by Theoni Pappas

World Wide Publishing/ Tetra
P.O. Box 476
San Carlos CA 94070

On page 150 of this book, I found an article about Fermat's Last
Theorem. Following is the statement of the theorem:

"To divide a cube into two cubes, a fourth power, or in general any
power whatever above second, into two powers of same denomination is
impossible."

restated :

If n > 2, there does not exist a,b,c such that

(a to the power of n) + (b to the power of n) =
(c to the power of n)

There is a way found to calculate numbers that contradict this
theorem:

1. Say x + y = z, where x,y,z are whole numbers
2. there exists a,b,c such that
a = n th root of x,
b = n th root of y,
c = n th root of z

and n > 2 and n is a whole number

3. from previous steps

( a to the power of n) + ( b to the power of n) =
( c to the power of n)

4. For every real number U,
there exists a whole number
such that U x ( 10 to the power of V) is whole number
and V is a whole number.
ex : 2.1 x ( 10 to the power of 1) = 21
2.1 in real number set
21 in whole number set.

5. So from previous principle,
there exists some whole numbers p,q,r
for a,b,c such that
a x ( 10 to the power of p)
b x ( 10 to the power of q)
c x ( 10 to the power of r)

6. Say s = maximum of p,q,r
then
a x ( 10 to the power of s),
b x ( 10 to the power of s),
c x ( 10 to the power of s)
are whole numbers.

7. say
e =    a x ( 10 to the power of s)
f =    b x ( 10 to the power of s)
g =    c x ( 10 to the power of s)
=> e,f,g are whole numbers.

8. Evaluate
( e to the power of n ) + ( f to the power of n )
=> ( ( a x 10 to the power of s ) to the power of n )
+ ( ( b x 10 to the power of s ) to the power of n )
=> ( a to the power of n x ( 10 to the power of ( s x n) ) )
+ ( b to the power of n x ( 10 to the power of ( s x n) ) )
=> ( a to the power of n + b to power of n ) x ( 10 to the power
of ( s x n ))
=> ( c to the power of n ) x  ( 10 to the power of ( s x n ))
( refer to step 3 above)
=>  ( ( c x 10 to the power of s ) to the power of n )
=>  ( g to the power of n )

so
( e to the power of n ) + ( f to the power of n ) =
( g to the power of n)

where e, f,g  are whole numbers.

So e,f,g are the numbers which contradict Fermat's theorm.

It would be difficult to find the numbers because the present
computing capabilities of computers will be exceeded if any numbers
are assumed for x,y,z values. But in the infinite space of numbers,
there obviously exist some numbers that can result in small number
combinations.

I would like to know what exactly is the problem, if there is any, in
this disproof of Fermat's Last Theorem.

With regards,

```

```
Date: 05/09/98 at 11:59:13
From: Doctor Ken
Subject: Re: Fermat's last theorm - a small issue.

Hello there,

You're right to suspect that there may be something wrong with this
disproof.  After all, this problem has been around for hundreds of
years, so we should be quite surprised to find a one-page refutation,
though it's certainly not impossible. What's more, as you're probably
aware, Fermat's Last Theorem has recently been _proved_, so a disproof
shouldn't be possible. Nevertheless, investigating apparent disproofs
is a good way to learn mathematics, so I commend your line of inquiry.

Now, as a matter of fact, there _is_ something wrong with your
disproof. It's in step 4:

>4. For every real number U,
>       there exists a whole number
>    such that U x ( 10 to the power of V) is whole number
>    and V is a whole number.
>    ex : 2.1 x ( 10 to the power of 1) = 21
>        2.1 in real number set
>        21 in whole number set.

This is not true. For instance, consider the number 1/3 = 0.333....
Its digits go on forever in decimal expansion. Therefore, we can never
multiply it by an integer power of 10 to get a whole number.

Well, perhaps all is not lost. Perhaps we could modify statement 4 to

>4. For every real number U,
>       there exists a whole number V and a whole number K
>    such that U x ( K to the power of V) is whole number.

Now 1/3 doesn't present a problem, since (1/3) x 3^1 = 1, which is a
whole number. It wouldn't be too hard to change the rest of your
argument to use this new statement. The problem is, it's still not
true. For instance, consider the number Pi = 3.14159....  Its digits
also go on forever in decimal expansion, BUT there is no whole number
we can multiply Pi by to get an integer. Numbers like these are called
"irrational" numbers because they cannot be expressed as a ratio of
two integers.

http://mathforum.org/dr.math/problems/fermat1.html
http://mathforum.org/dr.math/problems/saleh12.10.96.html
http://mathforum.org/dr.math/problems/li1.21.98.html
http://mathforum.org/dr.math/problems/gardner7.23.97.html
http://mathforum.org/dr.math/problems/renjen11.2.97.html

http://mathforum.org/dr.math/faq/faq.integers.html    .

By the way, I'm glad to hear you're reading Theoni Pappas' book.  I
think it's great.

-Doctor Ken,  The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Number Theory
High School Number Theory

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