Fermat's Last Theorem - Disproof?Date: 05/09/98 at 11:33:28 From: Venkata Pradyumna Kumar Muppirala 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, Pradyu Venkata Pradyumna Kumar M 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 read something like this: >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. For more information about Fermat's Last Theorem, see 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 For more information about irrational numbers, you may want to look at 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/ |
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