Fermat's Last Theorem

[jakuta]

It is said below.

 

Yes, that is correct.

 

But, you are trying to prove assuming aⁿ + bⁿ = cⁿ is true

 

that is the case that there are three positive integers a, b, c or you arrive at a contradiction.

 

So you cannot write a + k = c with k integer as k might be a real decimal number in order to satisfy aⁿ + bⁿ = cⁿ

You must prove k can only be an integer.

 

You cannot assume the truth of all parts of the theorem and then prove the theorem. That is vacuous implication.

By claiming, a + k = c and b + i = c, with k, i integer , you have already asserted there are three positive integers a, b, c .

 

Otherwise, for example, select k < -a < 0 and 0 < i < b. Then c < 0 and c > 0.

 

So, both k and i must be > 0. For the same reason, a, b, c > 0 or you arrive at a contradiction.

 

But, then you assume the theorem you are trying to prove.

[McPgr]

[ name=jakuta' timestamp='1288580098' post='301686]

 

 

You cannot assume the truth of all parts of the theorem and then prove the theorem. That is vacuous implication.

By claiming, a + k = c and b + i = c, with k, i integer , you have already asserted there are three positive integers a, b, c

 

Don't understand. If the statement says that aⁿ + bⁿ = cⁿ is not true for integers I must assume that it is true for integers and then find a contradiction compromising such assumtion. What can be proved by assumption that the equation is true for real decimal numbers? Only that the assumption is true.

 

 

 

So, both k and i must be > 0. For the same reason, a, b, c > 0 or you arrive at a contradiction

.

 

No objections

 

But, then you assume the theorem you are trying to prove.

 

Not at all. The assumption is contrary to the statement of the theorem (see above).

[phillip1882]

hey McPgr, i read through you proof, i can tell you put alot of time in it, so forgive my stupidity.

in lemma 5, you have the equation a= vp and b = wq

and state that...

a = uwv +v^n; b = uwv +w^n; c = uwv +v^n +w^n.

i'm not sure this is correct.

for example, let's take the simple 3,4,5 triangle. (n = 2)

what integer values for u, w, v work?

none of u, w ,v can be zero, nor can they be greater than 2.

this leaves nine possible solutions.

u = 2; v = 1; w = 1;

a = 2*1*1 +1^2 = 3 okay

b = 2*1*1 +1^2 = 3 nope.

u = 2; v = 2; w = 1;

a = 2*2*1 +2^2 = 8 nope

u = 1; v = 2; w = 1;

a= 1*2*1 +2^2 = 6 nope

u = 1; v = 1; w = 1

a = 1*1*1 +1^2 = 2 nope

and the other combinations are just as bad.

am i missing something?

edit:

okay reading through the proof again, that makes more sence now,

that equation is conditional on lemma 3 being true, which assumes f and k being coprime to n.

that is we have...

let a = 3 b = 4 c = 5 n = 2

then 3^2 +4^2 = (3+2)^2 = (4+1)^2 (k = 2, f =1)

skipping down to version B:

a = 2^g uwv +v^2

b = 2^g uwv +2^(2g -1)*w^2

c = 2^g uwv +v^2 +2^(2g -1)*w^2

g obviously has to be 1, as do u,w, and v,

a = 3

b = 4

c = 5.

[jakuta]

[ name=jakuta' timestamp='1288580098' post='301686]

 

 

 

 

Don't understand. If the statement says that aⁿ + bⁿ = cⁿ is not true for integers I must assume that it is true for integers and then find a contradiction compromising such assumtion. What can be proved by assumption that the equation is true for real decimal numbers? Only that the assumption is true.

 

 

 

.

 

No objections

 

 

 

Not at all. The assumption is contrary to the statement of the theorem (see above).

 

 

 

I am not communicating.

 

According to the Fermat's Last Theorem the equation (1)

 

aⁿ + bⁿ = cⁿ

cannot be true when a, b, c and n are positive integers and n > 2.

 

Let statement A be aⁿ + bⁿ = cⁿ

and statement B be a, b, c and n are positive integers and n > 2.

Therefore, the theorem says B → ~A.≡ ~B or A.

In other words if you assume aⁿ + bⁿ = cⁿ ≡ A, you must prove it is the not the case a, b, c and n are positive integers and n > 2 ≡ ~B to prove the theorem.

But, you are not doing this based on

The contradiction appears as result of transformations based on the assumption that the equation (1) is true and following considerations. It proves that the (1) is not true when the exponent n is a prime number.

You are therefore trying to prove A is not true which has nothing to do with proving ~B is true as required by the theorem when A is assumed true.

 

Otherwise, you have a vacuous implication as I said by assumming the the truth of A without proving ~B since B → ~A.≡ ~B or A. is always true by assuming A.

 

You must prove ~B.

[phillip1882]

jakuta,

B is a condition on the range of values for a, b, c and n. to prove ~B you would need to prove this condition false, which makes no sense.

if i said the following...

all apples are red

on the condition that the apple seeds the grew from were also red

it doesn't really make sense to prove that condition false, you simply need to take a red apple seed and grow a sufficient number of red apples to prove it true. or by proof of contradiction if you take every other color of apple seed and it never produces red, then you have a proof. now if one of them does produce a red, this doesn't necessarily disprove that only red apples come from red apple seeds, but it does prove that there are alternate ways of getting a red apple.

i hope this made sense.

[McPgr]

[ name=phillip1882' timestamp='1288810584' post='301750]

 

in lemma 5, you have the equation a= vp and b = wq

and state that...

a = uwv +v^n; b = uwv +w^n; c = uwv +v^n +w^n.

i'm not sure this is correct.

for example, let's take the simple 3,4,5 triangle. (n = 2)

Read the note after eq (13). It says that the latter does not work with n=2. The eq (26) is true for this case with u=1 and any w and v

[jakuta]

jakuta,

B is a condition on the range of values for a, b, c and n. to prove ~B you would need to prove this condition false, which makes no sense.

if i said the following...

all apples are red

on the condition that the apple seeds the grew from were also red

it doesn't really make sense to prove that condition false, you simply need to take a red apple seed and grow a sufficient number of red apples to prove it true. or by proof of contradiction if you take every other color of apple seed and it never produces red, then you have a proof. now if one of them does produce a red, this doesn't necessarily disprove that only red apples come from red apple seeds, but it does prove that there are alternate ways of getting a red apple.

i hope this made sense.

 

 

Nope, not how it works.

 

I need only take an assignment from the universe of positive integers for the B part of the theorem.

 

That assignment will make B true.

 

Actually, B is a formula B(a,b,c) with a,b,c free variables in the universe of positive integers.

 

Now, let an assignment of a,b,c be given. Then, A is impossible by the theorem.

 

OK, now consider this assigment a,b,c, then B is true. It is supposed to be the case that A is false.

 

Material implication (x→y) is logically equivalent to ~x or y. This is basic logic. Now, if you assume y is true, then ~x or y is automatically true. Again, this is basic logic.

 

Now, let's takes this theorem, it basically says,

( B→ ~A) ≡ ( ~B or ~~A) ≡ ( ~B or A).

Now, if you assume A is true as did the OP, then ( ~B or A) is vacuously true as I have been saying. This is elementary logic.

 

In other words, given ( ~B or A) and A is assumed true, you can prove anything since ( ~B or A) is automatically true.

 

[Edit] The statement above is a mistake. This is the corrected logic. ( B→ ~A) ≡ ( ~B or ~A) ≡ ~( B and A). and that is what I believe the OP is trying to show..

[McPgr]

Hello everybody.

 

The discussion on this Proof died and I'm not going to resurrect it.

But I think that found another way to prove FLT. The new version of the proof is based on the same expressions for a, b, c deduced in the previous one but is much shorter and in my opinion more consistent and easier to deal with.

The URL of the new version

To open with MS Office 2010

My link

 

To open with MS Internet Explorer

My link

 

The changes start with Lemma-9

 

Thanks to all interested.