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July 5th, 2019, 12:08 PM   #101
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Quote:
 Originally Posted by michaelcweir c/d = (2h +/- (2^2 - (4* h^2 * ( h^2) * e/f +2))^1/2) ----------------------------------------------------- 2h^2
this is wrong there is another b under square root

c/d = (-b + √(b^2 - 4ac))/2a

a = h^2
b = -2h
c = -h^2 × e/f + 2

the final result is:

c/d = (1 + √(h^2 × e/f - 1))/h

so now prove FLT

July 9th, 2019, 06:33 AM   #102
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Flt

Quote:
 Originally Posted by youngmath this is wrong there is another b under square root c/d = (-b + √(b^2 - 4ac))/2a a = h^2 b = -2h c = -h^2 × e/f + 2 the final result is: c/d = (1 + √(h^2 × e/f - 1))/h so now prove FLT

the final result is:

c/d = (1 + √(h^2 × e/f - 1))/h

So here's the proof
This term
√(h^2 × e/f – 1) I claim is an irrational number.
I prove this by saying
h^2 × e/f – 1 can be factored
into
(h x e/f^1/2 + 1) x ( h x e/f^1/2 -1)

If
h x e/f^1/2 -1 is a rational number then you can factor this term into
( h^1/2 x e/f^1/4 +1) x ( h^1/2 x e/f^1/4 -1)

If ( h^1/2 x e/f^1/4 -1) is not an irrational number, then we can factor
( h^1/2 x e/f^1/4 -1) again.

We can repeat this process an infinite number of times until the you get a factor that is irrational.

Once one of the terms is irrational, then the product of an irrational and rational is still an irrational number.
And the square root of an irrational number is still an irrational number.

And so finally c/d must be an irrational number because it is the sum of an irrational number and a rational number.

But c/d = (X^2 + Y^2 )/Z^2
So at least one or more of X, Y, or Z is an irrational number.

This is the proof of FLT.

Last edited by skipjack; July 9th, 2019 at 03:38 PM.

 July 9th, 2019, 07:43 AM #103 Senior Member   Joined: Mar 2019 From: iran Posts: 317 Thanks: 14 this is my last post in this thread if one of the terms is irrational the other term is also irrational and the product of two irrationals is not always irrational for example if (√h × √√c/d - 1) is irrational then (√h × √√c/d + 1) is also irrational this way is just playing with x,y,z √(e/f × h^2 - 1) where e/f = (x^2 + y^2)/z^2 and h = z/y √(e/f × h^2 - 1) = x/y and x/y assuming x and y are integers is rational c/d = (x + y)/z c/d = (1 + √(e/f × h^2 - 1))/h = (1 + x/y)/(z/y) = (x + y)/z then (x + y)/z = (x + y)/z it's just playing with x,y,z
July 9th, 2019, 03:53 PM   #104
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Quote:
 Originally Posted by michaelcweir We can repeat this process an infinite number of times until the you get a factor that is irrational.
There's no need to do that. Every non-zero real number (rational or irrational) can be expressed as the product of two irrational numbers. For example, 4 = √8 × √2.

July 9th, 2019, 06:04 PM   #105
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Quote:
 Originally Posted by michaelcweir But c/d = (X^2 + Y^2 )/Z^2 So at least one or more of X, Y, or Z is an irrational number.
I'm not following the discussion lately but isn't $\frac{3^2 + 4^2}{5^2} = 1$ rational?

If this was explained earlier, nevermind.

July 10th, 2019, 01:07 PM   #106
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perfect squares

Quote:
 Originally Posted by skipjack There's no need to do that. Every non-zero real number (rational or irrational) can be expressed as the product of two irrational numbers. For example, 4 = √8 × √2.
Yes, you are right as far as that goes. The example you shared is a version of perfect squares. The perfect squares are square root of 2 and 8 (8 happens to 2 cubed)

Youngmath provided the equation with this portion being the square root

(4* h^2 * ( h^2) * e/f +2))^1/2

When you factor that square root into

(h x e/f^1/2 + 1) x ( h x e/f^1/2 -1)

You should ask yourself 2 questions
1. Is this the right factoring?
If you multiply the 2 parts together you do get the original square root.

2. Are these two parts a perfect square?

I think that part the 1 is subtracted and in the other part has the 1 added. This would seem to indicate that this is not a perfect square.
(h x e/f^1/2 + 1) is not equal to (h x e/f^1/2 -1)

A further question you should ask yourself is this the end of factoring into smaller products?
(h x e/f^1/2 + 1) may or may not be able to be factored as there are unknowns in the part.

But ( h x e/f^1/2 -1) has a -1 in this part and it can be factored further, just like the first time. And if the part with the -1 is a positive number, we can factor it further. Because we can do this infinite times, there will be a part that is irrational.

And that part, when multiplied by a rational part, will result in an irrational number. Since there are no perfect squares, you can multiply up the chain of multiplicants.

This is a very startling result, and I think that this is what Youngmath was bothered about when he suggested I was just playing with numbers. It's hard to tell otherwise what he was talking about.

His response reminded me of the people saying the proof can't be right because of a lemma . It is only when someone confirmed that every line of the proof was not a p-act statement (the lemma only worked when every line was a p-act statement) that people finally shut up about that issue.

Do I need to quote the proof that square root of an irrational number is an irrational number? Or that an irrational number times an irrational number is irrational with very limited exceptions (none of which apply to this proof).

Last edited by skipjack; July 10th, 2019 at 05:00 PM.

July 10th, 2019, 05:06 PM   #107
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Quote:
 Originally Posted by michaelcweir The perfect squares are square root of 2 and 8 (8 happens to 2 cubed).
Square roots, such as √2, aren't perfect squares (in general).
Here's another example: 6 = (√7 + 1) × (√7 - 1).
There are plenty of other ways that the product of two irrational numbers can be rational.
Hence your method is fundamentally flawed.

Quote:
 Originally Posted by michaelcweir . . . with very limited exceptions (none of which apply to this proof).
That's just a vague assertion, as you haven't told us the exceptions or shown that they are the only exceptions, and you haven't shown that they can't arise in your alleged proof.

 July 11th, 2019, 09:15 AM #108 Member   Joined: Mar 2019 From: california Posts: 67 Thanks: 0 I was waiting for that question. Maschke brought it up with a special example. The answer is that there is an infinite descent involved. It doesn't matter if any one part is positive, you just take the next iteration until you get an irrational number. A quick example is the series 16, 4 , 2 , 2 ^1/2. Once the progression is an irrational number, every one that follows is also an irrational number. Last edited by skipjack; July 19th, 2019 at 08:22 AM.
July 11th, 2019, 09:25 AM   #109
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Quote:
 Originally Posted by michaelcweir I was waiting for that question. Maschke brought it up with a special example. The answer is that there is an infinite descent involved. It doesn't matter if any one part is positive, you just take the next iteration until you get an irrational number. A quick example is the series 16, 4 , 2 , 2 ^1/2. Once the progression is an irrational number, every one that follows is also an irrational number.
You don't address the issue. Every step of the descent produces TWO more numbers. Those two numbers might BOTH be irrational. And since the product of two irrational numbers might well be rational, you got no contradiction.

Last edited by skipjack; July 19th, 2019 at 08:22 AM.

 July 11th, 2019, 09:36 AM #110 Senior Member   Joined: Oct 2009 Posts: 841 Thanks: 323 Theorem: x^2 - 1 = y has no nonzero rational solutions x,y. Proof: If x^2 - 1 is rational, decompose it (x-1)(x+1). If x-1 is rational, decompose it (sqrt(x) - 1)(sqrt(x)+1) Go further with decomposing sqrt(x)-1 and do it over and over again. At one time you'll hit an irrational. The product of an irrational with rationals is irrational, hence y is irrational. QED I imagine you have no problem with this proof since it's exactly what you did?

 Tags analysis, diophantine, diopphantine, fermat, theorem

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