My Math Forum Fermat Last Theorem Diophantine Analysis

 Number Theory Number Theory Math Forum

 July 12th, 2019, 04:39 PM #111 Member   Joined: Mar 2019 From: california Posts: 74 Thanks: 0 No, it's not exactly what I did. So, if you think that you have not understood the proof. Take your statement x^2 - 1 = y If I change it to sqrt (x^2 - 1) = y, where both x,y are integer (or rational if you prefer), this is a better approximation of the formula. With your statement, it is possible to separate the two factors into 2 rational numbers, which when multiplied together give a whole number. For example when x = 3, x^2 is 16, and x^2 -1 = 15. y then would be 17. But is the sqrt (15) a rational number? And what can you say in general whether sqrt (x^2 - 1) is always a irrational number. Also when both x, y are integer. Can the equation possibly be true? While waiting for the answer, I would like to address an earlier comment you made. You were asking if there were alternate ways of validating the proof. Here's one. In an earlier post, I actually used the equation to find some values for the equation. I used the outside of the ranges of c/d, e/f, and h. The equality did not hold and there was a large disparity between the sides of the equation. What I did not show was when I took the midpoint of the range for all 3 variables. i.e. c/d = 1.5, e/f =1.2 and h=1.5 The LHS was 1.5 but the RHS was 1.563333 (something like that). It was close enough that I wondered if a different choice of values would produce a number that match on both sides. FLT says that this would never happen. But you could do a brute force calculation by selecting points on the range 0.1 units apart. But then my worry would be if there was a closer approximation that could be taken that would make the numbers on both sides closer than the 0.1 spacing and you would therefore have to make a smaller segmentation of the variables. It could be done, but at least an initial run would have to be done to see if you got a meaningful answer. That's outside my skill set, as I don't know how to program a computer. Last edited by skipjack; July 12th, 2019 at 05:22 PM.
July 12th, 2019, 05:48 PM   #112
Global Moderator

Joined: Dec 2006

Posts: 20,919
Thanks: 2203

Quote:
 Originally Posted by michaelcweir Take your statement x^2 - 1 = y If I change it to sqrt (x^2 - 1) = y, . . . For example when x = 3, x^2 is 16, and x^2 -1 = 15. y then would be 17.
Do you check anything you write? When x = 3, x² is 9, not 16. Even if x² were 16, that wouldn't mean that y would be 17; y would be 15 or √15, depending on which equation above were used.

If x = 1.25, √(x² - 1) = √(1.5625 - 1) = √0.5625 = 0.75, which is rational.

If 0.75 were written as the product of two factors, one of which is irrational, the other factor would necessarily be irrational as well.

 July 14th, 2019, 08:15 AM #113 Member   Joined: Mar 2019 From: california Posts: 74 Thanks: 0 It's so bad, it's almost as if I did it deliberately. If x = 1.25, √(x² - 1) = √(1.5625 - 1) = √0.5625 = 0.75, which is rational. is your equation But if you add the requirement that x can only be integer, then the equation can never be right. What does that do with the price of cheese? In a similar way Z^n = X^n + Y^n can always solved with an infinite number of values X,Y,Z EXCEPT FOR INTEGER VALUES. So, I'm glad you found an example that works for a non-integer, but it's not a counter-example. In the context of the requirement that x be an integer, what you wrote is also gibberish. Don't you even check your logic? Last edited by skipjack; July 14th, 2019 at 10:51 AM.
 July 14th, 2019, 08:24 AM #114 Senior Member   Joined: Oct 2009 Posts: 850 Thanks: 325 OK, I figured you wouldn't really understand my little digression. It's ok, it's my fault for bringing it up. The issue remains in post 106 that you'd have to prove first that you don't multiply two irrationals to get a rational. That is, if you factor x- 1 into (sqrt(x)-1)(sqrt(x)+1), you'd have to prove only ONE of these factors is irrational.
July 14th, 2019, 10:59 AM   #115
Global Moderator

Joined: Dec 2006

Posts: 20,919
Thanks: 2203

Quote:
 Originally Posted by michaelcweir In the context of the requirement that x be an integer, . . .
I didn't intend my x to be the same as the x (or X) used previously, so there was no requirement for it to be an integer. I was just giving an example of non-zero rationals whose squares differ by 1. Anyway, in the relevant context, such as posts 106 and 110, the squares used had to be rational, but weren't necessarily integers.

 July 16th, 2019, 04:55 PM #116 Member   Joined: Mar 2019 From: california Posts: 74 Thanks: 0 Hi Mcrm@ss I will reply to your query, as yours is the more general one than Skipjack's, which touches on the same issue. Yes, you are right that it could be possible to have 2 irrationals that when multiplied together produce a rational. The answer is that it doesn't matter. Every time you do an iteration, you end up with two factors X^1/n + 1 and X^1/n - 1. What you need to remember is that both of the these factors are encircled by the square root. So you take the factor outside the square root and take the next iteration of the X^1/n -1. So while the product of the 2 factors may be a rational number, the square root will never be a rational number. Let's see how you do with this. Last edited by skipjack; July 17th, 2019 at 12:04 PM.
July 16th, 2019, 10:41 PM   #117
Senior Member

Joined: Oct 2009

Posts: 850
Thanks: 325

Quote:
 Originally Posted by michaelcweir Hi Mcrm@ss I will reply to your query, as yours is the more general one than Skipjack's, which touches on the same issue. Yes, you are right that it could be possible to have 2 irrationals that when multiplied together produce a rational. The answer is that it doesn't matter. Every time you do an iteration, you end up with two factors X^1/n + 1 and X^1/n - 1. What you need to remember is that both of the these factors are encircled by the square root. So you take the factor outside the square root and take the next iteration of the X^1/n -1. So while the product of the 2 factors may be a rational number, the square root will never be a rational number. Let's see how you do with this.
The square root of a rational number might be rational.

Last edited by skipjack; July 17th, 2019 at 12:04 PM.

July 19th, 2019, 03:31 AM   #118
Member

Joined: Mar 2019
From: california

Posts: 74
Thanks: 0

It's turtles all the way down

Quote:
 Originally Posted by Micrm@ss The square root of a rational number might be rational.
It might be, but it still doesn't matter.

Even if the square root of a rational number is rational, in this instance the rational number is composed of two numbers being multiplied together.

First, let's agree that these two numbers are not a perfect square. We know this because the multiplied are separated by plus and minus 1.
Because these multiplicands are inside the square root, you can move the X^1/n + 1 outside the square (whether it is a rational or not) and you are left with

sqrt (X^1/n -1).

This you can now factor again into sqrt((X^1/n+1 + 1) x (X^1/n+1 -1))

So, is this sqrt now a rational number? If it is, factor out the +1 multiplicand and do it again.

Eventually, you will have an irrational number from the sqrt.

It's turtles all the way down.

Are we turtlely finished yet?

July 19th, 2019, 03:35 AM   #119
Senior Member

Joined: Oct 2009

Posts: 850
Thanks: 325

Quote:
 Originally Posted by michaelcweir It might be, but it still doesn't matter. Even if the square root of a rational number is rational, in this instance the rational number is composed of two numbers being multiplied together. First, let's agree that these two numbers are not a perfect square. We know this because the multiplied are separated by plus and minus 1. Because these multiplicands are inside the square root, you can move the X^1/n + 1 outside the square (whether it is a rational or not) and you are left with sqrt (X^1/n -1). This you can now factor again into sqrt((X^1/n+1 + 1) x (X^1/n+1 -1)) So, is this sqrt now a rational number? If it is, factor out the +1 multiplicand and do it again. Eventually, you will have an irrational number from the sqrt. It's turtles all the way down. Are we turtlely finished yet?

Yeah, we're finished.

July 19th, 2019, 08:44 AM   #120
Senior Member

Joined: Feb 2010

Posts: 711
Thanks: 147

Quote:
 Originally Posted by michaelcweir and you are left with sqrt (X^1/n -1). This you can now factor again into sqrt((X^1/n+1 + 1) x (X^1/n+1 -1))
sqrt((X^1/n+1 + 1) x (X^1/n+1 -1)) is $\displaystyle \sqrt{\left(\dfrac{x^1}{n}+1+1 \right)\left(\dfrac{x^1}{n}+1-1 \right)}$

Fractions, exponents, and radicals using the latex parser on this website can be learned by a bright 13 year old. Find one and have him instruct you on how to use it.

... OK so I guess (boy is that dangerous) that you meant $\displaystyle \sqrt{(x^{1/n}+1+1)(x^{1/n}+1-1)}$. However in no way does this multiply to sqrt (X^1/n -1) whether you mean $\displaystyle \sqrt{\dfrac{x^1}{n}-1}$ or $\displaystyle \sqrt{x^{1/n}-1}$.

On the other hand maybe you don't mean either of these.

Just curious ... why are you posting all this here? The real experts on this site (people like Skipjack, Micrm@ss, Maschke, etc.) might have the requisite knowledge to wade through your stuff and find your mistakes. I don't. But even if they don't find any, so what. You get absolutely nothing by posting here. The only way this works is if you submit this for publication in a refereed journal.

 Tags analysis, diophantine, diopphantine, fermat, theorem

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Lourie Number Theory 3 April 1st, 2017 12:37 AM McPogor Number Theory 5 December 7th, 2013 07:28 PM mathbalarka Number Theory 2 April 3rd, 2012 11:03 AM theomoaner Number Theory 29 November 26th, 2011 10:23 PM SnakeO Number Theory 10 September 25th, 2007 04:23 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top