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July 19th, 2019, 08:49 AM   #121
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Quote:
 Originally Posted by michaelcweir 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.
You've neglected the case where the square root is irrational.

For a factorization such as 6 = (√7 + 1) × (√7 - 1), your process of repeatedly factoring until an irrational factor is reached is unnecessary, because both factors are already irrational.

Hence your alleged proof fails, because 6 is rational.

July 21st, 2019, 04:14 PM   #122
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Quote:
 Originally Posted by skipjack You've neglected the case where the square root is irrational. For a factorization such as 6 = (√7 + 1) × (√7 - 1), your process of repeatedly factoring until an irrational factor is reached is unnecessary, because both factors are already irrational. Hence your alleged proof fails, because 6 is rational.
I really do appreciate your effoorts. It is necessry to exhaust all types of inquiry.

Here is the counter statement
sqrt(7 +1) x sqrt(7-1) is not equal to 6. The answer to 2 decimal places is 6.88. And the answer is irrational, not rational.

There is a vast difference between sqrt (X) +1 and sqrt (x +1)

July 21st, 2019, 04:41 PM   #123
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Quote:
 Originally Posted by michaelcweir I really do appreciate your effoorts. It is necessry to exhaust all types of inquiry. Here is the counter statement sqrt(7 +1) x sqrt(7-1) is not equal to 6. The answer to 2 decimal places is 6.88. And the answer is irrational, not rational. There is a vast difference between sqrt (X) +1 and sqrt (x +1)
Jeez Louise dude. He said $(\sqrt 7 + 1) \times (\sqrt 7 - 1) = 7 - 1 = 6$.

Just curious, how come you appreciate some people's efforts, but think I'm personally trying to steal your work? Why aren't they trying to steal your work? They must be more discerning than me.

July 21st, 2019, 06:18 PM   #124
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Quote:
 Originally Posted by michaelcweir There is a vast difference between sqrt (X) +1 and sqrt (x +1)
Of course, but there's a well-established convention that √x + 1 means √(x) + 1, not √(x + 1).

Using that convention, (√7 + 1)(√7 - 1) = (√(7) + 1)(√(7) - 1) = 6 and your "proof" remains flawed, as it fails to consider such a factorization of a rational as a product of two irrationals.

 July 22nd, 2019, 10:07 AM #125 Senior Member     Joined: Feb 2010 Posts: 711 Thanks: 147 Apparently he still has not found a bright 13 year old yet. Thanks from topsquark
July 24th, 2019, 06:02 PM   #126
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Quote:
 Originally Posted by skipjack You've neglected the case where the square root is irrational. For a factorization such as 6 = (√7 + 1) × (√7 - 1), your process of repeatedly factoring until an irrational factor is reached is unnecessary, because both factors are already irrational. Hence your alleged proof fails, because 6 is rational.
Let's begin by stating the obvious.
You claim that the 2 irrationals, when multiplied together must be a rational

(X^1/n+1 + 1) x (X^1/n+1 -1)

Let's assume that

So (X^1/n+1 + 1) is an irrational.

So let's assign a variable I to represent this relationship.

(X^1/n+1 + 1) = I if we square both sides it becomes

X^1/n + 2 (X^1/n+1 + 1) + 1 = I^2

(X^1/n+1 + 1) is irrational, remember.

So X^1/n is an irrational number too? Do I have to spell it out for you?

Replace X^1/n with h x e/f
So the assumption of the irrationals multiplied together brings forth the result

h x e/f + 1 is irrational. Thus one or more of X,Y,Z is irrational.

Do you have anything else?

Last edited by skipjack; July 24th, 2019 at 08:50 PM.

July 24th, 2019, 06:13 PM   #127
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Quote:
 Originally Posted by Maschke Jeez Louise dude. He said $(\sqrt 7 + 1) \times (\sqrt 7 - 1) = 7 - 1 = 6$. Just curious, how come you appreciate some people's efforts, but think I'm personally trying to steal your work? Why aren't they trying to steal your work? They must be more discerning than me.
i never thought you were going to steal my theory. I was just mad at how you were treeating my theory by dissing me, in effect. So I sent you the worse insult i could tihink of. It was suprising to see how Microm@ss took up qnd reinforced it.

So if you are going to behave, I will ask you again. What do you think of the so far?

July 24th, 2019, 06:38 PM   #128
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Quote:
 Originally Posted by michaelcweir Let's begin by stating the obvious. You claim that the 2 irrationals, when multiplied together must be a rational (X^1/n+1 + 1) x (X^1/n+1 -1) Let's assume that So (X^1/n+1 + 1) is an irrational. So let's assign a variable I to represent this relationship. (X^1/n+1 + 1) = I if we square both sides it becomes X^1/n + 2 (X^1/n+1 + 1) + 1 = I^2 (X^1/n+1 + 1) is irrational, remember. So X^1/n is an irrational number too? Do I have to spell it out for you? Replace X^1/n with h x e/f So the assumption of the irrationals multiplied together brings forth the result h x e/f + 1 is irrational. Thus one or more of X,Y,Z is irrational. Do you have anything else?
I don't poke in on this thread very often, but would it kill you to either learn some simple LaTeX or use parentheses when needed? I presume that X^1/n + 1 is supposed to be $\displaystyle X^{1/(n + 1)} + 1$? In plain-text, it would be X^(1/(n + 1)) + 1.

-Dan

Last edited by skipjack; July 24th, 2019 at 08:50 PM.

July 24th, 2019, 08:57 PM   #129
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Quote:
 Originally Posted by michaelcweir You claim that the 2 irrationals, when multiplied together must be a rational
No, I claimed that any non-zero real (whether rational or irrational) can be written as the product of two irrationals, which is quite different.

July 25th, 2019, 09:14 AM   #130
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Quote:
 Originally Posted by michaelcweir (X^1/n+1 + 1) = I if we square both sides it becomes X^1/n + 2 (X^1/n+1 + 1) + 1 = I^2
I don't know how YOU are interpreting X^1/n+1 + 1 but I can't see any interpretation that yields X^1/n + 2 (X^1/n+1 + 1) + 1 when you square it. Perhaps the 13 year old can help you with squaring polynomials as well as latex.

 Tags analysis, diophantine, diopphantine, fermat, theorem

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