Root of 2

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Icebreaker

How do I show that 2 has no rational roots?
 

krab

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First assume it does. Write square root of 2 = p / q, where p and q are integers. Then p^2 = 2 q^2, and so must be even... you can guess the rest. Eventually you come to a conclusion that contradicts one of your original assumptions.
 
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Write square root of 2 = p / q, where p and q are integers.
Where p and q are coprime integers...
 

HallsofIvy

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Note: you will need the fact that the square of an odd number is always odd:
(2n+1)2= 4n2+ 4n+ 1= 2(2n2+ 2n) + 1.
 
I

Icebreaker

No, I meant ANY rational root. That is,

[tex]2^{\frac{1}{n}}[/tex] is not rational for any positive integer n > 1.

I tried "extending" the square root of 2 proof, however, at some point,

[tex]2a^n=b^n[/tex]

If n is even, then it works. But if n is odd, then the argument breaks down.

Or does it, let me think....
 
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AKG

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I think the same proof applies. Suppose (p/q)n = 2 where gcd(p,q) = 1. Then

pn = 2qn

This tells us that pn is even, which tells us that p is even, which tells us that 2n|pn. This in turn tells us that qn is even, which in turn tells us that q is even. Both q and p are even, and thus aren't co-prime, contradicting our assumption that they were.
 
I

Icebreaker

Yup, that's what I thought too. Thanks everyone.

On a side note, [tex]x^x = 2[/tex] implies that x cannot be rational because [tex]x = 2^{1/x}[/tex] and 2 has 2 rational roots, as we've shown. Is that correct?
 

Zurtex

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Icebreaker said:
Yup, that's what I thought too. Thanks everyone.

On a side note, [tex]x^x = 2[/tex] implies that x cannot be rational because [tex]x = 2^{1/x}[/tex] and 2 has 2 rational roots, as we've shown. Is that correct?
Our assumption before was that 21/n where n was an integer greater than 1. However I don't think it is hard to extend if you let x = p/q.
 

AKG

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Icebreaker said:
Yup, that's what I thought too. Thanks everyone.

On a side note, [tex]x^x = 2[/tex] implies that x cannot be rational because [tex]x = 2^{1/x}[/tex] and 2 has 2 rational roots, as we've shown. Is that correct?
If x is rational, then there are coprime p and q such that x = p/q. We get:

(p/q)(p/q) = 2

(p/q)p = 2q

If p/q is not a whole number, then in general (p/q)n is not whole for natural n, and in particular when n = p. On the other hand, 2q is of course whole, so we have a contradiction unless p/q is a whole number. But it's easy to check 11 is not 2, 22 is not 2, etc. So we get a contradiction regardless, so x is irrational.
 

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