Is x Irrational if x² Is Irrational?

kmeado07
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Homework Statement



Prove that if x^2 is irrational then x must be irrational.

Homework Equations





The Attempt at a Solution



Maybe do proof by contradiction. I'm not really sure where to start.
 
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Proof by contradiction sounds good. What if x is rational?
 
"Suppose x is rational. Then x= __________"
 
so i let x= a/b

then obviously x^2 = a^2/b^2

im not sure how to continue to reach the contradiction
 
Assume x^2 is irrational and x is NOT irrational. You've shown that in such a case, x^2 is rational. That's a contradiction. x^2 can't be both rational and irrational. Therefore x must be irrational.
 
kmeado07 said:
so i let x= a/b

then obviously x^2 = a^2/b^2

im not sure how to continue to reach the contradiction

It is not enough just to say "let x= a/b" without saying what a and b are. A number is rational if and only if it can be expressed as a ratio of integers: a/b where a and b are integers (and b is not 0). If x is rational, the x= a/b where a and b are integers. You arrive at the fact that x2= a2/b2, also a ratio of integers. That itself contradicts the hypothesis, that x2 is irrational.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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