# Proof - irrational numbers

1. Nov 1, 2008

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

Maybe do proof by contradiction. I'm not really sure where to start.

2. Nov 1, 2008

### Office_Shredder

Staff Emeritus
Proof by contradiction sounds good. What if x is rational?

3. Nov 1, 2008

### HallsofIvy

Staff Emeritus
"Suppose x is rational. Then x= __________"

4. Nov 1, 2008

so i let x= a/b

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

im not sure how to continue to reach the contradiction

5. Nov 1, 2008

### Dick

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.

6. Nov 1, 2008

### HallsofIvy

Staff Emeritus
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.