1. ## Very easy proof

I have done this one before can't remember how I went about it.

Let x be a real number. Show that if $x^2$ is irrational, then so is x.

$x^{2}$ is irrational and x is rational.

A rational number squared won't be irrational but how do I show it?

Thanks.

2. Originally Posted by dwsmith
Show that if $x^2$ is irrational, then so is x.
Can you show that $\text{if }x\text{ is rational then }x^2\text{ is rational}~?$.
If so, what is the contrapositive ?

3. Originally Posted by Plato
Can you show that $\text{if }x\text{ is rational then }x^2\text{ is rational}~?$.
If so, what is the contrapositive ?
If x is rational, then $x=\frac{m}{n}, \ n\neq 0, \ m,n\in\mathbb{Z}$.

$x^2=\frac{m}{n}\cdot\frac{m}{n}$

And a rational times a rational number is rational. The contrapositive is my original statement.

Is that just it?

4. The fact that the product of 2 rationals is rational follows from the fact that the product of two integers is an integer, and the product of two nonzero integers is nonzero.

If x is rational, then $x=\frac{m}{n}, \ n\neq 0, \ m,n\in\mathbb{Z}$.
$x^2=\frac{m}{n}\cdot\frac{m}{n}$