# Very easy proof

• May 11th 2011, 10:50 AM
dwsmith
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 $\displaystyle x^2$ is irrational, then so is x.

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

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

Thanks.
• May 11th 2011, 10:59 AM
Plato
Quote:

Originally Posted by dwsmith
Show that if $\displaystyle x^2$ is irrational, then so is x.

Can you show that $\displaystyle \text{if }x\text{ is rational then }x^2\text{ is rational}~?$.
If so, what is the contrapositive ?
• May 11th 2011, 11:07 AM
dwsmith
Quote:

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

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

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

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

Is that just it?
• May 11th 2011, 11:24 AM
DrSteve
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.

• May 11th 2011, 11:35 AM
Plato
Quote:

Originally Posted by dwsmith
If x is rational, then $\displaystyle x=\frac{m}{n}, \ n\neq 0, \ m,n\in\mathbb{Z}$.
$\displaystyle x^2=\frac{m}{n}\cdot\frac{m}{n}$
And a rational times a rational number is rational.
The contrapositive is my original statement.

Quote:

Originally Posted by DrSteve