Question

Why is the product of a rational number and an irrational number irrational

Answer 1

A proof by contradiction.

Let assume that the product of a rational number and an irrational number is rational.

Let $\dfrac{a}{b}$ and $\dfrac{c}{d}$ be rational numbers, where $a,b,c,d\in \mathbb{Z} \wedge b,d\not=0$ and $x$ an irrational number.

Then

$\dfrac{a}{b}\cdot x=\dfrac{c}{d}\\ x=\dfrac{bc}{ad}$

Integers are closed under multiplication, therefore $bc$ and $ad$ are integers, making the number $x=\dfrac{bc}{ad}$ rational, which is contradictory with the earlier statement that $x$ is an irrational number.

Answer 2

Great question.  Let's let r be a rational number and s be irrational.  Note r has to be nonzero for this to work.   In other words, it's not true that when we multiply zero, a rational number, by an irrational number like π we get an irrational number.  We of course get zero.

The question is: why is the product

$p = rs$

irrational?

In math "why" questions are usually answered with an illuminating proof.  Here the indirect proof is enlightening.

Suppose p was rational.   Then

$s = \dfrac p r$

would be rational as well, being the ratio of two rational numbers, so ultimately the ratio of two integers.  

But we're given that s is irrational so we have our contradiction and must conclude our assumption that p is rational is false, that is, we conclude p is irrational.