# Proof by contrapositive

Prove that if the product $ab$ is irrational, then either $a$ or $b$ (or both) must be irrational.

How do I prove this by contrapositive? What is contrapositive?

• It would help us to help you if you explained which part of the Wikipedia article is unclear to you. Commented Mar 16, 2012 at 7:18

The statement you want to prove is:

If $ab$ is irrational, then $a$ is irrational or $b$ is irrational.

The contrapositive is:

If not($a$ is irrational or $b$ is irrational), then not($ab$ is irrational).

A more natural way to state this (using DeMorgan's Law) is:

If both $a$ and $b$ are rational, then $ab$ is rational.

This last statement is indeed true. Since the truth of a statement and the truth of its contrapositive always agree, one can conclude the original statement is true, as well.

If you have to prove an implication $A\Rightarrow B$, contrapositive means you want to prove the equivalent statement $\neg B\Rightarrow\neg A$. The fact that they are equivalent guarantees you that also $A\Rightarrow B$ holds.

In your case, $A=$ 'the product $ab$ is irrational', while $B=$ '$a$ or $b$ must be irrational'. So you just have to negate both $A$ and $B$ and prove the contraposition $\neg B\Rightarrow\neg A$, which is not hard in your case.

By the way, there even is a Wikipedia page with exactly the title of your post here ;)

If $a,b \in \Bbb{Z}^+$ and $a\neq b$, then $ax^2+bx+(b-a) = 0$ has no positive integer root (solution).

• Use $\LaTeX\text{}$
– user93957
Commented Nov 26, 2013 at 19:02