Rational Number plus Irrational Number is Irrational

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Rational number plus irrational number is irrational.

That is, let $x \in \Q$, $y \in \R \setminus \Q$ and $x + y = z$.

Then $z \in \R \setminus \Q$.


Aiming for a contradiction, suppose $z \in \Q$.

By definition of rational numbers:

$\exists a, b \in \Z, b \ne 0: x = \dfrac a b$
$\exists c, d \in \Z, d \ne 0: z = \dfrac c d$


\(\displaystyle x + y\) \(=\) \(\displaystyle z\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac a b + y\) \(=\) \(\displaystyle \dfrac c d\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle \dfrac {b c - a d} {b d}\) by rearrangement

This shows that $y$ is rational, which is a contradiction.

By Proof by Contradiction, $z$ is irrational.