Rational Number plus Irrational Number is Irrational
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Theorem
Rational number plus irrational number is irrational.
That is, let $x \in \Q$, $y \in \R \setminus \Q$ and $x + y = z$.
Then $z \in \R \setminus \Q$.
Proof
Aiming for a contradiction, suppose $z \in \Q$.
By definition of rational numbers:
- $\exists a, b \in \Z, b \ne 0: x = \dfrac a b$
- $\exists c, d \in \Z, d \ne 0: z = \dfrac c d$
Then:
\(\ds x + y\) | \(=\) | \(\ds z\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac a b + y\) | \(=\) | \(\ds \dfrac c d\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac {b c - a d} {b d}\) | by rearrangement |
This shows that $y$ is rational, which is a contradiction.
By Proof by Contradiction, $z$ is irrational.
$\blacksquare$