Rational Plus Irrational Is Irrational

Problem Statement

If $a$ is rational and $b$ is irrational, prove that $a + b$ is irrational.

Hints

Use proof by contradiction. Assume $a + b$ is rational, then solve for $b$ in terms of rational numbers. What property of rational numbers does this contradict?

Solutions

Proof by Contradiction:

Assume, for contradiction, that $a + b$ is rational.

Since $a$ is rational and $a + b$ is rational (by assumption), we can write:

\begin{aligned} a &= \frac{p}{q} \quad \text{where } p, q \in \mathbb{Z}, q \neq 0 \\ a + b &= \frac{r}{s} \quad \text{where } r, s \in \mathbb{Z}, s \neq 0 \end{aligned}

Solving for $b$ :

b = (a + b) - a = \frac{r}{s} - \frac{p}{q} = \frac{rq - ps}{sq}

Since $r, q, p, s$ are all integers and $sq \neq 0$ , this shows $b$ is expressible as a ratio of two integers.

Therefore, $b$ must be rational, which contradicts the given condition that $b$ is irrational.

Hence, our assumption was false, and $a + b$ must be irrational. $\blacksquare$

Takeaways

Reconstruct the full proof from the hint and the solution outline, and justify every transformation explicitly.
Check edge cases and verify where each assumption is used in the argument.

Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills:

HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
HSC Collections: https://vumaths.com/booklets/hsc-collections/
HSC Inequalities: https://vumaths.com/booklets/hsc-inequalities/

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