Rational Plus Irrational Is Irrational

Proof by contradiction is a cornerstone of the HSC Mathematics Extension 1 syllabus, especially when we reason about irrational numbers. This problem asks you to prove a classic result: adding a rational number to an irrational number can never produce a rational sum. By working through the argument carefully, you will strengthen your ability to structure contradiction proofs and manipulate the fraction-based definition of a rational number.

Problem Statement

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

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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?

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Solutions

Proof by Contradiction:

We begin by supposing the opposite of what we want to prove. If the claim were false, then $a+b$ would have to be rational. We will show this leads to an impossibility.

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

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

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

To isolate $b$, we subtract $a$ from the sum. Because $b = (a+b)-a$, we can compute $b$ directly using the fractional forms of the two rational numbers:

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

The numerator $rq-ps$ and denominator $sq$ are integers, and $sq \neq 0$ because both $s$ and $q$ are non-zero. Thus $b$ has been expressed 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. $\blacksquare$

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Takeaways

• Proof by contradiction: assume the negation and chase a contradiction – here, that an irrational number would be forced to behave like a rational one.
• The set of rational numbers is closed under subtraction; that property is what forces $b$ to appear rational, even though we started with an irrational $b$.
• This argument can be adapted to prove that the sum of any rational and any irrational is always irrational, and the same logic extends to other operations with caution.

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Further Readings

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