Product of Two Irrationals: Counterexample

Problem Statement

Prove or disprove: If $x$ and $y$ are irrational numbers with $x \neq y$ , then $xy$ is irrational.

Hints

The statement is false. Find a counterexample using multiples of $\sqrt{2}$ .

Consider $x = \sqrt{2}$ and $y = k\sqrt{2}$ for some rational $k \neq 1$ . What is $xy$ ? Is it rational or irrational?

Solutions

The statement is false. We disprove by counterexample.

Counterexample:

Let $x = \sqrt{2}$ and $y = 2\sqrt{2}$ .

Check $x$ is irrational: $\sqrt{2}$ is irrational (well-known).
Check $y$ is irrational: $2\sqrt{2}$ is the product of rational $2$ and irrational $\sqrt{2}$ , so it's irrational.
Check $x \neq y$ : Clearly $\sqrt{2} \neq 2\sqrt{2}$ since $1 \neq 2$ .
Compute $xy$ :

xy = (\sqrt{2})(2\sqrt{2}) = 2 \cdot (\sqrt{2})^2 = 2 \cdot 2 = 4

Check rationality of $xy$ : $4 = \frac{4}{1}$ is rational.

Since we found irrational numbers $x$ and $y$ with $x \neq y$ such that $xy$ is rational, the original statement is disproven. $\blacksquare$

Note: This shows that the irrationals are not closed under multiplication. The rationals are closed under multiplication, but the irrationals are not.

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 Induction: https://vumaths.com/booklets/hsc-induction/
HSC Vectors: https://vumaths.com/booklets/hsc-vectors/

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