Irrationality of Non-Perfect-Square Roots

The classic proof that $\sqrt{2}$ is irrational can be extended to any positive integer that is not a perfect square. This result sits squarely in the HSC Proofs topic and shows you how the Fundamental Theorem of Arithmetic can be used to force a contradiction. By working through the argument below, you’ll practise setting up a proof by contradiction and harnessing the uniqueness of prime factorisation—techniques that appear again and again in Extension 1 and 2 mathematics.

Problem Statement

Let $a$ be a positive integer. If $a$ is not a perfect square, prove that $\sqrt{a}$ is irrational.

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Hints

Attempt the proof independently first. Focus on the key theorem, algebraic transformation, or contradiction setup that links the hypothesis to the target conclusion.

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Solutions

Proof by Contradiction

We begin by supposing the opposite of what we want to show—that $\sqrt{a}$ is rational—and we’ll chase this assumption until it crashes into a fact we know is false. That crash tells us our starting assumption cannot hold, so $\sqrt{a}$ must be irrational.

Step 1: Assume the negation

Assume, for contradiction, that $\sqrt{a}$ is rational. Then we can write

$$\sqrt{a} = \frac{p}{q}$$

where $p, q \in \mathbb{Z}$, $q \neq 0$, and $\gcd(p,q) = 1$.

Step 2: Square both sides

Squaring gets rid of the radical and gives a relation between integers:

$$\begin{aligned} a &= \frac{p^2}{q^2} \\ p^2 &= aq^2 \end{aligned}$$

Step 3: Compare prime factorizations

Now we bring in the Fundamental Theorem of Arithmetic to look at the prime exponents on each side of $p^2 = a q^2$.

By the Fundamental Theorem of Arithmetic, every positive integer has a unique prime factorization.

When an integer is squared, every prime exponent in its factorization becomes even. So:

• in $p^2$, every prime appears with an even exponent;
• in $q^2$, every prime appears with an even exponent.

Now the equation

$$p^2 = aq^2$$

shows that $aq^2$ must also have only even prime exponents.

Since $q^2$ already contributes only even exponents, this is possible only if every prime appearing in $a$ also has an even exponent.

Step 4: Derive the contradiction

But if every prime in the factorization of $a$ has an even exponent, then $a$ is a perfect square.

This contradicts the hypothesis that $a$ is not a perfect square.

Conclusion

Therefore, our assumption was false, and $\sqrt{a}$ must be irrational.

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Takeaways

• Parity of Prime Exponents: A perfect square has only even exponents in its prime factorization
• Why the Method Works: Squaring forces all prime exponents in $p^2$ and $q^2$ to be even, so the same must be true for $a$
• Generalization: This extends proofs like $\sqrt{2}$ is irrational or $\sqrt{23}$ is irrational to any positive integer that is not a square
• Key Tool: The argument relies on the Fundamental Theorem of Arithmetic and uniqueness of prime factorization

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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 Functions, HSC Probability