Evenness Equivalence for x and x Squared

In HSC Mathematics Extension 1, the concept of 'if and only if' (biconditional) statements is central to the Proofs topic. This classic problem demonstrates how a biconditional claim about parity requires two separate arguments—one forward direction using a direct proof, and the reverse direction using the contrapositive. Understanding this pattern equips you with a fundamental tool for establishing properties of integers and will reappear in proofs of irrationality (such as √2).

Problem Statement

Prove that $x$ is even if and only if $x^2$ is even (for integer $x$).

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Hints

This is a biconditional (iff) statement requiring two proofs:

• Direction 1: If $x$ is even, then $x^2$ is even (direct proof).
• Direction 2: If $x^2$ is even, then $x$ is even (try contrapositive: if $x$ is odd, then $x^2$ is odd).

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Solutions

Direction 1 ($\Rightarrow$): If $x$ is even, then $x^2$ is even.

To prove the forward direction, we assume $x$ is even and aim to show $x^2$ is even. By definition, an even integer can be expressed as $x = 2k$ for some integer $k$. Substituting into the square:

$$x^2 = (2k)^2 = 4k^2 = 2(2k^2)$$

Since $2k^2$ is an integer, we have written $x^2$ as twice an integer, which means $x^2$ is even by definition.

Direction 2 ($\Leftarrow$): If $x^2$ is even, then $x$ is even.

Proving this directly can be challenging, so we prove the logically equivalent contrapositive: If $x$ is odd, then $x^2$ is odd.

Assume $x$ is odd. Then $x = 2k + 1$ for some integer $k$.

Substituting and expanding:

$$\begin{aligned} x^2 &= (2k+1)^2 \\ &= 4k^2 + 4k + 1 \\ &= 2(2k^2 + 2k) + 1 \end{aligned}$$

Now $2k^2 + 2k$ is an integer, so $x^2$ has the form $2m+1$ (with $m$ integer), which is exactly the definition of an odd number. Therefore $x^2$ is odd. By contrapositive, if $x^2$ is even, then $x$ must be even.

Conclusion: Both directions proven, so $x$ is even $\iff$ $x^2$ is even. $\blacksquare$

Note: This result is fundamental in many irrationality proofs (e.g., $\sqrt{2}$ is irrational). If $p$ is an odd prime and $p|x^2$, then $p|x$.

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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.
• Recognise that this evenness equivalence is the backbone of the classic proof that $\sqrt{2}$ is irrational—a key Extension 1 idea.

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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 Polys Ext 1, HSC Last Resorts