Proving the cube of an integer even implies even integer

Key takeaway: The claim about $z^3$ being even depends on what kind of number $z$ is, because “even” only applies to integers.

The statement needs a careful check

The claim given is: if $z\in\mathbb{R}$ and $z^3$ is an even number, then $z$ is an even number. As written, this statement is not correct, because the word even applies to integers, not to arbitrary real numbers. For real numbers like $\sqrt[3]{2}$, the cube is $2$, which is even as an integer, but $\sqrt[3]{2}$ itself is not an even integer.

So before proving anything, we must fix the statement. A mathematically valid version is: if $z$ is an integer and $z^3$ is even, then $z$ is even.

Proof by contrapositive

To prove the corrected statement, use the contrapositive. Instead of proving "if $z^3$ is even, then $z$ is even," prove the equivalent statement:

If $z$ is odd, then $z^3$ is odd.

Let $z$ be odd. Then there exists an integer $k$ such that

$z=2k+1.$

Cube both sides:

$z^3=(2k+1)^3.$

Expand the binomial:

$z^3=8k^3+12k^2+6k+1.$

Factor out 2 from the first three terms:

$z^3=2(4k^3+6k^2+3k)+1.$

This has the form $2m+1$ for an integer $m$, so $z^3$ is odd. Therefore, the contrapositive is true, and the corrected statement is proven.

Why the original attempt fails

The work shown in the prompt starts by assuming $z$ is even and then shows $z^3$ is even. That proves only one direction: even integers have even cubes. It does not prove the converse. To prove "if $z^3$ is even, then $z$ is even," you need either a contrapositive or a contradiction argument.

There is also a language issue. The line "$z\in\mathbb{R}$ and $z^3$ is an even number" mixes real numbers with an integer property. If a teacher expects a proof in number theory, the domain should usually be $z\in\mathbb{Z}$. If the domain stays real numbers, the statement is false.

What a complete answer should say

A strong final response should mention the counterexample to the original real-number claim and then present the corrected integer proof. That shows both logical precision and mathematical understanding. In proof writing, fixing the statement is often just as important as proving it.

So the right conclusion is: the statement is false as written for real numbers, but the integer version is true, and its proof is best done by contrapositive.

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Pitfalls the pros know 👇 The first place this proof can go wrong is accepting the phrase “$z$ is even” when $z$ is only assumed to be real. Even and odd are properties of integers, so the statement must be checked for a domain mismatch before any algebra begins. Another common mistake is trying to prove the converse by showing that even integers have even cubes; that only proves one direction and does not justify the implication from $z^3$ to $z$. Students also sometimes write "$z=2k$" without stating that $k$ is an integer, which weakens the proof. If you want a rigorous result, either change the hypothesis to integers or supply a counterexample showing the real-number version is false. Then use the contrapositive, since that route is shorter and cleaner than trying to reason directly from cube parity.

What if the problem changes? If the problem were rewritten as “If $z\in\mathbb{Z}$ and $z^3$ is even, then $z$ is even,” the statement becomes true and the proof is straightforward. A different variant would be: “If $z$ is odd, then $z^3$ is odd.” In that version, write $z=2k+1$ and expand $(2k+1)^3$ to show it equals $2m+1$ for some integer $m$. Another useful variant is to replace the cube with a fifth power: if $z^5$ is even, then $z$ is even for integers $z$. The same contrapositive idea works, because any odd integer raised to an odd power remains odd. These variants help show that the parity argument is really about odd and even integers, not about arbitrary real numbers.

Tags: proof by contrapositive, parity of integers, odd and even integers