Proof by contrapositive:

Let $x \in \mathbb{Z}$. Assume that $2 \nmid x$. Thus, $\forall k \in \mathbb{Z}$, $2k \neq x \Rightarrow (2k)^3 \neq x^3 \Rightarrow 8k^3 \neq x^3 \Rightarrow 2(4k^3) \neq x^3$. B/c $4k^3 \in \mathbb{Z}$, $2 \nmid x^3$. $\therefore 2 \nmid x \Rightarrow 2 \nmid x^3$, so the contrapositive $2 \ \vert \ x^3 \Rightarrow 2 \ \vert \ x$ is true.

I didn't realize that $2 \nmid x \Rightarrow x$ is odd when I did this proof, but is the above method valid?

  • $\begingroup$ "B/c $4k^3 \in \mathbb{Z}$, $2 \nmid x^3$. " I'm not following the reasoning here. I'm not convinced that $2 \nmid x^3.$ $\endgroup$ May 15 at 22:00

3 Answers 3


This method is not valid. You showed that $2(4k^3)\neq x^3$ for all $k\in\mathbb Z$ correctly, but but to show that $2\nmid x^3$, you need to show that $2n\neq x^3$ for all integers $n$, not just integers of the form $4k^3$ for some $k\in\mathbb Z$.

To illustrate why this is important, consider the following (false) statement: $4\mid x^2\Rightarrow 4\mid x$. Using the proof technique proposed we could proceed to "prove" this statement by contrapositive:

Suppose $4\nmid x$. Then $4k\neq x$ for any $k\in \mathbb Z$, and $16k^2=4(4k^2)\neq x^2$ for any $k\in\mathbb Z$.

But if we consider $x=6$, the flaw in the argument becomes clear: while $6^2=36=4\cdot 9$, $9$ cannot be written as $4k^2$ for any integer $k$. Thus, even though $4\nmid 6$, $4\mid 6^2$.

  • 1
    $\begingroup$ You can get the right formatting with \not\mid by the way $\endgroup$
    – pancini
    May 15 at 21:22
  • $\begingroup$ @pancini Thanks! you learn something new every day. $\endgroup$
    – Alex S
    May 15 at 21:25

There are a couple issues:

  • You didn't really show that $2$ does not divide $x^3$. Rather, you showed that, for all $k\in\Bbb Z$, $x^3\neq 2(4k^3)$. It's not clear why $x^3$ must be of the form $2(4k^3)$ if $2|x$.
  • The statement $2k\neq x\implies (2k)^3\neq x^3$ is true, but only because $x\mapsto x^3$ is injective. (Note that $2k\neq x$ does not imply $(2k)^2\neq x^2$, for example.)

Here's a hint for another approach: a non-unit $p\in\Bbb Z$ is prime $\iff$ $p|ab$ implies $p|a$ or $p|b$ for all $a,b\in\Bbb Z$.


Suppose $2 \mid x^3$.

Assume $2 \nmid x$,

then $\exists r \in \mathbb{N}$ such that $x = 2r + 1$.

Then $$x^3 = (2r + 1)^3$$ $$=8r^3 + 12r^2 + 18r + 1$$

Since $2 \mid x^3$, then it follows that $2 \mid (8r^3 + 12r^2 + 18r + 1)$, But, while $2 \mid 8r^3$, $2 \mid 12r^2$ and $2 \mid 18r$, $$2 \nmid 1$$

Now we arrive at a contradiction such that:

$$x^3 = 8r^3 + 12r^2 + 18r + 1$$ $$2 \mid x^3$$ but $$2 \nmid (8r^3 + 12r^2 + 18r + 1)$$

Therefore, the original assumption is not correct, and that it is indeed true that $2 \mid x$.

$$2 \mid x^3 \implies 2 \mid x$$


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