Prove: For any integer $n$, if $n^3+5$ is even, then $n$ is odd.

Question

Is this proof good?

Proof:

This will be proven by contrapositive. Let $n$ belong to all integers such that $n$ is even. Then $n=2k$ where $k \in \mathbb Z$ Thus $n^3+5=(2k)^3+5$ which is odd.

Your proof has a good structure but could be more detailed/explanatory, as Dominique says in her answer, which I agree with. How detailed you are/ want to be in a proof depends on the background and context upon which the statement was being asked to prove, but I think that in most contexts, Dominique's level of detail is standard. — Adam Rubinson, May 11 '23 at 08:09

score 5 · Answer 1 · edited May 11 '23 at 10:07

Although your answer is correct, you might need to consider one thing:

An even number can be written as $2a$ where $a$ is an integer number.
An odd number can be written as $2a+1$ where $a$ is an integer number.

Your statement:

$n^3+5=(2k)^3+5$, which is odd.

You might elaborate this into $n^3+5=(2k)^3+5=2 \cdot (4k^3+2) + 1$, which has the “general” formulation of an odd number.

score 1 · Answer 2 · answered May 11 '23 at 07:29

The proof is correct. However, another way to show it is to suppose that $n^3+5$ is even, then $n^3$ is odd because $5$ is odd. But the product of even numbers is even. Therefore $n$ is odd. This is the same proof you provide; write more structured way.

score 0 · Answer 3 · answered May 11 '23 at 07:47

0

Another approach, you can use direct proof.

$n^3$ is equal to something odd that implies $n$ must be odd.Why? Assume n should be even and get a contradiction.

answered May 11 '23 at 07:47

nozalp10

108

score 0 · Answer 4 · answered May 11 '23 at 09:03

if you are a little bit familiar with modular arithmetic, the proof can be written as follows

$$n^3+5\equiv 0 \text{ mod }2\Rightarrow n^3\equiv -5\equiv 1\text{ mod } 2,$$ so we have, $$n\equiv 1 \text{ mod }2,$$ because $0^3\equiv 0 \text{ mod }2$.

score -1 · Answer 5 · answered May 11 '23 at 07:58

Your proof is correct, although, as Dominique wrote in their answer, you might want to write $(2k)^3 + 5 = 2(4k^3+2)+1$ and say that since $k$ was assumed to be integer, then so is $4k^3+2$.

Let me show another way to prove it. First note that $n$ is even if and only if $n^3$ is even. One direction is clear. For the other, you may use that $2$ is prime, so $$2\mid n^3 \implies 2\mid n \text{ or } 2\mid n^2 \implies 2\mid n \text{ or } 2\mid n\text{ or } 2\mid n \implies 2\mid n,$$

or you can assume that $n$ is odd, so $n = 2k+1$ and $n^3 = 8k^3+12k^2+6k+1 = 2(4k^3+6k^2+3k)+1$.

Now, if $n^3+5$ is even, then $$n^3+5 = 2k \implies n^3 = 2(k-3)+1$$ so $n^3$ is odd and therefore $n$ is odd.

Prove: For any integer $n$, if $n^3+5$ is even, then $n$ is odd.

5 Answers5