# Are all prime numbers greater than 2 odd?

Are all prime numbers greater than 2 odd?

I wasn't allowed to assume this was true.

AFSOC that there exists an even prime $$n$$. Then by definition of even, $$n = 2q$$, where $$q$$ is a positive integer. But if $$n$$ is even, it is divisible by $$2$$, hence that contradicts $$n$$'s prime-ness. QED.

What? Is there some wacky prime that can't be assumed odd?

• Yes. Any even number is divisible by $2$, and hence not prime. – David G. Stork Jul 29 '19 at 19:14
• Yes, it's true. Presumably, your teacher wanted you to prove it in order to use use it. – saulspatz Jul 29 '19 at 19:15
• You are correct: the only prime that can be divisible by $2$ is $2$ itself. – Robert Israel Jul 29 '19 at 19:16
• You may need something like "If $q=1$ then $n=2$ which we know is prime. Otherwise $q \gt 1$ and $n=2q$ is the product of two integers greater than $1$ and thus composite" – Henry Jul 29 '19 at 19:18
• @DavidG.Stork Except 2. – Servaes Jul 29 '19 at 19:20

That is correct, all prime numbers in $$\mathbb Z$$ which are greater than 2 are odd. Likewise all prime numbers less than $$-2$$ are also odd.

A prime number is a non-unit number (that is, not $$-1$$ or 1) which is divisible only by itself, the units and the associates of itself (the number times a unit). So, for example, $$-7$$ is prime, 14 is not.

And no, there is no wacky prime that can't be assumed odd. Except maybe $$-2$$ and 2. Those two are kind of wacky, in my opinion.

The thing is that in math (or "maths," if you prefer), if you haven't proven an assertion, that assertion might be false. If you make an assertion without proof, someone might challenge you to prove it.

Fortunately, the assertion that all primes greater than 2 are odd is an easy assertion to prove. So, if anyone challenges you on that, you can just quickly state the proof and move on.

Yes, all prime numbers greater than $$2$$ are odd.

That's because any even number greater than $$2$$ can be expressed as $$2$$ times a number greater than $$1$$,

so it is composite (not prime).

(As noted in comments to the question, $$2$$ itself is prime.)

Yes---

For if they they weren't, then they would be even; and hence, divisible by $$2$$; and hence, not prime.