# There is an even prime greater than $2$ [closed]

(NOTE: I am new to proof construction. Don't panic if your heart beat increase.)

Proof 1:

suppose $x$ is even and prime, then there is $k$ in $\mathbb N$ such that

$x = 2k$

But x is only divisible by itself and not 2.

$\frac{x}{2} != k$

$x$ can not be even.

Proof 2: $x$ is prime

$x = pq$ then either $p=1$ or $q=1$

$x$ is even

$x = 2k$

$pq = 2k$

$(2)\frac{k}{pq} = 1$

which is false

• The contradiction tells you exactly that it cannot exist. I'm sorry but I didn't get your point. Could you explain better what are you asking for? Commented May 13, 2017 at 9:05
• It isn't clear (at least for me) what you're even trying to prove. Are you trying to prove the claim is false or else to prove it is true? This is the first thing that must be crystal clear before even reading your "proof". Commented May 13, 2017 at 9:06
• I have a hard time following this. It sort of looks like you are arguing that if $x>2$ is both even and prime, then you get a contradiction. Commented May 13, 2017 at 9:07
• @Mankind You are right, this is exactly what I am trying to do Commented May 13, 2017 at 10:06
• Your mistake (after the recent edit) is in the line $x=2pq$. All you know is that $x=2k$ for some number $k$. There aren't two numbers $p$ and $q$ to deal with. Please read the two day old correct answer below. Commented May 15, 2017 at 13:52

Here's a proof that $2$ is the only even prime.

Let $a \in \mathbb{N}$ be prime and even. (We know this is okay to do, as there is at least one even prime)

Assume $a > 2$

Since $a$ is even, it can be written as $a = 2k$ for some $k \in \mathbb{N}$. By assumption, $k > 1$.

$\frac{a}{2} = k$, which implies that $a$ is divisible by $2$.

Since $a$ is prime, its only divisors are $1$ and $a$.

However this is contradicted by the fact that $2 | a$ and $a > 2$.

Thus the assumption that there is an even prime larger than $2$ is false.

NOTE: This proof was much more verbose than necessary, but it seemed you were in need a fully stated proof by contradiction argument.

What?

Whatever you're trying to do, it gets completely lost because you're not explaining what you're trying to accomplish, but just stating some things.

And for the mathematical content, I gave up after:

If $x>n$ then $\frac{x}{n}$ does not belong to $\mathbb N$.

That's simply not true: $4>2$ and $\frac{4}{2}=2\in\mathbb N$.

• I am to prove that the claim is false. Commented May 13, 2017 at 9:34
• @user2240414 Make that perfectly clear in the body of your question before it gets closed... Commented May 13, 2017 at 9:51