# Sets and element of.

I am doing the following question, which is quite basic, but for some reason I can't work it out.

$\{x \in \mathbb N : x + x= x ^2\} \cup \{x \in \mathbb N: 3 x = x^2\}$

Am I right in saying that for $\{x ∈ \mathbb N : x + x= x ^2\}$ the answer is $2$ because $2+2 = 4$ and $2^2 = 4$?

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No. $\{ x \in \mathbb{N} : x+x=x^2 \}$ is a set of numbers, not a single number. –  Chris Eagle May 5 '12 at 14:41
By the way, you should clarify if $0$ is in $\mathbb{N}$ for you. –  Chris Eagle May 5 '12 at 14:41
You also have to show that $2$ is the only element of $\mathbb N$ which satisfies these conditions. –  Dylan Moreland May 5 '12 at 14:43

If $2x = x^2$ then we have that $2x - x^2 = x(2 - x) = 0$ Hence $x = 0$ or $x =2$. But $0$ is not a natural number so that the only natural number that satisfies this is $x = 2$. For the other set on the right, as
$$3x = x^2$$
therefore you have that $x(3 - x) = 0$. By similar reasoning only $x=3$ can satisfy this. Hence
$$\{x \in \mathbb N : x + x= x ^2\} \cup \{x \in \mathbb N: 3 x = x^2\} = \{2,3\}.$$
Now I address your question. You are not right because all you have shown is that $2 \in \{x \in \mathbb N : x + x= x ^2\}$ (tell me why all you have shown is this). So you need to show that in fact if you have any element $x$ in $\{x \in \mathbb N : x + x= x ^2\}$, you have that $x \in \{2\}$ which is what I showed above.