0
$\begingroup$

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 $2^2 = 4$?

  • 0
    No. $\{ x \in \mathbb{N} : x+x=x^2 \}$ is a *set* of numbers, not a single number.2012-05-05
  • 0
    By the way, you should clarify if $0$ is in $\mathbb{N}$ for you.2012-05-05
  • 2
    You also have to show that $2$ is the _only_ element of $\mathbb N$ which satisfies these conditions.2012-05-05

1 Answers 1

1

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.

  • 0
    Oh yes i should have shown 2 is the only element of N: thanks for displaying that so clearly2012-05-05