Is $\mathbb{N}\rightarrow \mathbb{N\times N},f(x)=(x,x)$ onto?

Question

I am not sure how to tell. Say $b\in N\times N$ this means the codomain is all the different combinations of the natural numbers.

But the domain is only the natural numbers multiplied, like $3\times 7$

Oh I see there cannot be a n for (3,14) because n is a single natural number that when multiplied by itself will make one natural number. Not a pair is that the reason. — Fernando Martinez, Jul 12 '14 at 21:29
@FernandoMartinez There is no multiplication in this scenario. The $\times$ in $\mathbb{N} \times \mathbb{N}$ denotes the Cartesian product of the two sets. — Austin Mohr, Jul 12 '14 at 21:31
@FernandoMartinez In André's comment, $(3,14)$ means the pair: first element $3$, second element $14$. — egreg, Jul 12 '14 at 21:32
I see so it means the pair cannot be in the range. Because N in the domain is a single natural and not two different naturals. — Fernando Martinez, Jul 12 '14 at 21:35
What is $\left(x,x\right)$? is it an open interval? The notation is bad.. — Elimination, Jul 12 '14 at 22:14
The notation is not bad, because (,) is standard notation for an element of a Cartesian product, which we know it has to be by context. — DavidButlerUofA, Jul 13 '14 at 02:50

score 3 · Accepted Answer · answered Jul 12 '14 at 21:51

Firstly, $\mathbb{N} \times \mathbb{N}$ denotes the set of pairs of natural numbers. That is, $$ \mathbb{N} \times \mathbb{N} = \{ (a,b) \mid a, b \in \mathbb{N} \}. $$ It does not denote the set of products of numbers in $\mathbb{N}$.

Secondly, the definition of "onto" is that all elements of the codomain are in the range. That is, for any $(a,b) \in \mathbb{N}\times\mathbb{N}$ it is possible for $f(x) = (a,b)$ for some $x$.

If $a \neq b$, this is not possible, so it's not onto.

Is $\mathbb{N}\rightarrow \mathbb{N\times N},f(x)=(x,x)$ onto?

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