Prove that $h(x,y)=xy$ is surjective

Question

Prove that the function $h(x,y)=xy$, where $h: \mathbb{R^2} \to \mathbb{R}$ is surjective.

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In all my examples, it has only been in one dimension and I am not sure how to go about proving this.

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In one dimension, I know if a function is surjective then for $h: X \to Y, \forall b \in B$ $\exists$ $a\in A$ such that $h(a)=b$.

But in two dimension, I suppose I will have to prove the following?

$$h: \mathbb R \times\mathbb R \to \mathbb R, \forall\ x,y \in \mathbb{R^2}\ \exists \ x,y\in \mathbb R$$.

Answer 1

You need to prove that for all $u\in\mathbb{R}$ there exists $v\in\mathbb{R}^2$, where $v=(v_1,v_2)$, such that $f(v)=u$. Take $v_1=u$ and $v_2=1$. Then $f(v)=f(v_1,v_2)=v_1\cdot v_2=u\cdot 1=u$. Thus $f$ is onto.

Answer 2

Choose $x\in\mathbb{R}$. You have $h(x,1) = x.$

Answer 3

For surjectivity in two dimension you nee to show: $$h: \mathbb R \times\mathbb R \to \mathbb R, \forall\ z \in \mathbb{R}\ \exists \ x,y\in \mathbb R\text { such that $xy=z$}$$

In order to prove that you may simply let $x=z$ and $y=1$ to obtain $xy=z$

Answer 4

A function $f:A\rightarrow B$ is surjective if for every $b\in B$, there exists an $a\in A$ such that $f(a) = b$.

According to formalism, functions "of multiple arguments" are just functions whose input is a tuple. In your case, $h:\mathbb{R}^2\rightarrow \mathbb{R}$ sends pairs of numbers $\langle x,y\rangle$ to their product $z \equiv xy$.

To prove that $h$ is surjective, you must show:

For every $b\in \mathbb{R}$, there's an $a\in \mathbb{R}^2$ such that $h(a) = b$.

or in other words,

For every $z\in \mathbb{R}$, there's an $\langle x,y\rangle$ in $\mathbb{R}^2$ such that $h(\langle x,y\rangle) = z$.

But $h(x,y) = xy$, so all you have to show is:

For every $z\in \mathbb{R}$, there's an $\langle x,y\rangle$ in $\mathbb{R}^2$ such that $xy = z$.

which you can do readily: given $z\in \mathbb{R}$, you can use the pair $\langle z, 1\rangle \in \mathbb{R}^2$ as an example.