I'm attempting to prove that the dot product function is surjective (onto). I am given that f: $f((x,y),(u,v)) = xu + yv$, which is simply the dot product function. I believe that this function is surjective, as intuitively it seems any number in $\mathbb R$ can be made from the sum of two products. However, I am having trouble writing the actual proof.
Here's what I have so far:
Let $a \in \mathbb R$. If f is onto, then for all $a \in f(x)$, $a = xu + yv$.
After that, I'm stuck. I tried to solve the equation $a = xu + yv$ for x, u, y, and v then substitute it back into the function to show that $f((x,u),(y,v)) = a$, but it just got very messy and didn't seem to lead anywhere.
Any help would be appreciated! (Edit: had some notation mistakes, fixed now. Thanks to SK19 for pointing it out)
Proposition. The function $f:\mathbf{R}^2\times\mathbf{R}^2\to \mathbf{R}$ defined by $f((x,y),(u,v)) = xu+yv$ is surjective.
Proof. Let $\alpha\in\mathbf{R}$ then by appealing to the above definition of $f$ we may observe that $$f\left((\frac{\sqrt{\alpha}}{\sqrt{2}},\frac{\sqrt{\alpha}}{\sqrt{2}}),(\frac{\sqrt{\alpha}}{\sqrt{2}},\frac{\sqrt{\alpha}}{\sqrt{2}})\right) = \frac{\sqrt{\alpha}}{\sqrt{2}}\cdot\frac{\sqrt{\alpha}}{\sqrt{2}}+\frac{\sqrt{\alpha}}{\sqrt{2}}\cdot\frac{\sqrt{\alpha}}{\sqrt{2}} = \alpha$$ consequently since the $\alpha$ in question was an arbitrary real number it follows that $f$ is surjective.