Functions & Relations I have to determine if the given relations are functions, and if they are prove if it is injective or not. Can someone please help me understand them. Here $x$ , $y$, and $z$ belong to real numbers.


*

*$(x,y)\,R\,z \Longleftrightarrow (z - x)^2 = -x^2y^2$

*$(x,y)\,R\,z \Longleftrightarrow (z-y)^2 = (x+y)^2$


For the second one,  when we solve it:
$$z-y = x+y$$
$$z = x + 2y$$
So, $(0,0)\,R\,0$, $(1,1)\,R\,3$, so I guess it is a function.
$D_m =\{(x,y) \mid x,y\in\mathbb{R}\}$
And can you tell me how could I prove it is 1-1 or no?
Thank you!
 A: Question 1. Since the $(x-z)^2\ge0$ and $-x^2y^2\le0$, the equality is only possible when both are equal to $0$. This implies two things:


*

*From $-x^2y^2=0$, we deduce that either $x=0$ or $y=0$, i.e. all pairs $(x,y)$ that appear on the left side of this relation have to satisfy this condition.

*From $(z-x)^2=0$ we deduce that $z=x$.


So is it a function, i.e. can we view this as a function $z=f(x,y)$ whenever $(x,y)\,R\,z$? It's tempting to say yes, because of item 2 above, which essentially gives us the formula for this function: $z=f(x,y)=x$. For each input $(x,y)$ there's precisely one output $z=x$, so it sounds like a function. But…


*

*If we need this hypothetical function to be defined on all pairs $\{(x,y) \mid x,y\in\mathbb{R}\}$, then it's not a function on this domain, because according to item 1 above, only some special pairs $(x,y)$ satisfy the given relation $R$.

*But if we're allowed to restrict the domain, then this is a function on the domain $D=\{(x,y) \mid x,y\in\mathbb{R},xy=0\}$ (which is a shorthand for saying "either $x=0$ or $y=0$").


In the latter case, if we accept this interpretation, then we have a function, but it's not injective. Quick counterexample: $(0,1)\,R\,0$ and $(0,2)\,R\,0$, which in function notation says that $f(0,1)=f(0,2)=0$.
Question 2. From the given equation we deduce that $z-y=\pm(x+y)$, so either $z=x+2y$ or $z=-x$. This immediately shows that this is not a function since for almost any input $(x,y)$ we would have two different outputs $z$. For example, both $(1,1)\,R\,3$ and $(1,1)\,R\,(-1)$.
