Determine if R is a function from A to R

Question

For the given subset Ai of Real numbers and the relation Ri (1 ≤ i ≤ 3) from Ai to Real numbers, determine whether Ri is a function from Ai to Real numbers.
(a) A1 = Real numbers, R1 = {(x, y) : x ∈ A1, y = 4x − 3}
(b) A2 = [0,∞), R2 = {(x, y) : x ∈ A2, (y + 2)^2 = x}
(c) A3 = Real numbers, R3 = {(x, y) : x ∈ A3, (x + y)^2 = 4}

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I know that (a) is a function because y=4x-3 is bijective for all x in the real numbers. Note that (y+2)^2=x == y=sqrt(x)-2, so (b) is also bijective when x is from 0 to infinity. For (c), note that (x+y)^2 = 4 == y=2-x so this is bijective for all x in the real numbers.

Am I doing this correctly?

Answer 1

$(a)$ is a function and bijection, since if $x_1 = x_2$ then $4x_1-3 = 4x_2-3$, and if $4x_1 - 3 = 4x_2 - 3$ then $x_1 = x_2$.

$(b)$ is not a bijection or function. If $x = 9$, then $y$ can be $1$ or $-5$, i.e. both $(9, 1)$ and $(9,-5)$ in $R_2$.

$(c)$ is not a bijection or function. If $x = 0$, then $y$ can be $2$ or $-2$, i.e. both $(0, 2)$ and $(0, -2)$ in $R_3$

Answer 2

First of all, there exist functions which are not bijective. Moreover, I assume that you are talking about functions with a single value (there exist also multivalued functions, e.g. $f(x)= \pm \sqrt{x}$, with $f : \mathbb{R}^{+} _{0} \rightarrow \mathbb{R}$). Your third example is not a single-valued function, because you can write it as: $ f(x)= - x \pm 2 $ (not only $y=2-x$). This is an example of multivalued function (with two values).