Implicit Function/Examples/x^2 + y^2 - 25 = 0/Proof

Proof
Solving for $y$, we obtain:
 * $y = \pm \sqrt {25 - x^2}$

As it stands, $(1)$ does not define a real function, because:
 * $\text{(a)}: \quad y$ is not defined for values of $x$ outside the range $-5 \le x \le 5$
 * $\text{(b)}: \quad$ For all $x$ within the range $-5 \le x \le 5$, there are two possible values of $y$ that can be taken.

This can be corrected by:
 * $\text{(a)}: \quad$ Specifying the domain to be the closed interval $\closedint {-5} 5$ (or a subset thereof)
 * $\text{(b)}: \quad$ Specifying which value of $y$ that is to be chosen, for example:


 * $\forall x \in \closedint {-5} 5: y = \sqrt {25 - x^2}$ (where $\sqrt {}$ in this context specifically means the positive square root)


 * $\forall x \in \closedint {-5} 5: y = -\sqrt {25 - x^2}$


 * $\forall x \in \closedint {-5} 5: y = \begin {cases} \sqrt {25 - x^2} & : x \le 0 \\ -\sqrt {25 - x^2} & : x > 0 \end {cases}$

Thus we have:
 * $y = \map f {x, \map g x}$

where:
 * $\map f {x, \map g x} = x^2 + \paren {\sqrt {25 - x^2} }^2 - 25 = 0$

thereby demonstrating that $y$ is an implicit function of $x$.