Going from $\sqrt A = \sqrt A$ to $\sqrt A= \sqrt{A\cdot(-1)^2}$ to $\sqrt A = -\sqrt A$

Question

Assume A is a positive real number. If $$\sqrt A = \sqrt A,$$ then $$ \begin{aligned} \sqrt A &= \sqrt{A\cdot(-1)^2}\\ \text{or},\ \ \ \sqrt A &= -\sqrt A. \end{aligned} $$ How does this work?

Answer 1

For real positive numbers, the squareroot is defined as $$\sqrt{a^2}=|a|$$ that is, always positive. If we defined it as $$\sqrt{a^2}=a$$ we would get a problem, namely $$\sqrt{1}=\sqrt{(-1)^2}\Rightarrow 1=-1$$ This problem arises because $f:\Bbb R\rightarrow \Bbb R$, $f(x)=x^2$ is not injective, that is, $x^2=y^2$ does not imply $x=y$.

Answer 2

What you are trying to do is that you are trying to conclude $\sqrt{A}=-\sqrt{A}$ from $A=A\cdot(-1)^2$. While one can simply note that squaring is non-injective as a map from $\mathbb{R}$ to $\mathbb{R}$, we can also give a simpler argument with less mathematical jargon involved. We can argue as follows.

$A=A\cdot(-1)^2$

$\implies (\sqrt{A})^2=(-\sqrt{A})^2$

$\implies (\sqrt{A})^2-(-\sqrt{A})^2=0$

$\implies (\sqrt{A}-\sqrt{A})(\sqrt{A}+\sqrt{A})=0$

$\implies 0(\sqrt{A}+\sqrt{A})=0$

which is a mathematical identity for real numbers. So there is no clash!