Explanation of uniqueness of square root

Question

Let $c$ be a positive number. Then there is a unique positive number whose square is $c$. That is, $x^2=c$

Start:

• Suppose $a$ and $b$ are numbers whose square is $c$.
• then $a^2=c$ and $b^2=c$
• $c-c=0 = a^2-b^2 = (a-b)(a+b)$

We know $(a+b) > 0$ because $a$ and $b$ are positive numbers. For some reason my textbook concludes $a=b$ and thus the positive number whose square is $c$ is unique. I'm not seeing this.

Answer 1

Since $a+b>0$, it is particularly not zero. The only way for a product of two reals to give zero is if at least one of them is zero. Since $a+b\neq 0$, but $0 = (a-b)(a+b)$, it must be the case that $a-b=0$, i.e. $a=b$.

Answer 2

Since $a+b \neq 0$ we can multiply $(a-b)(a+b) =0$ across by ${1 \over a+b}$ to get $a-b=0$, that is, $a=b$.