We show that there is only one non-negative root of $a$. Suppose $b$ and $c$ are both roots. Then we have $b^2=a=c^2$, hence $b^2-c^2=0$.

From this follows that $(b-c)\cdot(b+c)=0$, from which again follows that $bc=0$ or $b+c=0$. Hence $b=c$ or $b=-c$. But if they are both non-negative, $b=c$ should be true. This proves that there is only one root of $a$ is.

If $a$ is a negative real number, then there is no real number $b$ with $b^2 = a$. After all, if $b$ would exist, then $b^2$ would have to be a non-negative number (because squares are never negative). But, since $a=b^2$, this is in contradiction with the assumption that $a$ is negative.

The first equality of the first line follows directly from the definition: if $c=\sqrt{a}$, then the definition says $c$ is non-negative and $c^2=a$. If we fill in $\sqrt{a}$ for $c$ once again, we find $\left(\sqrt{a}\right)^2=a$.

The second equality of the first line follows from the definition of $d= \sqrt{a^2}$. After all the definition says $d$ is non-negative and satisfies $d^2=a^2$. But from this follows $d=a$ or $d=-a$; because $d$ and $a$ neither are negative, it must mean that $d=a$, which is proven by $a=\sqrt{a^2}$.

Finally, if $e=\sqrt{a}\cdot\sqrt{b}$, then $e^2 = \left(\sqrt{a}\cdot\sqrt{b}\right)^2 = \left(\sqrt{a}\right)^2\cdot \left(\sqrt{b}\right)^2= a\cdot b$. Hence we see that $e^2=a\cdot b$. Because $e$ is non-negative, $e=\sqrt{a\cdot b}$. Accordingly $\sqrt{a}\cdot\sqrt{b}=\sqrt{a\cdot b}$. This proves the equality in the second line.

Proof: use the notation of $a$ as a product of primes. If $p$ is prime that divides $a$, then the question is what the highest power of $p$ is that divides $a$. If it $p$ itself, the prime $p$ corresponds with $c$. If it is an even exponent, for example, $p^6$, then the root thereof is $p^3$; we add that factor to $b$. If it is an odd exponent, eg $p^5$, we add $p^2$ to $b$, and $p$ to $c$. So we get integers $b$ and $c$, in such a way that no factor of $c$ is a square. From $a=b^2c$ and the rules of calculation rules stated above it follows that $\sqrt{a} = \sqrt{b^2}\cdot\sqrt{c} = b\sqrt{c}$.