Square root of a sum

Question

Can anyone explain to me the reasoning behind this proof? Specifically up until the system of equations. So why $\sqrt{a + b} = \sqrt{c} + \sqrt{d}$? And $a = c + d$ and $b = 2\sqrt{cd}$?

Answer 1

The goal of this is to express $\sqrt {a+b}$ as the sum of two square roots, that is, as $\sqrt c+\sqrt d$. But\begin{align}\sqrt{a+b}=\sqrt c+\sqrt d&\iff a+b=\left(\sqrt c+\sqrt d\right)^2\\&\iff a+b=c+d+2\sqrt{cd}\end{align}and therefore it will be enough to find numbers $c$ and $d$ such that $c+d=a$ and that $2\sqrt{cd}=b$.

Answer 2

It starts with: Let $\sqrt{a + b}$ be expressed as $\sqrt{c} + \sqrt{d}$ and of course you can do that for every number. As an example, lets say $a = b =2$ then we would get $c=d=1$. You can always find such numbers, and the proof shows how to find those numbers. The next part is probably a guess. We can try to set $a=c+d$ and $b = 2\sqrt{cd}$ and see if we can solve this system. Luckily, it was a good guess. You could for instance, have also tried $a = c$ and $b=d+2\sqrt{cd}$, but it would have been hard to solve for d, right?

Answer 3

The proof there is invalid - it doesn't really prove that:

$$\sqrt{a+b}=\sqrt{\frac{a}{2}+\frac{\sqrt{a^2-b^2}}{2}}+\sqrt{\frac{a}{2}-\frac{\sqrt{a^2-b^2}}{2}}$$

for $a,b\in\mathbb R, a\ge b$. What it does prove is that, if $\sqrt{a+b}$ can be expressed as a sum of two square roots $\sqrt{c}$ and $\sqrt{d}$ then $c$ and $d$ can be chosen to be the two numbers $\frac{a}{2}+\frac{\sqrt{a^2-b^2}}{2}$ and $\frac{a}{2}-\frac{\sqrt{a^2-b^2}}{2}$.

This is really more of a motivation how you write the statement in the first place (i.e. what should the right-hand side look like), rather than a proof that this statement is valid.

In fact, the statement itself is invalid - it is valid only for $a\ge b\ge 0$. Otherwise, note that the right side has $b$ only as a square, so the right side does not change when you replace $b$ with $-b$. The left side obviously changes, and nobody can convince me that e.g. $\sqrt{5+3}$ is the same as $\sqrt{5-3}$.

To actually prove the statement, with the additional condition $b\ge 0$, and knowing (from the motivation given in the alleged proof) what the right-hand side should look like, you need to reason as follows:

• All values under the root sign are non-negative (given this additional condition $b\ge 0$), so you can get an equivalent statement by squaring the left- and right-hand side.
• Once squared, the left side is $a+b$ and the right side (after some calculation) ends up being $a+|b|$, and so they are equal assuming $b\ge 0$.
• Thanks to the equivalence established in the first step, then, the original two expressions are equal.