# Condition for Denesting of Square Root

Jump to navigation
Jump to search

## Theorem

Let $a, b \in \Q_{\ge 0}$

Then:

- $\exists p, q \in \Q: \sqrt {a + \sqrt b} = \sqrt p + \sqrt q$

iff:

- $\exists n \in \N: a^2 - b = n^2$.

## Proof

The proof is split into the necessary condition:

If

- $\exists p, q \in \Q: \sqrt {a + \sqrt b} = \sqrt p + \sqrt q$

then

- $\exists n \in \N: a^2 - b = n^2$

and the sufficient condition:

If

- $\exists n \in \N: a^2 - b = n^2$

then

- $\exists p, q \in \Q: \sqrt {a + \sqrt b} = \sqrt p + \sqrt q$

### Necessary condition

### Sufficient condition

Let $n^2 = a^2 - b$.

As $a^2 = b + n^2$ it follows that:

- $a \ge \sqrt b$

Then:

\(\displaystyle \sqrt {a + \sqrt b}\) | \(=\) | \(\displaystyle \sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b} } 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b} } 2}\) | Square Root of Sum as Sum of Square Roots | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sqrt {\dfrac a 2 + \dfrac {\sqrt {n^2} } 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {n^2} } 2}\) | substituting $n^2 = a^2 - b$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sqrt {\dfrac a 2 + \dfrac n 2} + \sqrt {\dfrac a 2 - \dfrac n 2}\) |

By Rational Addition is Closed and Rational Subtraction is Closed:

- $\dfrac a 2 + \dfrac n 2$ and $\dfrac a 2 - \dfrac n 2$

are rational.

$\blacksquare$