Condition for Denesting of Square Root

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Theorem

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

Suppose $\sqrt b \notin \Q$.


Then:

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

if and only if:

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


Proof

Lemma

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

Suppose $\sqrt b \notin \Q$.

Then:

$\sqrt {a + \sqrt b} = \sqrt {c + \sqrt d} \implies a = c, b = d$

$\Box$


Necessary condition

\(\displaystyle \sqrt {a + \sqrt b}\) \(=\) \(\displaystyle \sqrt p + \sqrt q\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {p + q + \sqrt {4 p q} }\) Sum of Square Roots as Square Root of Sum
\(\displaystyle \leadsto \ \ \) \(\displaystyle a\) \(=\) \(\displaystyle p + q\) by Lemma
\(\displaystyle b\) \(=\) \(\displaystyle 4 p q\) by Lemma
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0\) \(=\) \(\displaystyle b - 4 p q\)
\(\displaystyle \) \(=\) \(\displaystyle b - 4 p \paren {a - p}\)
\(\displaystyle \) \(=\) \(\displaystyle b - 4 a p + 4 p^2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle p\) \(=\) \(\displaystyle \frac {4 a \pm \sqrt {16 a^2 - 16 b} } 8\) Quadratic Formula
\(\displaystyle \) \(=\) \(\displaystyle \frac a 2 \pm \frac {\sqrt {a^2 - b} } 2\)

This shows that, in order for $p \in \Q$, we must have $\sqrt {a^2 - b} \in \Q$ as well.

The result follows from $\paren {\sqrt {a^2 - b} }^2 = a^2 - b$.

$\Box$


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.

$\blacksquare$