Condition for Denesting of Square Root
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Theorem
Let $a, b \in \Q_{\ge 0}$.
Suppose $\sqrt b \notin \Q$.
Then:
- $\exists p, q \in \Q: \sqrt {a + \sqrt b} = \sqrt p + \sqrt q$
- $\exists n \in \Q: a^2 - b = n^2$.
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
\(\ds \sqrt {a + \sqrt b}\) | \(=\) | \(\ds \sqrt p + \sqrt q\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {p + q + \sqrt {4 p q} }\) | Sum of Square Roots as Square Root of Sum | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(=\) | \(\ds p + q\) | by Lemma | ||||||||||
\(\ds b\) | \(=\) | \(\ds 4 p q\) | by Lemma | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0\) | \(=\) | \(\ds b - 4 p q\) | |||||||||||
\(\ds \) | \(=\) | \(\ds b - 4 p \paren {a - p}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b - 4 a p + 4 p^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds p\) | \(=\) | \(\ds \frac {4 a \pm \sqrt {16 a^2 - 16 b} } 8\) | Quadratic Formula | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac a 2 \pm \frac {\sqrt {a^2 - b} } 2\) |
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:
\(\ds \sqrt {a + \sqrt b}\) | \(=\) | \(\ds \sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b} } 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b} } 2}\) | Square Root of Sum as Sum of Square Roots | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\dfrac a 2 + \dfrac {\sqrt {n^2} } 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {n^2} } 2}\) | substituting $n^2 = a^2 - b$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\dfrac a 2 + \dfrac n 2} + \sqrt {\dfrac a 2 - \dfrac n 2}\) |
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.
$\blacksquare$