Condition for Denesting of Square Root/Lemma

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Theorem

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

Suppose $\sqrt b \notin \Q$.

Then:

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


Proof

\(\displaystyle \sqrt {a + \sqrt b}\) \(=\) \(\displaystyle \sqrt {c + \sqrt d}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a + \sqrt b\) \(=\) \(\displaystyle c + \sqrt d\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a - c\) \(=\) \(\displaystyle \sqrt d - \sqrt b\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sqrt d - \sqrt b\) \(\in\) \(\displaystyle \Q\) Rational Subtraction is Closed


Aiming for a contradiction, suppose $b \ne d$.

Then:

\(\displaystyle \sqrt d - \sqrt b\) \(\ne\) \(\displaystyle 0\)
\(\displaystyle d - b\) \(=\) \(\displaystyle \paren {\sqrt d - \sqrt b} \paren {\sqrt d + \sqrt b}\) Difference of Two Squares
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sqrt d + \sqrt b\) \(=\) \(\displaystyle \dfrac {d - b} {\sqrt d - \sqrt b} \in \Q\) Rational Division is Closed
\(\displaystyle \leadsto \ \ \) \(\displaystyle 2 \sqrt b\) \(=\) \(\displaystyle \paren {\sqrt d + \sqrt b} - \paren {\sqrt d - \sqrt b} \in \Q\) Rational Subtraction is Closed
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sqrt b\) \(\in\) \(\displaystyle \Q\) Rational Division is Closed

But this contradicts our assertion that $\sqrt b \notin \Q$.

Hence our supposition that $b \ne d$ must be false.

Therefore we must have $b = d$.


Consequently:

$a - c = \sqrt d - \sqrt b = 0$

This implies $a = c$.

$\blacksquare$