Condition for Denesting of Square Root/Lemma

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

\(\ds \sqrt {a + \sqrt b}\)

\(=\)

\(\ds \sqrt {c + \sqrt d}\)

\(\ds \leadsto \ \ \)

\(\ds a + \sqrt b\)

\(=\)

\(\ds c + \sqrt d\)

\(\ds \leadsto \ \ \)

\(\ds a - c\)

\(=\)

\(\ds \sqrt d - \sqrt b\)

\(\ds \leadsto \ \ \)

\(\ds \sqrt d - \sqrt b\)

\(\in\)

\(\ds \Q\)

Rational Subtraction is Closed

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

Then:

\(\ds \sqrt d - \sqrt b\)

\(\ne\)

\(\ds 0\)

\(\ds d - b\)

\(=\)

\(\ds \paren {\sqrt d - \sqrt b} \paren {\sqrt d + \sqrt b}\)

Difference of Two Squares

\(\ds \leadsto \ \ \)

\(\ds \sqrt d + \sqrt b\)

\(=\)

\(\ds \dfrac {d - b} {\sqrt d - \sqrt b} \in \Q\)

Rational Division is Closed

\(\ds \leadsto \ \ \)

\(\ds 2 \sqrt b\)

\(=\)

\(\ds \paren {\sqrt d + \sqrt b} - \paren {\sqrt d - \sqrt b} \in \Q\)

Rational Subtraction is Closed

\(\ds \leadsto \ \ \)

\(\ds \sqrt b\)

\(\in\)

\(\ds \Q\)

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 $a = c$.

$\blacksquare$