Square Root of Number Plus Square Root/Proof 2

Theorem

Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.

Then:

$\ds \sqrt {a + \sqrt b} = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}$

Proof

\(\ds \paren {\sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2} }^2\)

\(=\)

\(\ds \dfrac {a + \sqrt {a^2 - b} } 2 + \dfrac {a - \sqrt {a^2 - b} } 2 + 2 \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}\)

multiplying out

\(\ds \)

\(=\)

\(\ds a + \sqrt {a + \sqrt {a^2 - b} } \sqrt {a - \sqrt {a^2 - b} }\)

simplifying

\(\ds \)

\(=\)

\(\ds a + \sqrt {a^2 - \paren {a^2 - b} }\)

Difference of Two Squares

\(\ds \)

\(=\)

\(\ds a + \sqrt b\)

simplifying

\(\ds \leadsto \ \ \)

\(\ds \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}\)

\(=\)

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

taking square root of both sides

This needs considerable tedious hard slog to complete it. In particular: Report on the matter of the signs and magnitudes of $a$ and $b$ according to the constraints given To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

$\blacksquare$