Square Root of Number Minus Square Root/Proof 1
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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
We are given that $a^2 - b > 0$.
Then:
- $a > \sqrt b$
and so $\ds \sqrt {a - \sqrt b}$ is defined on the real numbers.
Let $\ds \sqrt {a - \sqrt b} = \sqrt x - \sqrt y$ where $x, y$ are (strictly) positive real numbers.
Observe that:
- $\ds 0 < \sqrt {a - \sqrt b} = \sqrt x - \sqrt y \implies x > y$
Squaring both sides gives:
\(\ds a - \sqrt b\) | \(=\) | \(\ds \paren {\sqrt x - \sqrt y}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x + y - 2 \sqrt {x y}\) |
Set $x + y = a$ and $\sqrt b = 2 \sqrt {x y}$
From $\sqrt b = 2 \sqrt {x y}$ we get:
\(\ds \sqrt b\) | \(=\) | \(\ds 2 \sqrt {x y}\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds b\) | \(=\) | \(\ds 4 x y\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x y\) | \(=\) | \(\ds \frac b 4\) |
By Viète's Formulas, $x$ and $y$ are solutions to the quadratic equation:
- $z^2 - a z + \dfrac b 4 = 0$
From Solution to Quadratic Equation:
- $z_{1, 2} = \dfrac {a \pm \sqrt {a^2 - b} } 2$
where $a^2 - b > 0$ (which is a given)
Because we have that $x > y$:
\(\ds x = z_1\) | \(=\) | \(\ds \dfrac {a + \sqrt {a^2 - b} } 2\) | ||||||||||||
\(\ds y = z_2\) | \(=\) | \(\ds \dfrac {a - \sqrt {a^2 - b} } 2\) |
Subsituting into $\ds \sqrt {a - \sqrt b} = \sqrt x - \sqrt y$:
\(\ds \sqrt {a - \sqrt b}\) | \(=\) | \(\ds \sqrt x - \sqrt y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} - \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}\) |
$\blacksquare$