User:Ihri/Sandbox

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

Then:

Proof
We are given that $a>0$ and $b>0$.

Then:
 * $a + \sqrt b > 0$

and so $\ds \sqrt {a + \sqrt b}$ is defined on the real numbers.

Consider the quadratic equation:


 * $z^2 - a z + \dfrac b 4 = 0$

with discriminant $(- a) ^ 2 - 4 \dfrac b 4 = a ^ 2 - b$ which is a (strictly) positive real number (a given).

Let $x, y$ be the solutions of the above equation.

From Solution to Quadratic Equation, :

with $x > y$. Note $x > 0$ because $ a > 0$.

By Sum of Roots of Quadratic Equation, $x$ and $y$ satisfy:


 * $x + y = a$

By Product of Roots of Quadratic Equation and the conditions $x > 0, b >0$ we get:


 * $x y = \dfrac b 4 > 0 \implies y > 0$

which implies $\sqrt x$ and $\sqrt y$ are defined on the real numbers.

Furthermore, from $x y = \dfrac b 4$ we get:

Finally, substituting into $\sqrt {a + \sqrt b}$: