Square Root of Number Plus Square Root
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 1
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.
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 $\sqrt b = 2 \sqrt {x y}$
This article, or a section of it, needs explaining. In particular: How do you know that the $a$ and $b$ which are $x + y$ and $2 \sqrt {x y}$ are the same $a$ and $b$ that you started with? $x$ and $y$ are free to choose. They were introduced by hand, and can be set to any value provided that they satisfy the constraint above. This is similar to the proof of Cardano's Formula. In that case it needs to be explained. As it stands, it looks as though $x$ and $y$ are pulled out of thin air, with no actual indication that having picked them, they bear the relations given to $a$ and $b$ as presented. It's incredibly confusing for beginners, and others whose abilities and understanding are limited, like me.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. 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 {{Explain}} from the code. |
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)
This article, or a section of it, needs explaining. In particular: the notation $z_{1, 2}$ You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. 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 {{Explain}} from the code. |
\(\ds x = z_1\) | \(=\) | \(\ds \dfrac {a + \sqrt {a^2 - b} } 2\) | ||||||||||||
\(\ds y = z_2\) | \(=\) | \(\ds \dfrac {a - \sqrt {a^2 - b} } 2\) |
Subsituting into $\ds \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$:
\(\ds \sqrt {a + \sqrt b}\) | \(=\) | \(\ds \sqrt x + \sqrt y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}\) |
$\blacksquare$
Proof 2
\(\ds \paren {\sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2} }^2\) | \(=\) | \(\ds \dfrac {a + \sqrt {a^2 - b} } 2 + \dfrac {a - \sqrt {a^2 - b} } 2 + 2 \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}\) | 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. |
$\blacksquare$