Prove the $\frac {2r+5}{r+2}$ is always a better approximation of $\sqrt {5}$ than $r$.

Question

Prove the $\frac {2r+5}{r+2}$ is always a better approximation of $\sqrt {5}$ than $r$.

SOURCE : Inequalities (Page Number 4 ; Question Number 207)

I tried a lot of approaches, but without success.

I rewrote $\frac {2r+5}{r+2}$ as $2 + \frac {1}{r+2}$.

$\sqrt {5} \approx2.2360679775$

Equating $\frac {1}{r+2}$ and $0.2360679774$ , I get $r=2.23606797929$.

So, $\frac {2r+5}{r+2}$ is a still a better approximation than $r$.

How to proceed ?

Any hints/ideas/pointers ?

Answer 1

$$\frac{2r+5}{r+2} - \sqrt{5} = \frac{2r+5-\sqrt{5}r-2\sqrt{5}}{r+2} = \frac{(2-\sqrt{5})r - (2-\sqrt{5})\sqrt{5}}{r+2} = \frac{2-\sqrt{5}}{r+2}(r-\sqrt{5})$$

That means as long as $$\left|\frac{2-\sqrt{5}}{r+2}\right| < 1$$ holds, the proposition is true. If we assume $r > 0$ as reasonable for an approximation of a square root of a positve integer, then the proposition always holds.

Answer 2

Observe if $r<\sqrt{5}$, then there exists $\epsilon>0$ such that \begin{align} r+\epsilon<\sqrt{5} \ \ \Leftrightarrow& \ \ (r+\epsilon)^2 <5\\ \Leftrightarrow& \ \ r^2-5+2\epsilon r+ \epsilon^2=P(\epsilon)<0 \end{align} which is true if we sketch $P(\epsilon)$ as a function of $\epsilon$. In particular, we see that the positive root of $P(\epsilon)$ is \begin{align} \epsilon = \frac{-2r+\sqrt{20}}{2} = -r+\sqrt{5}= \frac{-r^2+5}{r+\sqrt{5}}>\frac{-r^2+5}{r+2}=\epsilon_0>0 \end{align} since $\sqrt{5}>2$. Hence \begin{align} r<r+\epsilon_0 = r+\frac{-r^2+5}{r+2} = \frac{r^2+2r-r^2+5}{r+2} = \frac{2r+5}{r+2}. \end{align}

Remark: The idea is to find $\epsilon_0>0$ rational such that $(r+\epsilon_0)^2<5$.

Answer 3

The function $$T:\quad z\mapsto T(z):={2z+5\over z+2}\qquad(z\in\bar{\mathbb C})$$ is a Moebius transformation with the two fixed points $\pm\sqrt{5}$. One computes $$T'\bigl(\sqrt{5}\bigr)={-1\over\bigl(\sqrt{5}+2\bigr)^2}=-0.05573\ .$$ It follows that the fixed point $\sqrt{5}$ is attracting, and more: For all initial points $z_0\ne-\sqrt{5}$ we have $\lim_{n\to\infty} z_n=\sqrt{5}$, whereby in the final stage the error is multiplied by about $0.055$ at each step.

Answer 4

Rather than doing manipulations with $\sqrt{5}$ it is better to use just the rationals. We can see that $$\left(\frac{2r+5}{r+2}\right)^{2}-5=\frac{4r^{2}+20r+25-5r^{2}-20r-20}{(r+2)^{2}}=\frac{5-r^{2}}{(r+2)^{2}}$$ Since $r>0$ it follows that $(2r+5)/(r+2)$ is a better approximation to $\sqrt{5}$ than $r$, but in a different direction.

The real benefit comes from iterating this procedure two times to get $$\dfrac{2\cdot\dfrac{2r+5}{r+2}+5}{\dfrac{2r+5}{r+2}+2}=\frac{9r +20}{4r+9}$$ so that $(9r+20)/(4r+9)$ is a better approximation to $\sqrt{5}$ and that too in same direction.