prove $ 1 < \varphi = \frac{1 + \sqrt{5}}{2} < 2$ Prove $ 1 < \varphi =  \frac{1 + \sqrt{5}}{2} < 2$
Just want to see if my "reasoning is sound.
1) Showing $1 < \frac{1 + \sqrt{5}}{2}$
Consider $\frac{1}{2}$:
$$\frac{1}{2} < \frac{1}{2} + \frac{1}{2} < \frac{1}{2} + \frac{\sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2} $$
2) Showing $\frac{1 + \sqrt{5}}{2} < 2$
We know $$\sqrt{5} < 3 \\ \Rightarrow 1 + \sqrt{5} < 4 \\ \Rightarrow \frac{1 + \sqrt{5}}{2} < 2$$
Therefore: $ 1 < \varphi =  \frac{1 + \sqrt{5}}{2} < 2$

What about:
$4 < 5 < 9$
$2< \sqrt 5 < 3$
$3 < 1 + \sqrt{5} < 4$
$\frac 32 < \frac {1+\sqrt{5}}2 < 2$.
$1 < \frac 32 < \frac{1+\sqrt{5}} 2 < 2$.
.....
I suppose maybe a more direct way of solving would be
$n < \frac {1+\sqrt 5}2 < n+1 \iff$
$2n < 1+\sqrt 5 < 2n + 2\iff$
$2n-1 < \sqrt 5 < 2n+1 \iff$
$(2n-1)^2 < 5 < (2n+1)^2$.
So our task is to:  Find to consecutive odd numbers so that $5$ is between the squares.  Those odd numbers are $1$ and $1^2 < 5$ and $3$ and $3^2 > 5$ and so $2n -1 = 1$ and $2n+1 = 3$
And so $n =1$.
