prove $ 1 < \varphi = \frac{1 + \sqrt{5}}{2} < 2$

Question

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$

Answer 1

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 $n =1$.

Answer 2

Another way

As $\varphi>0,$

$\varphi=1+\dfrac1{\varphi}>1$

$\implies1+\dfrac1{\varphi}<1+1$