Question

Where is `sqrt(15)` on the real number line?

Answer 1

It is a point a little to the left of `4`...

Explanation:

The square root of `15` is a little less than the square root of `16`, which is `4` (since `4^2 = 16`).

How much less?

If `a` is an approximation to the square root of `n`, then a better approximation is:

`(a^2+n)/(2a)`

So putting `n=15` and `a=4`, we find a better approximation:

`(4^2+15)/(2*4) = (16+15)/8 = 31/8 = 3.875`

If you want a better approximation you can repeat this to find the approximation:

`((31/8)^2+15)/(2*(31/8)) = 1921/496 ~~ 3.872984`

So we can draw this as a point on the real number line something like this:

graph{(x-sqrt(15))^2+y^2-0.003 = 0 [-3.686, 6.314, -2.54, 2.46]}

Here I have used the graphing tool, so we have a `y` axis too, but just look at the `x` axis and the marked point.