Proof for $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$ can never be the terms of a single arithmetic progression with non zero common difference.

Question

Proof for $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$ can never be the terms of a single arithmetic progression with non zero common difference.

Here is what I have tried.

Let:

$$ax +c=\sqrt{2}$$

$$bx +c=\sqrt{3}$$

$$dx +c=\sqrt{5}$$

$$\frac{\sqrt2-c}{a}=\frac{\sqrt3-c}{b}=\frac{\sqrt5-c}{d}$$

What property can I use next?

Answer 1

We have \begin{eqnarray*} a= \sqrt{2} \\ a+nd= \sqrt{3} \\ a+md= \sqrt{5} \\ \end{eqnarray*} multiply the second equation by $m$, the third equation by $n$ and subtract.

We have \begin{eqnarray*} (m-n)\sqrt{2} = m\sqrt{3}-n \sqrt{5}. \\ \end{eqnarray*} Now square this equation & conclude that $\sqrt{15}$ cannot be irrational.