Question

Prove that the numbers $\sqrt{2},\sqrt{3},\sqrt{5}$ cannot be the terms of a single A.P. with non-zero common difference.

Answer 1

$T_p=\sqrt{2},T_q=\sqrt{3},T_r=\sqrt{5}$
$\therefore \sqrt{3}-\sqrt{2}=T_1-T_p=(q-p)d$,
$\sqrt{5}-\sqrt{3}=(r-p)d$
$\therefore \frac{\sqrt{3}-\sqrt{2}}{\sqrt{\left(5\right)}-\sqrt{\left(3\right)}}=\frac{q-p}{r-p}=k$, say ..$\left(1\right)$
As p, q, r are $+$ive integers so k is a rational number in $\left(1\right)$.
Squaring $\left(1\right)$, we get $5-2\sqrt{6}=k^2(8-2\sqrt{15})$
or $\sqrt{15}{k}^2-\sqrt{6}=(8k^2-5)/2=$s say .$\left(2\right)$
Here s is again a rational number. Squaring again
$\therefore 15k^4+6-2\sqrt{90}{k}^2=s^2$
or $15k^4+6-s^2=6\sqrt{10}{k}^2$
or $(15k^4-s^2+6)/6k^2=\sqrt{10}$ $\left(3\right)$
L.H.S. of $\left(3\right)$ is rational whereas R.H.S. $\sqrt{10}$ is irrational which is not possible. Hence $\sqrt{2},\sqrt{3},\sqrt{5}$ cannot be three terms of an A.P.