Show that the numbers \(\sqrt 2, \sqrt 3, \sqrt 5\) can never be the terms (not necessarily consecutive) of a single Arithmetic Progression.

**Generalized Form:** Show that the numbers \(\sqrt p, \sqrt q, \sqrt r\) can never be the terms (not necessarily consecutive) of a single Arithmetic Progression, where \(p,q,r\) are three distinct prime numbers.

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TopNewestSuppose \(a_{i} = \sqrt{2}, a_{j} = \sqrt{3}, a_{k} =\sqrt{5}\) are terms in an arithmetic sequence \(\{a_{n}\}\) with common difference \(d,\) where \(i,j,k\) are not necessarily consecutive. Then

\(\sqrt{2} = a_{1} + (i - 1)d, \sqrt{3} = a_{1} + (j - 1)d\) and \(a_{k} = a_{1} + (k - 1)d.\)

We can then subtract these equations from one another successively to find that

\(\sqrt{3} - \sqrt{2} = (j - i)d\) and \(\sqrt{5} - \sqrt{3} = (k - j)d\)

\(\Longrightarrow \dfrac{k - j}{j - i} = \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{3} - \sqrt{2}} = \sqrt{15} + \sqrt{10} - \sqrt{6} - 3.\)

But then the LHS of this last equation is rational and the RHS irrational, and so our initial assumption that \(\sqrt{2}, \sqrt{3}, \sqrt{5}\) are terms of the same AP is false.

For the generalized form with primes \(p,q,r\) the RHS of the last equation would become

\(\dfrac{\sqrt{r} - \sqrt{q}}{\sqrt{q} - \sqrt{p}} = \dfrac{1}{q - p}(\sqrt{rq} + \sqrt{rp} - \sqrt{pq} - q),\)

which is, as before, irrational, (since none of \(rq, rp, pq\) are perfect squares, nor do the roots cancel).

(Note that they cannot be terms of the same geometric progression either.)

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Greetings, Brian! Just to play the devil's advocate: How do you know that \(\sqrt{15}+\sqrt{10}-\sqrt{6}\) isn't rational?

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The most rational reasoning is to use the standard method of rationalising and finding a contradiction. This should help

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Greetings, Otto! Yes, I was afraid that someone was going to ask that. I was just about to write up a proof, but I notice now that the approach I was going to use is the same as in Sualeh's link, so I think that reference will suffice. :)

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