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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.

Note by Sandeep Bhardwaj
2 years, 9 months ago

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Suppose $$a_{i} = \sqrt{2}, a_{j} = \sqrt{3}, a_{k} =\sqrt{5}$$ 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.)

- 2 years, 9 months ago

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

- 2 years, 9 months ago

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

- 2 years, 9 months ago

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. :)

- 2 years, 9 months ago

Thank you both!

- 2 years, 9 months ago