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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.)

Log in to reply

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

Log in to reply

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

Log in to reply

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. :)

Log in to reply

Log in to reply