Intuition can be one way!

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

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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.)

Brian Charlesworth - 2 years, 9 months ago

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?

Otto Bretscher - 2 years, 9 months ago

Log in to reply

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

Sualeh Asif - 2 years, 9 months ago

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. :)

Brian Charlesworth - 2 years, 9 months ago

Log in to reply

@Brian Charlesworth Thank you both!

Otto Bretscher - 2 years, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...