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        riddles >> medium >> A condition for rationality?
        
(Message started by: Michael_Dagg on Feb 3rd, 2008, 9:03pm)
Title: A condition for rationality?
Post by Michael_Dagg on Feb 3rd, 2008, 9:03pm
Prove/disprove that a real number  q  is rational iff there are three distinct integers
r1, r2, r3  such that   q + r1 , q + r2 , q + r3   forms a geometric progression.
Title: Re: A condition for rationality?
Post by SMQ on Feb 4th, 2008, 5:44am
Half of it is easy: w.l.o.g. choose r1 < r2 < r3, then (q + r1)(q + r3) = (q + r2)2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif q = (r22 - r1r3)/(r1 - 2r2 + r3), so given integers r1, r2 and r3 clearly q is rational.  That leaves the trickier question of whether or not the range of f: http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif3 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif, r1 < r2 < r3, f(r1, r2, r3) = (r22 - r1r3)/(r1 - 2r2 + r3) is http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif or only a subset thereof.

--SMQ
Title: Re: A condition for rationality?
Post by Grimbal on Feb 4th, 2008, 6:25am
That was half of the easy half.  You still have to make sure (r1 - 2r2 + r3) is not zero.

In fact, you can show that if it is zero, then you must have r1=r2=r3.
Title: Re: A condition for rationality?
Post by SMQ on Feb 4th, 2008, 6:32am

on 02/04/08 at 06:25:38, Grimbal wrote:
That was half of the easy half.

Ahh, yes, darn.

Quote:
You still have to make sure (r1 - 2r2 + r3) is not zero.  In fact, you can show that if it is zero, then you must have r1=r2=r3.

Eh?  What about, for instance, -2, 1, 4...

--SMQ
Title: Re: A condition for rationality?
Post by Grimbal on Feb 4th, 2008, 6:59am
(q+r1)/(q+r2) = (q+r2)/(q+r3)
=> (q+r1)(q+r3) = (q+r2)2
=> q·r1 + q·r3 + r1·r3 = 2q·r2 + r22
=> q·(r1 - 2r2 + r3) = (r22 - r1·r3)

If (r1 - 2r2 + r3)!=0 we have q rational.

If not, (r22 - r1·r3) is also zero
we have:
  2r2 = r1 + r3
and
  r22 = r1·r3

From there, (r1+r3)2 = 4r22 = 4r1·r3
=> (r1-r3)2 = 0
=> r1 = r3

This with
  2r2 = r1 + r3
implies r1 = r2 = r3, which is against the conditions.  (r1=r3 already was).

So, in fact, (r1 - 2r2 + r3) is never zero and q is always rational.
Title: Re: A condition for rationality?
Post by SMQ on Feb 4th, 2008, 7:09am

on 02/04/08 at 06:32:41, SMQ wrote:
What about, for instance, -2, 1, 4...



on 02/04/08 at 06:59:43, Grimbal wrote:
If not, (r22 - r1·r3) is also zero

???

r1 = -1, r2 = 1, r3 = 4.  r1 < r2 < r3; r1 - 2r2 + r3 = -2 - 2(1) + 4 = 0; r22 - r1r3 = 12 - (-2)(4) = 9.

Am I missing something obvious...or are you? ;)

--SMQ
Title: Re: A condition for rationality?
Post by Grimbal on Feb 4th, 2008, 7:38am

on 02/04/08 at 07:09:56, SMQ wrote:
Am I missing something obvious...or are you? ;)

q·(r1 - 2·r2 + r3) = (r22 - r1·r3)
Title: Re: A condition for rationality?
Post by SMQ on Feb 4th, 2008, 8:06am
Ahh, I see.  I'm saying: "Given three integers r1 < r2 < r3..."; you're saying: "Given three integers such that for some q, q(r1 - 2r2 + r3) = (r22 - r1r3)...".  We were just talking past each other.

--SMQ
Title: Re: A condition for rationality?
Post by Hippo on Feb 4th, 2008, 11:56am

on 02/04/08 at 06:59:43, Grimbal wrote:
we have:
  2r2 = r1 + r3
and
  r22 = r1·r3


Actually r2 must be both arithmetic and geometric mean of r1,r3. And they are equal only on constant sample ...
Title: Re: A condition for rationality?
Post by Eigenray on Feb 4th, 2008, 8:59pm
Write r,s,t for r1,r2,r3.

For the other direction, it suffices to consider the case q=1/n (why?).  In fact, for n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif -1, we can find a solution with r=1: [hide](r,s,t) = (1, n+2, n2+3n+3), or (1, s, 1-2s+2s2) if n=-2[/hide].

More generally, setting q=m/n, we find

rt = r(ns2 + 2ms - mr)/(m+nr)
= s2 - m(s-r)2/(m+nr).

If we pick s-r = c(m+nr) for some integer c, then rt at least will be an integer, and in fact we get the solution

(r,s,t) = ( r, r+c(m+nr), r+c(m+nr)(2+cn) ),

which is valid as long as r http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif -1, c http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif 0, and cn http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif -2.  (But of course there are more solutions.)
Title: Re: A condition for rationality?
Post by Hippo on Feb 5th, 2008, 3:49am
May be, I am wrong again :( ... but is
(n+1/n),((n+1)^2/n),((n+1)^3/n) geometric progression?
Title: Re: A condition for rationality?
Post by rmsgrey on Feb 5th, 2008, 8:31am

on 02/05/08 at 03:49:08, Hippo wrote:
May be, I am wrong again :( ... but is
(n+1/n),((n+1)^2/n),((n+1)^3/n) geometric progression?

It looks like one - each term is (n+1) times the previous...
Title: Re: A condition for rationality?
Post by Grimbal on Feb 5th, 2008, 8:53am

on 02/05/08 at 03:49:08, Hippo wrote:
May be, I am wrong again :( ... but is
(n+1/n),((n+1)^2/n),((n+1)^3/n) geometric progression?

((n+1)/n),((n+1)^2/n),((n+1)^3/n) is one
Title: Re: A condition for rationality?
Post by Eigenray on Feb 5th, 2008, 12:13pm

on 02/05/08 at 03:49:08, Hippo wrote:
May be, I am wrong again :( ... but is
(n+1/n),((n+1)^2/n),((n+1)^3/n) geometric progression?

That's a very elegant solution!  So another characterization is:

q is rational iff there exists an infinite sequence of distinct integers {ri} such that {q+ri} is a geometric series!
Title: Re: A condition for rationality?
Post by Hippo on Feb 5th, 2008, 2:19pm
So q_i=m/n(n+1)^i is the infinite geometric series ... where (n+1)^i=K_in+1 for a whole K_i, we get q_i=mK_i+m/n.

If it looks like my geniality ... I am sorry, I was only confused and asked a stupid question. I only translated Eigenray's solution ...


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