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wu :: forums    riddles    medium (Moderators: Grimbal, towr, Eigenray, william wu, ThudnBlunder, Icarus, SMQ)    A condition for rationality? « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: A condition for rationality?  (Read 566 times) Michael Dagg Senior Riddler     Gender: Posts: 500 A condition for rationality?   « on: Feb 3rd, 2008, 9:03pm » Quote Modify 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. IP Logged Regards, Michael Dagg SMQ wu::riddles Moderator Uberpuzzler     Gender: Posts: 2084 Re: A condition for rationality?   « Reply #1 on: Feb 4th, 2008, 5:44am » Quote Modify Half of it is easy: w.l.o.g. choose r1 < r2 < r3, then (q + r1)(q + r3) = (q + r2)2 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: 3 , r1 < r2 < r3, f(r1, r2, r3) = (r22 - r1r3)/(r1 - 2r2 + r3) is or only a subset thereof.   --SMQ « Last Edit: Feb 4th, 2008, 5:56am by SMQ » IP Logged --SMQ Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: A condition for rationality?   « Reply #2 on: Feb 4th, 2008, 6:25am » Quote Modify 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. IP Logged SMQ wu::riddles Moderator Uberpuzzler     Gender: Posts: 2084 Re: A condition for rationality?   « Reply #3 on: Feb 4th, 2008, 6:32am » Quote Modify on Feb 4th, 2008, 6:25am, 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 IP Logged --SMQ Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: A condition for rationality?   « Reply #4 on: Feb 4th, 2008, 6:59am » Quote Modify (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. IP Logged SMQ wu::riddles Moderator Uberpuzzler     Gender: Posts: 2084 Re: A condition for rationality?   « Reply #5 on: Feb 4th, 2008, 7:09am » Quote Modify on Feb 4th, 2008, 6:32am, SMQ wrote: What about, for instance, -2, 1, 4...   on Feb 4th, 2008, 6:59am, 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 IP Logged --SMQ Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: A condition for rationality?   « Reply #6 on: Feb 4th, 2008, 7:38am » Quote Modify on Feb 4th, 2008, 7:09am, SMQ wrote: Am I missing something obvious...or are you? q·(r1 - 2·r2 + r3) = (r22 - r1·r3) IP Logged SMQ wu::riddles Moderator Uberpuzzler     Gender: Posts: 2084 Re: A condition for rationality?   « Reply #7 on: Feb 4th, 2008, 8:06am » Quote Modify 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 « Last Edit: Feb 4th, 2008, 8:06am by SMQ » IP Logged --SMQ Hippo Uberpuzzler     Gender: Posts: 919 Re: A condition for rationality?   « Reply #8 on: Feb 4th, 2008, 11:56am » Quote Modify on Feb 4th, 2008, 6:59am, 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 ... IP Logged Eigenray wu::riddles Moderator Uberpuzzler     Gender: Posts: 1948 Re: A condition for rationality?   « Reply #9 on: Feb 4th, 2008, 8:59pm » Quote Modify 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 -1, we can find a solution with r=1: (r,s,t) = (1, n+2, n2+3n+3), or (1, s, 1-2s+2s2) if n=-2.   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 -1, c 0, and cn -2.  (But of course there are more solutions.) IP Logged Hippo Uberpuzzler     Gender: Posts: 919 Re: A condition for rationality?   « Reply #10 on: Feb 5th, 2008, 3:49am » Quote Modify May be, I am wrong again ... but is   (n+1/n),((n+1)^2/n),((n+1)^3/n) geometric progression? IP Logged rmsgrey Uberpuzzler     Gender: Posts: 2873 Re: A condition for rationality?   « Reply #11 on: Feb 5th, 2008, 8:31am » Quote Modify on Feb 5th, 2008, 3:49am, 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... IP Logged Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: A condition for rationality?   « Reply #12 on: Feb 5th, 2008, 8:53am » Quote Modify on Feb 5th, 2008, 3:49am, 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 IP Logged Eigenray wu::riddles Moderator Uberpuzzler     Gender: Posts: 1948 Re: A condition for rationality?   « Reply #13 on: Feb 5th, 2008, 12:13pm » Quote Modify on Feb 5th, 2008, 3:49am, 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! IP Logged Hippo Uberpuzzler     Gender: Posts: 919 Re: A condition for rationality?   « Reply #14 on: Feb 5th, 2008, 2:19pm » Quote Modify 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 ... « Last Edit: Feb 5th, 2008, 2:23pm by Hippo » IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy => medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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