# Math Q&A Game

by Gokul43201
Tags: game, math
P: 352
 Quote by CRGreathouse "What is six multiplied by nine"? ...at least according to Douglas Adams.
Ye, I read something of that sort. But why 6 and why 9? Why not 6+9? Why not $$\frac {6} {9}$$
 P: 127 Thanks matt, I'll post a hint if anyone's interested. The problem is a bit out of the range of most high school math students but anyone with a bit of university experience has the tools to solve it. I hope nobody is scared off by it. Steven
 P: 127 The first of the big hints. This is a linear algebra problem but the vector space you want isn't over the reals. Think about what would be a useful basis and what would be in the span of that basis.
 P: 127 Does anybody want to post any work in progress or thoughts of a solution? That way at least a next questioner can be chosen. I apologize if I bogged this game down with a poor question . I still have feel someone out there can solve it. I solved it as a junior undergraduate and I wasn't that amazing of a junior undergrad.
 P: 58 Let’s say you have an irrational x in A which is a paired set. You’re sure of 0,x and 1 being in the set. Now, obviously x-0 != 1-x .So there must be at least one additional point in the set (or two points p,q distinct from x such that p-q=x-0) .Now this point is irrational( in the other case, at least one of p & q is irrational). If we go on this way…we get infinitely many irrationals in the set. That , I guess ,amounts to saying that a paired set containing an irrational point contains infinitely many irrationals. All of this seems too simple to be correct….and what’s infinitely worse is that I don’t see any use of the hint given by snoble...and lastly, why are the biggies like matt and gokul not coming up with smart ideas,like they always do? i guess the odds aren't in my favour...THERE HAS TO BE SOMETHING WRONG IN THIS!!
 Sci Advisor HW Helper P: 9,396 I guessed the first and thus set the last one; I haven't even bothered to think about this one really, other than to clarfiy something with the setter.
 P: 127 mansi you are on the right line of thought but as always the devil is in the details. If you can formalize "If we go on this way" you've basically proven it. Try even just take it to the next stage... why can't there be only two irrationals in a paired set... why can't there be only n irrationals.
 P: 58 i'm writing down the proof...i'm onto it ,head on... but there's something i'd like to clarify... in a paired set ,is it possible to have 3 pairs of elements having the same distance between them....or do we consider it to be exactly two pairs?
 P: 127 3 of the same distance is totally cool. That's why {0,1/5,2/5,4/5,1} is a paired set even though 1/5 shows up 3 times.
 P: 58 Ok…so I think I’ve got something (not a proof) here… Let A be a paired set containing an irrational point, say x. 0,x,1 belong to A and x-0 != 1-x, so there exists at least one additional point which is irrational or two points,at least one of which is irrational. In the first case, say , the irrational point is p. so you have 0,x,p,1 belonging to A. p –x !=p-0 and p-x !=1-x. The only possibility is that p-x =1-p and p-0=1-x but that contradicts the fact that x is irrational. Therefore, there exists at least one more irrational point in the set….and so on. In the second case , say you’ve 2 points p and q and WLOG p is irrational. i.e. we have x-0= p-q . If q is rational…and if q=1/2 then you don’t have another rational point but taking all the possible combinations of the distances between rational and irrational , you have to have an irrational… If q != ½ then there’s definitely another rational point z in the set. Consider the difference x-z. We need a pair of elements giving us this difference. x- z !=p-z (contradicts the fact that z is rational) and x-z !=1-p and x-z !=p-0. Therefore , we’re sure that there’s another irrational in the set… So , now you have 3 irrationals in the set…x,p and q(say). Consider the difference between the two irrationals q and x. Picking up all possible Pairs, you get to know that another irrational exists in the set . Finally,as I said earlier…existence of an irrational in a paired set implies the existence of infinitely many irrationals in the set… I think this is more like an observation, not a proof in the typical sense…
 P: 58 so...am i supposed post a new question,or not??
 Emeritus Sci Advisor PF Gold P: 11,155 Give snoble a day or two to respond...
 P: 127 Unless somebody minds how about we wait until tomorrow morning to see if anyone wants to fill in the details. Otherwise I'll say it's all yours mansi (and I'll post a proof myself for why there are no finite paired sets containing an irrational number) Steven
 P: 127 go ahead and ask a new question mansi. I'll post a solution to my problem some time soon (today or tomorrow)
 P: 58 Hi! Here is a problem I've been struggling with,so it appears real tough to me.just for the record...i haven't (yet) been able to write down a proof for it... prove that Z + 2^(1/2)Z is dense in R. (in words the given set is the set of integers + (square root of 2) times the set of integers and R is is the real line)
 P: 197 I am thinking you only need to consider the interval [0,1] and study if sqrt(2)*z (mod 1) is dense in the interval. Then, it will look like a circle (by identifying 1 with 0). Explore the consequence if it is not dense (i.e. prove by contradiction).

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