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Topic: Unsolvable subsets (Read 687 times) |
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Earendil
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Unsolvable subsets
« on: Mar 13th, 2004, 11:58pm » |
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Divide the set of positive real numbers into three subsets, such that the equation "a + b = 5c" is not solvable in any of the subsets, i.e. the equation does not hold, for any three numbers a, b, c, that come from the same subset. Can this be done for the equation "a + b = 3c", maybe by using more subsets?
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Barukh
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Re: Unsolvable subsets
« Reply #1 on: Mar 19th, 2004, 11:59pm » |
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[smiley=blacksquare.gif]
Let q = [sqrt]5, and define the subsets as follows: Si = [smiley=bigcup.gif] [q3k+i, q3k+i+1), k [in] [bbz]. To put it differently, x [in] Si if [ logqx ] = i mod 3.
By symmetry considerations, we may assume a, b [in] S0 and a [ge] b. Then a = q3k+e, e [in] [0,1), and c > a/5 = aq-2 = q3k+e-2 [ge] q3k-2,c [le] 2a/5 = 2q3k+e-2 < 2q3k-1 < q3k,
which implies that c [in] S1 or c [in] S2.
For the second question, I think the answer is yes: take q = 31/3, and divide into 4 subsets.
[smiley=blacksquare.gif]
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Icarus
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Re: Unsolvable subsets
« Reply #2 on: Mar 21st, 2004, 2:27pm » |
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on Mar 13th, 2004, 11:58pm, Earendil wrote:| Can this be done for the equation "a + b = 3c", maybe by using more subsets? |
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As stated, it can easily be done for any equation. Just make each set a singleton! But I suppose you meant do it with a finite number of sets.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Barukh
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Re: Unsolvable subsets
« Reply #3 on: Mar 21st, 2004, 11:18pm » |
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on Mar 21st, 2004, 2:27pm, Icarus wrote:| As stated, it can easily be done for any equation. Just make each set a singleton! |
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Except the equation 'a+b=2c', of course! 
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Icarus
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Re: Unsolvable subsets
« Reply #4 on: Mar 22nd, 2004, 4:07pm » |
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Okay - I'll grant that one! So, what's true is that setting each set to be a singleton will work for any equation which has no solution with all the variables equal. And for equations which have at least one such solution, every partition will have at least one set containing a solution.
Still, the real challenge is: Can it be done with a finite partition?
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Barukh
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Re: Unsolvable subsets
« Reply #5 on: Mar 23rd, 2004, 12:38am » |
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on Mar 22nd, 2004, 4:07pm, Icarus wrote:| Still, the real challenge is: Can it be done with a finite partition? |
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Consider the equation a+b = kc, k [in] [bbr]+. If the partitions are made as in my proposed solution, it can be shown that q must satisfy the following conditions:
k1/(P-1) [le] q [le] k/2,
where P is the number of partitions. So, for k > 2 it suffices to make P = log(k2/2) / log(k/2) partitions.
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| « Last Edit: Mar 23rd, 2004, 12:39am by Barukh » |
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