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Jun 28, 2013 at 1:09 vote accept Maaz-ul-Haq
Jun 27, 2013 at 23:41 answer added David E Speyer timeline score: 4
Jun 27, 2013 at 19:53 comment added Maaz-ul-Haq @domotorp exactly.
Jun 27, 2013 at 19:20 history edited Maaz-ul-Haq CC BY-SA 3.0
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Jun 27, 2013 at 16:39 comment added domotorp @Maarten: I think he means to find min x st for all a1, a2, exists k, l, n.
Jun 27, 2013 at 10:23 history edited Maaz-ul-Haq CC BY-SA 3.0
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Jun 27, 2013 at 10:06 comment added Maarten Derickx I have some problems with parsing this question. Could you please add "for all" and "there exists" quantifiers to your question so that it becomes unambiguous?
Jun 27, 2013 at 1:10 comment added Gerry Myerson Previously posted to, and deleted from, m.se: math.stackexchange.com/questions/429474/…
Jun 26, 2013 at 20:15 comment added Douglas Zare Cases where $n=3^-$ is necessary include $b_1 \in (9/10,15/14), b_2 = 3/2^+$ or $a_1 \in (9/5,15/7), a_2 = 3^+$.
Jun 26, 2013 at 20:00 comment added Douglas Zare Let $b_i = a_i/2$. The conditions are that $b_i \ge 1/2$ and that $[nb_i]$ is odd, where $[x]$ is the nearest integer to $x$, with ties broken in favor of odd integers.
Jun 26, 2013 at 19:06 comment added Maaz-ul-Haq In the case $a_1=a_2=4$ we select $n=\frac{5}{4}$ and that would render both $a_1, a_2 \in [5,7]$. The question is that no matter what $a_1$ and $a_2$ you go for, I can always find an $1\le n\le 3$ such that $4k−3≤na_1≤4k−1$ and $4l−3≤na_2≤4l−1$ where $k,l \in \mathbb{N}$.
Jun 26, 2013 at 19:00 comment added Stephen Sturgeon I don't think your question make's sense in its current form. Let a_1=a_2=4. Then 12 does not satisfy your inequalities, so 3 does not "ensure" what you are asking.
Jun 26, 2013 at 18:37 answer added The Masked Avenger timeline score: 0
Jun 26, 2013 at 18:16 history asked Maaz-ul-Haq CC BY-SA 3.0