Wikipedia:Reference desk/Archives/Mathematics/2016 July 29

= July 29 =

A question related to indices
If a^x=bc, b^y=ca , c^z=ab then show that xyz=x+y+z+2 — Preceding unsigned comment added by Sahil shrestha (talk • contribs) 02:08, 29 July 2016 (UTC)


 * Hint: Logarithm. Let us know if you are still stuck. -- ToE 05:08, 29 July 2016 (UTC)


 * Hey, that's cute. —Tamfang (talk) 05:38, 30 July 2016 (UTC)
 * I don't want the answer using logarithm. Rather I would like to get the answer by the indices rules, exponential equations & Whatever!! But not logarithm!! — Preceding unsigned comment added by Sahil shrestha (talk • contribs) 05:49, 30 July 2016 (UTC)
 * Well, the rules of logarithms are really just the rules of indices in disguise. But if you want to avoid using logarithms,, start by showing that:
 * a^(xyz) = ((b^y)^z)((c^z)^y)
 * Gandalf61 (talk) 07:09, 30 July 2016 (UTC)
 * Why don't you want to use logarithms? Were you bitten by a logarithm as a young child? -- Meni Rosenfeld (talk) 07:07, 31 July 2016 (UTC)
 * Or perhaps they were frightened by the sight of adders multiplying on a log table. -- ToE 00:00, 1 August 2016 (UTC)
 * The question is missing quantifiers, and from some reasonable choices of quantifiers the result is false. In particular, if a = b = c = 1 then x, y, z can be arbitrary.  --JBL (talk) 16:47, 30 July 2016 (UTC)
 * "If, for any a,b,c positive reals, (...)" is the only choice that sounds reasonable to me (given the symmetry of the problem, and what you mentioned). Tigraan Click here to contact me 18:44, 30 July 2016 (UTC)
 * You don't need this to hold for every $$a,b,c$$, just for some particular $$a,b,c$$ with $$a\neq1$$. -- Meni Rosenfeld (talk) 07:07, 31 July 2016 (UTC)
 * It suffices to require that $$a,b,c>0$$ and that $$a\neq1$$. Following Gandalf's hint, we can straightforwardly show that $$a^{xyz}=a^{x+y+z+2}$$, and then $$a \neq 1$$ gives $$xyz=x+y+z+2$$. -- Meni Rosenfeld (talk) 07:07, 31 July 2016 (UTC)

Idempotents generating the same principal right ideal
Does there exist a ring R with two distinct idempotents e and f generating the same principal right ideal (i.e. eR = fR)? GeoffreyT2000 (talk) 04:14, 29 July 2016 (UTC)


 * Yes. 2x2 matrices, with idempotents $$\begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}1&1\\0&0\end{bmatrix}$$.   Sławomir Biały  (talk) 11:55, 30 July 2016 (UTC)

Sequential simultaneous games
Given the existence of simultaneous games and sequential games, what (distinctive) features has their combination, a sequential simultaneous game?--82.137.9.125 (talk) 15:52, 29 July 2016 (UTC)


 * See Repeated game and in particular Repeated game. Loraof (talk) 20:00, 29 July 2016 (UTC)

Combinations between game and queuing theory
What possibilities of combining game theory and queuing theory are there?--82.137.9.125 (talk) 15:59, 29 July 2016 (UTC)


 * Here's a paper that approached a problem in queuing theory by formulating it as a game and applying game theoretic techniques . SemanticMantis (talk) 18:29, 29 July 2016 (UTC)
 * Thanks. How about the other perspective, of applying queuing theory techniques in game theory of multiplayer games? What is the degree of equivalence of the opposite approaches?--82.137.9.125 (talk) 17:47, 29 July 2016 (UTC)