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January 17[edit]

Find the values that satisfy the condition...[edit]

Hello, all. I'm not very good at this sort of thing, so I'd like to submit it for your attention, and hopefully I can come away with a better understanding for this sort of problem.

Find all the values for ABC, such that AB'C = BC, where A, B > 0.

For one, how do I interpret B'? And then of course there's the whole combinatoric mess in solving it after that point. Thanks, although I think plenty here would find this a piece of cheesecake.Qsc246 (talk) 02:05, 17 January 2012 (UTC)[reply]

I think you are going to have to supply some more information about this problem. Are A, B, B' and C integers? Are there any other restrictions on their values? Are these maybe vertices of triangles? What's the context of this question? Rckrone (talk) 05:10, 17 January 2012 (UTC)[reply]

The context is the problem as it is stated. All values; there are no restrictions. The formalism need not be dictated by a "context" from what I can tell.Qsc246 (talk) 05:42, 17 January 2012 (UTC)[reply]

I did a short dig and found something that may be relevant to uncovering a solution. If anyone thinks this is relevant and yields a path to the solution, please chime in: http://cseweb.ucsd.edu/classes/wi06/cse140/solution1.pdf .Qsc246 (talk) 05:46, 17 January 2012 (UTC)[reply]

OK, that PDF is all about Boolean logic, is that the context of this question ? StuRat (talk) 05:52, 17 January 2012 (UTC)[reply]

I don't have the complete "context". But I'm beginning to think it is so... I'm going to test out some logic gate scenarios. Qsc246 (talk) 05:55, 17 January 2012 (UTC)[reply]

I just finished setting up a circuit that represents the equations. The only way it holds is where A and B are "on" (or 1>0) and C is "off" (or 0), such that ABC=0, AB'C=BC=0. No other solution exists... However, I'm not entirely convinced that this is how I should approach the problem. Any ideas out there? Qsc246 (talk) 06:10, 17 January 2012 (UTC)[reply]

Since C > 0 wasn't specified, you could just make C = 0, then the values of A, B, and B' don't matter. StuRat (talk) 05:23, 17 January 2012 (UTC)[reply]

This occurred to me as well; however, simple observation shows that we need to provide all values for "ABC", not just "AB". C is the only truly free variable, so arbitrarily restricting it to 0 seems to invite a narrow, non-general result (i.e. a non-solution).Qsc246 (talk) 05:42, 17 January 2012 (UTC)[reply]

ABC could be any real number. As you have observed, C is a free variable. Sławomir Biały (talk) 10:57, 17 January 2012 (UTC)[reply]

Yes, but there is a limiting condition: AB'C = BC must hold for those values. Does it sound correct to regard B' as the "opposite" or "negation" of B as in the Boolean logic case; if we take that logic to hold, then A and B can indeed by any value, so long as they aren't 0, C must be 0, such that the equation will hold for all values, in turn meaning that, no matter what, ABC will always be 0? Qsc246 (talk) 17:33, 17 January 2012 (UTC)[reply]

But this imposes no condition at all on C. Sławomir Biały (talk) 00:42, 18 January 2012 (UTC)[reply]

Notation is most reminiscent of Boolean logic, linear algebra (matrices), or geometry. If it's boolean logic, conditions "A, B > 0" are strange (they just mean "A=B=1"). But at least they are valid. In linear algebra, you're not supposed to write "A>0". In geometry, "AB'C = BC" is meaningless (left hand side is a triangle, right hand side is a line). It's either the Boolean logic solution you found above, or the context is something exotic like multilevel gate logic (or worse). --Itinerant1 (talk) 23:01, 17 January 2012 (UTC)[reply]

Human in a looked room[edit]

Would he be able to reconstruct all the field of mathematics? (provided infinite time, discounted the symbolic part (like knowing that = is the equal sign)). — Preceding unsigned comment added by 80.58.205.34 (talk) 18:53, 17 January 2012 (UTC)[reply]

No, because one person couldn't keep up with the untold millions still developing the subject, and it seems unlikely that mathematics is finite.←109.151.41.148 (talk) 23:34, 17 January 2012 (UTC)[reply]

This sounds a bit like what Whitehead and Russell attempted in Principia Mathematica. They tried to derive all mathematical truths from a well-defined set of axioms and inference rules in symbolic logic. There are some big problems.--Salix (talk): 23:47, 17 January 2012 (UTC)[reply]

Yes quite easily given infinite time and resources, see Infinite monkey theorem or The Library of Babel. The equal sign will be used properly in the correct texts. If the universe is infinite there's probably someone just like me somewhere except they forgot to put a full top at the end of this sentence Dmcq (talk) 00:01, 18 January 2012 (UTC)[reply]

depending on the person, he mmight not even be able to multiply all the integers from 1 to 100 together, given infinite time.--80.98.112.4 (talk) 23:31, 21 January 2012 (UTC)[reply]

Jacobi ellipsoid[edit]

Resolved

Is a Jacobi ellipsoid a special case of scalene ellipsoid? It is just a scalene ellipsoid of a solid/plastic body under rotation? It's frequently mentioned in hydrodynamics, and is pertinent to many of our astronomy articles. — kwami (talk) 21:59, 17 January 2012 (UTC)[reply]

Answered at talk:ellipsoid. — kwami (talk) 23:04, 20 January 2012 (UTC)[reply]