Let's discuss Mathematics
- Thread starter ParadigmShifter
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Now wouldn't it be really something if this was a test question and indeed they wanted -1? ^^
Does that mean that technically the lowest common multiple of any two numbers is minus infinity (or aleph null or whatever)?
The common definition of "lowest common multiple" states that it's the lowest common positive multiple, but if you don't take that part into account I would say there is no lowest common multiple. You define a multiple of A as a number B where there exists another integer K for which A times K = B. Therefore minus infinity is not a multiple of whatever number you chose.
But proof by intimidation is typically a stronger form of proof by authority.
I have another geometry question (this time it is one about sole unequal sides' comparison to their corresponding angles).
It seems to me that the following (which is generally the schoolbook's proof) is a bit too overkill (the actual proof is of all cases, but also bypasses working on the specific triangle with m)
Isn't there a way to work on the triangle with angles m1,m2 to be compared, without relying on the far more general theorem (theorem 1 as I named it here)? (as always, angles of triangle =2 right angles is not part of the progression of theorems to be used in a proof; and goes without saying that this implies that the theorem equating the external angle to the two non-suplementary to it internal ones isn't allowed).
If it was an allowed part, one could just do something like this:
And argue that as b>c, a1>a2, and all three angles=2L, obviously m2<m1. But it is not allowed, so got any ideas?
It seems to me that the following (which is generally the schoolbook's proof) is a bit too overkill (the actual proof is of all cases, but also bypasses working on the specific triangle with m)
Isn't there a way to work on the triangle with angles m1,m2 to be compared, without relying on the far more general theorem (theorem 1 as I named it here)? (as always, angles of triangle =2 right angles is not part of the progression of theorems to be used in a proof; and goes without saying that this implies that the theorem equating the external angle to the two non-suplementary to it internal ones isn't allowed).
If it was an allowed part, one could just do something like this:
And argue that as b>c, a1>a2, and all three angles=2L, obviously m2<m1. But it is not allowed, so got any ideas?
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Hm, if the question is verbatim stated as "draw 12 circles, so that each of them is tangential to exactly 5 of them", wouldn't the following be a perfectly sufficient construction? (I know that others, including very different ones, exist; just asking about this one)
Inferred that in this construction each of the two identical groups of 6 circles have 1 common point.
Book only had one construction of the many possible ones, and it made an impression on me that they went with one which would only be forced under a very specific limitation (not in the question). In the spoiler:
Inferred that in this construction each of the two identical groups of 6 circles have 1 common point.
Book only had one construction of the many possible ones, and it made an impression on me that they went with one which would only be forced under a very specific limitation (not in the question). In the spoiler:
Spoiler :
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What about the question in the spoiler?
Is that the only arrangement where each point of tangency is unique?I am shocked
Due to my type of personality I'd take it as implying that I am starting to ask decent/non-simple questions ^^Well, it is an interesting question, and I would like to know. It's also imo potentially far easier/faster to use such constructions to establish relations than (the analogue in complication) with algebra. Provided, of course, that the question is translatable to such in the first place.
The thing is, non-coordinate geometry is always much harder. Also "being the only possibility" in geometry is particularly hard to prove.
In such situations, either you have the answer or you don't. It's much rarer to have an idea without knowing how it goes with problems like these, in algebra you usually have basic methods to move forward before declaring that you don't know something.
In such situations, either you have the answer or you don't. It's much rarer to have an idea without knowing how it goes with problems like these, in algebra you usually have basic methods to move forward before declaring that you don't know something.
danjuno
Cole Phelps, Badge 1247
Greatest math quote I've ever read!
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