-
We are currently performing site maintenance, parts of civfanatics are currently offline, but will come back online in the coming days. For more updates please see here.
Let's discuss Mathematics
- Thread starter ParadigmShifter
- Start date
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?
Last edited:
Similar threads
- Locked
- Locked
