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Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Skewbrow:
--- Quote from: pwhodges on 20 Oct 2013, 15:55 ---
--- Quote from: snalin on 20 Oct 2013, 05:48 ---Maths is, right? I mean, maths is written in plural form, but it is still just a single thing, right? Or is it maths are sexy?
--- End quote ---
Is. "Maths" is not the plural of "math"; it's a different abbreviation of "mathematics". Many words end in s without being plural.
--- End quote ---
Depends. It's "Maths" in the UK, and "Math" on the other side of the pond (or should I say elsewhere?).
But looks like you guys started having fun while I was gone from this subforum. I don't have anything against doing it here, but I also might recommend that we take it to Math.Stackexchange. There is a real-time editable TeX-interpreter (MathJax) there, so we don't need to use graphics. I'm sure Paul might be able to hack us into the cloud running MathJax (or whatever it is), but we won't have hundreds or thousands of qualified people answering here.
Take it from someone who typed the plainTeX-souce to his dissertation 23 years ago. With the VI-editor that came with Unix.
ankhtahr:
Currently working on this sheet (sorry for using this shitty owncloud-website, I don't have SSH-access to my server, so I can only use that if I don't want to upload it at any strange service.
Don't really think anyone here can help me with it, if I don't give a translation. I don't have the time to give a translation right now.
Loki:
You are doing weird stuff.
Re 2b)
--- Quote ---Let G be a group and H1, H2 subgroups of G.
Show that H1 union H2 is a subgroup or find a counterexample
--- End quote ---
Should be true, imho. The neutral element of G is in H1uH2. For every element x of H1, -x in in H1 (H1 is subgroup), and therefore in H1uH2. The same for H2.
Associativity should be, uh, trivial.
Do you know where to start on the others?
Skewbrow:
--- Quote from: Loki on 21 Nov 2013, 08:10 ---You are doing weird stuff.
Re 2b)
--- Quote ---Let G be a group and H1, H2 subgroups of G.
Show that H1 union H2 is a subgroup or find a counterexample
--- End quote ---
--- End quote ---
This is false. Let G be the group S_3 of symmetries of an equilateral triangle. Let H1 and H2 be groups generated by a single reflection (or single transposition, if you are doing it in terms of permutations rather than symmetries of a triangle). There union then contains two distinct reflections, but not their composition (which is rotation or a 3-cycle again depending which way you view this group).
In terms of the group axioms it is the zeroth axiom that fails. This set (the union of H1 and H2) does not have an operation.
It is a "standard" exercise to show that the union of two subgroups is a subgroup, if and only if one of the subgroups is contained in the other. The idea is the same. If x is an element of H1 but not an element of H2, and y is an element of H2 but not of H1, then xy cannot be an element of either H1 or H2. Therefore it is not in the union.
Skewbrow:
--- Quote from: ankhtahr on 21 Nov 2013, 06:50 ---Don't really think anyone here can help me with it, if I don't give a translation. I don't have the time to give a translation right now.
--- End quote ---
If the topic of the sheet is something like the first course in abstract algebra (judging from Loki's post), I am qualified. I don't fully trust my mathematical German, but I have read a number of German math books (though it was a while ago). The problem is that my browser is convinced that following up your link is a bad idea, risky even.
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