HL Math Paper 3 Sets option- 24 hours later...
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HL Math Paper 3 Sets option- 24 hours later...Okay, our whole grade found the paper 3 (sets, relations, groups guy here) confusing. Perhaps not difficult, but we all made silly errors we want to hurt ourselves about. For example- not counting the empty set as a subset of R* in one of the questions... and also the last question about matrices.
What did you think? Be honest, I'm trying to get an idea of where the 7 boundary is gonna be... where do you think it's gonna be? -
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Re: HL Math Paper 3 Sets option- 24 hours later...Yeah we all found it strange as well, never have seen a question on the power set, but guess what, it came up haha, wrote down the proof for it instead of the right thing ...doh...
The matrices question/ last two questions were quite horrible and strange, the last one was just plain confusing.
Really don't have a clue where it would be, but I would guess around 40? -
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Re: HL Math Paper 3 Sets option- 24 hours later...I actually liked this paper a lot. I found that most of the questions were unusual but doable.
The first question was relatively straightforward, it was just about recalling (Abelian) group axioms, although the equation at the end was a bit tedious, to say the least, since the operation wasn't associative. i think I got that every element except p was a solution of the equation.
I really wasn't expecting to see power sets on the exam, but fortunately I remembered that the power set has 2^n elements, the empty group being one of them.
The equivalence relation on question 3 was nice. I especially enjoyed proving that some of the equivalence classes were the same.
Question 4 was very nice as well, especially part (b). Question 5 started off very well but halfway through I realised that it wasn't as easy as I had thought. The key thing, though, was that some of the elements of the matrices in the group could be complex numbers. Also, since the order of h was 2 and the order of the unknown matrices was 4, we knew that the square of those unknown matrices had to equal h. There were actually two such pairs of matrices.
I was surprised, though, that there wasn't a single question on isomorphism on Paper 3, and only a very short one in one of the Further Maths papers. I thought it was pretty much guaranteed to see an epic isomorphism question, but it seems not. -
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Re: HL Math Paper 3 Sets option- 24 hours later...I just realized this now. For the last question, were the two matrices:(Original post by Draconian28)
I actually liked this paper a lot. I found that most of the questions were unusual but doable.
The first question was relatively straightforward, it was just about recalling (Abelian) group axioms, although the equation at the end was a bit tedious, to say the least, since the operation wasn't associative. i think I got that every element except p was a solution of the equation.
I really wasn't expecting to see power sets on the exam, but fortunately I remembered that the power set has 2^n elements, the empty group being one of them.
The equivalence relation on question 3 was nice. I especially enjoyed proving that some of the equivalence classes were the same.
Question 4 was very nice as well, especially part (b). Question 5 started off very well but halfway through I realised that it wasn't as easy as I had thought. The key thing, though, was that some of the elements of the matrices in the group could be complex numbers. Also, since the order of h was 2 and the order of the unknown matrices was 4, we knew that the square of those unknown matrices had to equal h. There were actually two such pairs of matrices.
I was surprised, though, that there wasn't a single question on isomorphism on Paper 3, and only a very short one in one of the Further Maths papers. I thought it was pretty much guaranteed to see an epic isomorphism question, but it seems not.
( i 0 )
( 0 -1 )
And:
( -i 0 )
( 0 -1 )
?
How did you prove gh=hg also? -
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Re: HL Math Paper 3 Sets option- 24 hours later...Yes, that was one of the pairs of matrices that worked. The other pair had i in the place of -i and -i in the place of i.(Original post by Saymyname!)
I just realized this now. For the last question, were the two matrices:
( i 0 )
( 0 -1 )
And:
( -i 0 )
( 0 -1 )
?
How did you prove gh=hg also?
About the proof, they told us that h had order 2 and we had to prove that ghg^-1 also had order 2. However, since they told us that there is only one element of order 2 in the group, then those two element must be equal. Therefore, h=ghg^-1, which implies that hg=gh, if we multiply both sides by g on the right.
