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# PMA112: Algebra (& PAS152: Statistical Inference) watch

1. Latest exam at the Uni of Sheffield in the Pure Maths department, was sat today at Sheffield Wednesday Football Club.

Anyone wanting to see what sort of thing gets asked at Uni level exams, the exam is attached in Adobe form.
Attached Images
2. 2004.PDF (52.5 KB, 149 views)
3. Sweet f***. Hope you did well in that one!
4. Which questions did you answer?
5. Wowee, looks taxing, can't wait for it! (Uni maths that is) I think I could probably answer a widgey bit of the groups question, or at least understand the solution. But I'm not going to
Hope you get a good score.
7. (Original post by JamesF)
Q2, Q3, Q4 and Q5.

(Original post by XTinaA)
Sweet f***. Hope you did well in that one!
It went alright actually.
8. Well that's 2 hours for 4 questions. And this is only the Algebra paper. I don't know what I'm more nervous about, the independence or the exams!
9. (Original post by XTinaA)
Well that's 2 hours for 4 questions. And this is only the Algebra paper. I don't know what I'm more nervous about, the independence or the exams!
Well now attached is my Statistical Inference paper (PAS152) sat this session - the main complaint in this was that we ran out of time.
Attached Images
10. PAS152 2004.pdf (23.7 KB, 60 views)
11. (Original post by Expression)
Latest exam at the Uni of Sheffield in the Pure Maths department, was sat today at Sheffield Wednesday Football Club.

Anyone wanting to see what sort of thing gets asked at Uni level exams, the exam is attached in Adobe form.
Out of interest, can you check my answer for the first part or two of q.2?

2)

i) Decomposes as:

1 4 5 6
4 5 6 1

2 7 9
9 2 7

3 8
8 3.

The order is 12 (lcm(4,3,2)).

ii)

a) a(6):

1 6
6 1

2 5
5 2

3 4
4 3

a(5):

1 5
5 1

2 4
4 2

3
3.

a(6) is even as its number of odd cycles is 0, and a(5) has 1 odd cycle so is odd.

b) Okay now i get stuck, but surely a(6) can be written as 3 distict permutations. But any permutation squared is only going to give a number of permutations that is a multiple of 2, so there is no permutation?

For c^3 try c = (1,6,2,5,3,4)? for d^2 try d = (1,5,4,2,3)? actually i have no idea really i'm just guessing...
12. (Original post by theone)
Out of interest, can you check my answer for the first part or two of q.2?

2)

i) Decomposes as:

1 4 5 6
4 5 6 1

2 7 9
9 2 7

3 8
8 3.

The order is 12 (lcm(4,3,2)).
2) (i) Cycle decomposition is (1 4 5 6)(2 9 7)(3 8), order (2, 3, 4) = 12

(Original post by theone)
ii)

a) a(6):

1 6
6 1

2 5
5 2

3 4
4 3

a(5):

1 5
5 1

2 4
4 2

3
3.

a(6) is even as its number of odd cycles is 0, and a(5) has 1 odd cycle so is odd.
alpha-6 = (1 6)(2 5)(3 4)

Made up of 3 transpositions, and so is ODD.

alpha-5 = (1 5)(2 4)

Made up of 2 transpositions, and so is EVEN.

(Original post by theone)
b) Okay now i get stuck, but surely a(6) can be written as 3 distict permutations. But any permutation squared is only going to give a number of permutations that is a multiple of 2, so there is no permutation?

For c^3 try c = (1,6,2,5,3,4)? for d^2 try d = (1,5,4,2,3)? actually i have no idea really i'm just guessing...
alpha-6

=

( 1 2 3 4 5 6 )
( 6 5 4 3 2 1 )

c^3 = alpha-6

=>

gamma =

( 1 2 3 4 5 6 )
( 2 3 6 1 4 5 )

alpha-5 = d^2

( 1 2 3 4 5 )
( 5 4 3 2 1 )

==> delta =

( 1 2 3 4 5 )
( 2 5 3 1 4 )
13. Now for q1 hopefully.

Surjective means mapping one set onto another set, such that every value in the image has a value in the domain.

Injective is a one-to-one mapping, so for every value in the image, there is a unique value in the domain.

if g_b(a) = 0 => a^3 -ba^2 -4b + 4 = 0.

=> (a^3+4) = b(a^2+4) => a^3+4/a^2+4 = b => f(a) = b. Now the cubic in b has f(0) = 4, so a solution < 0 (for a > 0, the argument is easily reversed for a<0), and for a = b, g_b (b) = 4-4b. Now if b>1 this will be <0, hence there will be 3 solutions to the cubic for b>1, and so 3 values of a which give f(a) = b. Hence f is surjective.

Now If f(a) = 1 => a^3 = a^2, i.e a = 1 or 0. hence f is not injective.

For the array, surely it's only neccessary to show that n, n^1/2, n^1/3... is a countable set, as the union of countable sets with the same cardinality, is countable with the same cardinality?

To show n, n^1/2, n^1/3 is countable, we take the kth power of the kth term, and then multiply it through by k, to get n, 2n, 3n, 4n.. which is a bijection to 1,2,3,4... and so is countable?

As for the last part, i don't know the terminology...
14. (Original post by Expression)
2) (i) Cycle decomposition is (1 4 5 6)(2 9 7)(3 8), order (2, 3, 4) = 12

alpha-6 = (1 6)(2 5)(3 4)

Made up of 3 transpositions, and so is ODD.

alpha-5 = (1 5)(2 4)

Made up of 2 transpositions, and so is EVEN.

alpha-6

=

( 1 2 3 4 5 6 )
( 6 5 4 3 2 1 )

c^3 = alpha-6

=>

gamma =

( 1 2 3 4 5 6 )
( 2 3 6 1 4 5 )

alpha-5 = d^2

( 1 2 3 4 5 )
( 5 4 3 2 1 )

==> delta =

( 1 2 3 4 5 )
( 2 5 3 1 4 )
Ah thanks, i didn't know how you wrote down the breakdowns of the cycles...
15. And for 4 part ii, 9397 = 4 mod 31.

Hence 9397^3 = 2 mod 31, hence 9397^15 = 1 mod 31, hence 9397^9390 = 1 mod 31, hence 9397^9397 = 4^7 mod 31 = 16 mod 31?
16. (Original post by theone)
Now for q1 hopefully.

Surjective means mapping one set onto another set, such that every value in the image has a value in the domain.

Injective is a one-to-one mapping, so for every value in the image, there is a unique value in the domain.

if g_b(a) = 0 => a^3 -ba^2 -4b + 4 = 0.

=> (a^3+4) = b(a^2+4) => a^3+4/a^2+4 = b => f(a) = b. Now the cubic in b has f(0) = 4, so a solution < 0 (for a > 0, the argument is easily reversed for a<0), and for a = b, g_b (b) = 4-4b. Now if b>1 this will be <0, hence there will be 3 solutions to the cubic for b>1, and so 3 values of a which give f(a) = b. Hence f is surjective.

Now If f(a) = 1 => a^3 = a^2, i.e a = 1 or 0. hence f is not injective.

For the array, surely it's only neccessary to show that n, n^1/2, n^1/3... is a countable set, as the union of countable sets with the same cardinality, is countable with the same cardinality?

To show n, n^1/2, n^1/3 is countable, we take the kth power of the kth term, and then multiply it through by k, to get n, 2n, 3n, 4n.. which is a bijection to 1,2,3,4... and so is countable?

As for the last part, i don't know the terminology...
I haven't sat and worked through this question as it was the one I deliberately left, as I knew that I'd take the easy marks from Fermat's Little Theorem.

"Find the remainder when 9397^9397 is divided by 31."

For question 1, if byb3 comes online he might have done that question.

Question 5 is a lovely set of marks to pick up too.
17. (Original post by theone)
And for 4 part ii, 9397 = 4 mod 31.

Hence 9397^3 = 2 mod 31, hence 9397^15 = 1 mod 31, hence 9397^9390 = 1 mod 31, hence 9397^9397 = 4^7 mod 31 = 16 mod 31?
And so we conclude that the remainder is 16.

Doesn't use Fermats Little Theorem BUT does use modular arithmetic, which is a fair enough arguement.
18. (Original post by Expression)
And so we conclude that the remainder is 16.

Doesn't use Fermats Little Theorem BUT does use modular arithmetic, which is a fair enough arguement.
5 looks fairly nice, i'll try and work through it tomorrow, the last bit is out of reach though. The first part of 4 doesn't look too bad either, if you know the terminology?
19. (Original post by theone)
5 looks fairly nice, i'll try and work through it tomorrow, the last bit is out of reach though. The first part of 4 doesn't look too bad either, if you know the terminology?
5 is quite pretty you'll enjoy that if you have a geometric head on you from the offset.

The most part of 4 is about left cosets, which yeah if you know whats going off with them if marks to be picked up.

But as you say it's all about knowing the lingo.
20. (Original post by Expression)
5 is quite pretty you'll enjoy that if you have a geometric head on you from the offset.

The most part of 4 is about left cosets, which yeah if you know whats going off with them if marks to be picked up.

But as you say it's all about knowing the lingo.
So you think you did okay then by the look of it?

What about statistical inference? Looked just like recalling the tests and number crunching, was it hard or not?
21. (Original post by theone)
So you think you did okay then by the look of it?

What about statistical inference? Looked just like recalling the tests and number crunching, was it hard or not?
Well this 112 I reckon I did enough to grab a 2:1 at least.

The Statistical Inference was OK and although the material wasn't hard, applying it in the format for reporting hypothesis tests that our lecturer uses can be time consuming.

Also, take a look at the whole of Question 1 on that Stats paper, bearing in mind that the exam was only two hours long, that first question is really time consuming which added a nasty level of time pressure.
22. (Original post by Expression)
Well this 112 I reckon I did enough to grab a 2:1 at least.

The Statistical Inference was OK and although the material wasn't hard, applying it in the format for reporting hypothesis tests that our lecturer uses can be time consuming.

Also, take a look at the whole of Question 1 on that Stats paper, bearing in mind that the exam was only two hours long, that first question is really time consuming which added a nasty level of time pressure.
Yeah that's a bit of a time waster i'd agree, esp with the hypothesis test thrown in and the sketch graph, depends how much you want to check your figures

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