MAT Prep Thread - 2nd November 2016

Original post by furmat

There was no need to try 10 different permutations of A and B to find c. In the first part you showed that the composition of functions A and B is commutative. Function composition is also associative, so any composition with 2 As and 3 Bs would be equivalent to AABBB. You could then check it for this value and be satisfied it was the only solution. Sadly I must have made an arithmetic blunder because I don't think I got 107 when I did that - I can't believe I made such a mistake, but I gave all my reasoning so if they're reasonable marking it maybe they'll only take off a mark for the arithmetic slip, given I explained my answer in full.

How did you know that about function composition?

Reply 1483

11234

8

How did everyone find q3 and q4

Reply 1484

anony-mousse

10

Original post by RuairiMorrissey

How did you know that about function composition?

Yeah it's not the most obscure knowledge but it's certainly not in C1 or C2. I think there must be a C1/C2 way of figuring out there's only 1 value of c

Reply 1485

sillygoose123

Original post by furmat

There was no need to try 10 different permutations of A and B to find c. In the first part you showed that the composition of functions A and B is commutative. Function composition is also associative, so any composition with 2 As and 3 Bs would be equivalent to AABBB. You could then check it for this value and be satisfied it was the only solution. Sadly I must have made an arithmetic blunder because I don't think I got 107 when I did that - I can't believe I made such a mistake, but I gave all my reasoning so if they're reasonable marking it maybe they'll only take off a mark for the arithmetic slip, given I explained my answer in full.

Could you explain this because the way i see it, I dont think proving that A(B(x)) = B(A(x)) implies commutativity.

I may be wrong but how do you know that A(B(B(x))) = B(A(A(x))) etc..

Reply 1486

11234

8

when do they start givin out interviews

Reply 1487

m0.4444

7

Original post by 11234

How did everyone find q3 and q4

I found q3 strange but easy. I didnt do q4 cuz i did q6 instead

Reply 1488

11234

8

Original post by m0.4444

I found q3 strange but easy. I didnt do q4 cuz i did q6 instead

would u say this paper was harder than last years?

Reply 1489

collmr

Original post by sillygoose123

could you explain this because the way i see it, i dont think proving that a(b(x)) = b(a(x)) implies commutativity.

I may be wrong but how do you know that a(b(b(x))) = b(a(a(x))) etc..

ab = ba

abb = bab

bab = baa

Reply 1490

Original post by m0.4444

I messed that one up too. Everyone else is saying n has to be even

I got no solutions : (

Reply 1491

sillygoose123

Original post by collmr

ab = ba

abb = bab

bab = baa

Ahhh i see now i feel like a sillygoose

Reply 1492

AmanGottu

2

Original post by RuairiMorrissey

I got no solutions : (

I'm pretty sure it's no solutions...

Reply 1493

furmat

3

Original post by sillygoose123

Could you explain this because the way i see it, I dont think proving that A(B(x)) = B(A(x)) implies commutativity.

I may be wrong but how do you know that A(B(B(x))) = B(A(A(x))) etc..

A(B(x)) = B(A(x)) is kind of the definition of commutativity for these two functions, if you treat the functions as elements and function composition as a binary operation that is associative (which is the case). The way I'm saying it makes it sound more obscure than it really is. I still made an error in the actual calculation for c :colondollar:

Reply 1494

m0.4444

7

Original post by RuairiMorrissey

I got no solutions : (

Thats what i put

Reply 1495

sillygoose123

Original post by RuairiMorrissey

I got no solutions : (

Basically you show that if n is even, you can do difference of squares and play around with the algebra until you get a factor of x^2 + 1 which therefore proves works for all even n. To prove that it doesnt work for odd n, I just tested it with random values of odd n and found counter examples. :smile:

Reply 1496

Original post by sillygoose123

Basically you show that if n is even, you can do difference of squares and play around with the algebra until you get a factor of x^2 + 1 which therefore proves works for all even n. To prove that it doesnt work for odd n, I just tested it with random values of odd n and found counter examples. :smile:

I just subbed in x = i

: I

Reply 1497

boombox111

Original post by RuairiMorrissey

How did you know that about function composition?

if AB = BA...
AAB = A(BA) = (BA)A
AABB = A(BA)B = (BA)AB = AB(BA) = (BA)(BA) = B(BA)A
etc.
basically for any two combinations of A and B, as long as the number of A and number of B are constant, they come out as the same answer

Reply 1498

anony-mousse

10

Original post by RuairiMorrissey

I just subbed in x = i

: I

Yeah I think that's the simplest way to do it. You can even avoid using i by just subbing in x^2 = -1 but this requires extreme reliance on the factor theorem if you haven't been taught about imaginary numbers lol

Reply 1499