A Summer of Maths (ASoM) 2016

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What's the expected method for Q8 pg 29 in the beardon book? : " Show that for some integer m,

3^{m} = 1

\mathbb{Z}_{7}

. Deduce that 7 divides

1+3^{2001}

".

I just checked a couple of m values and pretty easy to see it works for m=6. So then I said

3^{2001} = 3^{2001-333(6)}=3^{3}

(mod 7) so

3^{2001} +1 = 27+1 = 0 mod 7

. Is there a more general way without having to specifically work out m?

Reply 341

Gregorius

Original post by krishdesai7

Yeah lol. I spend so much time trying to figure out what exactly the difference between co-domain and range is before I finally gave up.

Ooooh, get used to it, there's a lot of that coming your way :smile:

Got to have coimages and kernels and cokernels yet!

Reply 342

username1162972

Original post by EnglishMuon

What's the expected method for Q8 pg 29 in the beardon book? : " Show that for some integer m,

3^{m} = 1

\mathbb{Z}_{7}

. Deduce that 7 divides

1+3^{2001}

".

I just checked a couple of m values and pretty easy to see it works for m=6. So then I said

3^{2001} = 3^{2001-333(6)}=3^{3}

(mod 7) so

3^{2001} +1 = 27+1 = 0 mod 7

. Is there a more general way without having to specifically work out m?

Just FLT.

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Reply 343

EnglishMuon

Original post by physicsmaths

Just FLT.

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oh yea lol for the first bit

Reply 344

drandy76

Original post by EnglishMuon

What's the expected method for Q8 pg 29 in the beardon book? : " Show that for some integer m,

3^{m} = 1

\mathbb{Z}_{7}

. Deduce that 7 divides

1+3^{2001}

".

I just checked a couple of m values and pretty easy to see it works for m=6. So then I said

3^{2001} = 3^{2001-333(6)}=3^{3}

(mod 7) so

3^{2001} +1 = 27+1 = 0 mod 7

. Is there a more general way without having to specifically work out m?

I don't see this as being too different from any expected method tbh

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Reply 345

EnglishMuon

Original post by drandy76

I don't see this as being too different from any expected method tbh

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kk, thanks. Sometimes I just have no idea if im being really dumb as its feels strange not to check every solution you do against some ms :tongue:

Reply 346

drandy76

Original post by EnglishMuon

kk, thanks. Sometimes I just have no idea if im being really dumb as its feels strange not to check every solution you do against some ms :tongue:

I've started to implement a mental checklist, if at every stage I can feasibly follow my working without any undue assumptions, I can be reasonably convinced that my solution is correct

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Reply 347

EnglishMuon

Is that a typo on q6 page 41 of Beardon? In mine it says az+ (b bar) z but i think its meant to say az + b (z bar)

Reply 348

krishdesai7

Original post by Gregorius

Ooooh, get used to it, there's a lot of that coming your way :smile:

Got to have coimages and kernels and cokernels yet!

Oh god no. Why you do this to me. This is downright torture. Aren't kernels bad enough that one must throw cokernals at me too.

Reply 349

username1162972

Aren't kernels the popcorn thing.
One thing i learnt from edexcel physics. Qed.

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Reply 350

EnglishMuon

Original post by physicsmaths

Aren't kernels the popcorn thing.
One thing i learnt from edexcel physics. Qed.

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nah m8, ur thinking of potato batteries. kernels are wat k is for in kfc.

Reply 351

krishdesai7

Original post by physicsmaths

Aren't kernels the popcorn thing.
One thing i learnt from edexcel physics. Qed.

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Yeah kernels are the popcorn thing. But they're the nice kernels. (Don't judge me but I actually like eating them from the bottom of the popcorn bowl) But no; the kernals I've been talking about are the members of the input (domain?) that map to zero under a given transformation. At least that's what I've understood. It's also very plausible that I'm really stupid and Dr Cowley was in fact talking about the popcorn thing

Reply 352

drandy76

Original post by EnglishMuon

nah m8, ur thinking of potato batteries. kernels are wat k is for in kfc.

Close but not quite, Kernel is the guy who started Kfc, the K stands for Kool

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Reply 353

krishdesai7

Um guys, could I have a bit of help with this question?

Beardon book page 11, EX 1.3, Q5
Suppose that the permutation ρ of {1,...,n} satisfies

\rho^3 = I

. Show that ρ is a product of 3-cycles, and deduce that if n is not divisible by 3 then ρ fixes some k in {1,...,n}

Reply 354

EnglishMuon

Original post by krishdesai7

Um guys, could I have a bit of help with this question?

Beardon book page 11, EX 1.3, Q5
Suppose that the permutation ρ of {1,...,n} satisfies

\rho^3 = I

. Show that ρ is a product of 3-cycles, and deduce that if n is not divisible by 3 then ρ fixes some k in {1,...,n}

Think about the orbit decomposition of any permutation, e.g. why can't we have 2 cycles or 4+ cycles in this case? Also remember that 1-cycles fix elements so by definition any power of one cycles is the identity so they don't effect anything.

Further hint:

Spoiler

Reply 355

krishdesai7