IB math HL may 2008 no 5, thanks millions

johaneshalim · 2009-03-22 04:30:39

let p =( 2^k ) + 1, k elements of positive integers, be a prime number and let G be the group of integers 1, 2,..., p-1 under multiplication defined modulo p.
a). by first considering the elements 2^1 , 2^2 , 2^3, ..., 2^k and then the elements 2^(k+1), 2^(k+2), ... show that the order of the element 2 is 2k.
b). deduce that k = 2^n for n elements of N(natural numbers).

JaneFairfax · 2009-03-22 07:41:16

johaneshalim · 2009-03-22 22:34:37

thanks miss Jane for the help,thanks a lot, and how to deduce that k = 2^n

JaneFairfax · 2009-03-22 22:46:05

Oops, I didnt see the last part.

The last part is easy. The order of an element must divide the order of the group, i.e.

johaneshalim · 2009-03-22 23:29:25

thanks a lot, but for the second, i meant there is must be a way to show the works, right master Jane? please forgive my bad math knowledge, can i just give examples by substitute any values of k and n consecutively? then in the end i write ur answer? thx again

JaneFairfax · 2009-03-23 00:04:08

The second part is straightforward.

The first part is more important. Did you understand my proof? I skimmed through bits of the proof, partly to save time typing but also partly hoping for you to ask questions. You are reading and making sure you understand what Ive written, are you?

johaneshalim · 2009-03-23 00:53:37

i did read the first part, and i think i understand....woo hoo thanks jane, thanks millions. i am new to this subject, havent got enough basic to do this, and now am reading and studying it , thanks for ur help. GOD bless you...see ya

Jacob10 · 2009-03-24 05:06:33

Hi, I am also really stuck on this question, I cant remember ever understanding a maths question less! I understand the second part, but I don't compleatly understand your proof for the first part, Jane. I understand that
-The order of G is 1, and as

if there is no number, m, lower than 2k for which

then 2k is the order of 2. However, I don't understand the proof from there onwards. Could you write that part out with more steps/reasoning shown?
Thanks.

JaneFairfax · 2009-03-24 11:05:57

Yes, I was expecting some questions to be asked.

I was doing a proof by contradiction. I assumed that the order of 2 was some positive integer m < 2k.

So either or .

would lead to

Since

is prime, either or .

Both are impossible since both

and are less than .

The other case

leads to a similar contradiction, with having to divide one of two numbers smaller than itself.

Hence

for any

I hope this makes it clearer.

Math Is Fun Forum

#1 2009-03-22 04:30:39

IB math HL may 2008 no 5, thanks millions

#2 2009-03-22 07:41:16

Re: IB math HL may 2008 no 5, thanks millions

#3 2009-03-22 22:34:37

Re: IB math HL may 2008 no 5, thanks millions

#4 2009-03-22 22:46:05

Re: IB math HL may 2008 no 5, thanks millions

#5 2009-03-22 23:29:25

Re: IB math HL may 2008 no 5, thanks millions

#6 2009-03-23 00:04:08

Re: IB math HL may 2008 no 5, thanks millions

#7 2009-03-23 00:53:37

Re: IB math HL may 2008 no 5, thanks millions

#8 2009-03-24 05:06:33

Re: IB math HL may 2008 no 5, thanks millions

#9 2009-03-24 11:05:57

Re: IB math HL may 2008 no 5, thanks millions

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