Question

what is teh slope of 3x+4y=1

Answer 1

hi

Answer 2

you can close this

Answer 3

ok

Answer 4

how is ur cat

Answer 5

hes fine. Just had to get boosters

Answer 6

wife got stuck at work

Answer 7

i see. ah. so u had to take him. ur a good cat dad.

Answer 8

you have no idea

Answer 9

lol

Answer 10

thanks for helping out. the hw over the weekend almost made me want to quit math. lol

Answer 11

lol

Answer 12

but i feel better about it now that everyone else struggled too

Answer 13

that's how it should be:)

Answer 14

what you working on?

Answer 15

why?? shouldn't math be roses and cotton candy??

Answer 16

:)

Answer 17

cptr 10...orders...the proofs. i cant' seem to find the logic to create a proof

Answer 18

yeah that's a tough section

Answer 19

not to create...but to make the proof work out

Answer 20

do u have the book?

Answer 21

yeah

Answer 22

E2. i tried it over and over but couldn't make something happen

Answer 23

m and n relative prime one?

Answer 24

yes.

Answer 25

i don't know how to integrate divisors with orders

Answer 26

there is a disconnect in my mind of how the two relate

Answer 27

give me a few to think about it

Answer 28

k.

Answer 29

Let 0 < i < m และ 0 < j < n
suppose ai = bj some i,j
e = am = ei = (am)i = ami = (ai)m = (bj)m =bjm
so n | mj but gcd(m,n) = 1
n | j
It's conflict because j<n
ai bj for all integer i,j which not 0

Answer 30

its chinese to me

Answer 31

http://pirun.ku.ac.th/~fscisut/417321/ex/order/EngPage4.html

Answer 32

ok

Answer 33

assume by contradiction there is a j<m and k<n such that a^j = a^k
the reason j<m is because every thing after a^m is just a repeat of some multiple of a, call it a^j

Answer 34

does this make sense?

Answer 35

if there is a v > m st a^v is defined then that same element is a^j for some j < m

Answer 36

so m is the smallest positive integer?

Answer 37

smallest positive intiger st a^m = e

Answer 38

ok

Answer 39

the only important thing is to understand why we can say j<m and K < n

Answer 40

its just because if they were greater, we could find that same element before with hit e

Answer 41

say a^3 = e
we have
a,a^1,a^2,a^3,a^4
note that a^4 = a^1

Answer 42

yes, i understand that

Answer 43

ok pellet, my dad is here I just have to drop him off about 4 blocks away. can you stay?

Answer 44

10 mins at most

Answer 45

yes, no worries. i will be here.

Answer 46

take ur time

Answer 47

still here

Answer 48

yes. i'm here. i reviewed what u wrote...and it makes sense so far

Answer 49

assume by contradiction there is a j<m and k<n such that a^j = a^k

\[e=a^m\]
since m is the order of a, we know that a raised to any multiple of m is also = e
\[e=a^m = a^{mi}\]

Answer 50

k

Answer 51

\[e = a^m = a^{mi}= (a^i)^m\]
since a^i = b^j
we have
\[e = a^m = a^{mi}= (a^i)^m = (b^j)^m\]

Answer 52

so assume by contradiction that a^i = b^j
\[e= a^m = a^{mi}= (a^i)^m= (b^j)^m\]

Answer 53

sorry about that

Answer 54

this make sense?

Answer 55

ok. maybe a stupid question...when was it established that a^i = b^j

Answer 56

ohhh

Answer 57

its a contradiction

Answer 58

the thing we are trying to prove is that if ord(a) = m, and ord(b) = n then a^i=b^j cant be possible when (a,m) = 1
so yes we assume by contradiction that a^i = b^j for some I,j

Answer 59

(n,m)=1?

Answer 60

i get it...but it just seems to far fetched for us group theory students to figure out on our own without proper training.

Answer 61

*too

Answer 62

(n,m) = 1 just means n,m are relative prime

Answer 63

ill show you when we are done the login into coming up with it on your own

Answer 64

yes, just confirming (n,m)=1 not (a,m)=1

Answer 65

o yeah:)

Answer 66

so far we have

PF:
let G be a group
let a,b be in G s.t. ord(a) = m and ord(b) = n where n,m are relatively prime
assume by contradiction there exists I,j in Z s.t. a^i = b^j
then there is i<m and j<n st a^i - b^j
then
\[e=a^m = a^{mi} = (a^i)^m = (b^j)^m\]

Answer 67

\[(b^j)^m = b^{jm}\]
so we have
\[e = b^{jm}\]

Answer 68

right?

Answer 69

yes

Answer 70

got it

Answer 71

since n is the order of b, n is the smallest number such that b^n = e
so we know that jm is a multiple of n because b^(jm) = e
so n divides jm
n|jm
we know n does not divide m because they are relative prime, and we know n does not divide j because n<j, so we have a contradiction that n|jm

Answer 72

so far we have

PF:
let G be a group
let a,b be in G s.t. ord(a) = m and ord(b) = n where n,m are relatively prime
assume by contradiction there exists I,j in Z s.t. a^i = b^j
then there is i<m and j<n st a^i - b^j
then
\[e=a^m = a^{mi} = (a^i)^m = (b^j)^m = b^{jm}\]
that is jm is a multiple of ord(b) = n
so n|mj
but (n,m) = 1 and j<n
by assumption this is a contradiction and no such i,j exist.
qed.

Answer 73

ok. i follow the logic...but most of the kids in my class couldn't figure this out. i feel like it requires more proof experience...as we have not learned by contradiction yet

Answer 74

as far as trying to come up with this.
first thing you ask yourself is
what do I need to show?
well we need to show that a^i = b^j for some I,j given the conditions we have
how do we show this is true^^^^^ THE HELL IF I KNOW
this screams contradiction. The fact that I cant even figure out how to say what we need to show screams contradiction because how do we show something for all I,j without induction ( im sure there is an inductive proof).

Answer 75

gerardo assigned it adn it was one of the problems due to day. i did the best i could but eventually just wrote him a short note that i was stuck with a frowny face

Answer 76

ah. ok. so its a clue that i may have to use contradiction when i am pulling my hair out

Answer 77

then we say
ok how do we start the contradiction proof
assume the thing a^i = b^j
and write one thing you know for sure
a^m = e (note we could have started with b^n = e and done the same thing)

Answer 78

so far we have had to come up with nothing..
so
here is the tricky part
a^(im) = e
once we have that, the proof is all natural

Answer 79

yeah..that is how i started...but never finished. its ok. its a learning experience

Answer 80

yeah for sure

Answer 81

thank u for ur help...gotta run...husband getting off of work soon.

Answer 82

this order stuff has the most ambiguous proofs, they are odd but really they are all alike. I think if you understand this one well. you will feel much better with the others

Answer 83

np

Answer 84

ill be around later whenever you want. just text

Answer 85

i will redo it over teh next couple of days

Answer 86

i have an exam in grp thry on thursday

Answer 87

ill show you the other "method" for order proofs with m = qn+r

Answer 88

thx zach...much appreciated!!

Answer 89

np sorry about today

Answer 90

don't worry...not a big deal. talk to u soon!!

Answer 91

bye