Question

Prove that an even number cannot divide an odd number.
Please, help

Answer 1

Let m is an odd number then m can be expressed as m = 2k +1 for some k in Z
n is an even number, then n can be expressed as n = 2s for some s in Z
then we need prove \(2s\cancel{|} 2k+1\)

Answer 2

@FibonacciChick666

Answer 3

hey I think I can do this!!!!!

Answer 4

Please

Answer 5

ok what does the prime factorization of an even number always include?

Answer 6

2

Answer 7

ok what primes will be included in an odd number's prime factorization?

Answer 8

3

Answer 9

or all odd primes :)

Answer 10

what do we know about prime numbers?

Answer 11

yup

Answer 12

gcd (a, b) =1

Answer 13

ok so you have an odd prime factorization and an even prime factorization, what number will never ever have a match?

Answer 14

I lost. :)

Answer 15

lol it's ok, you'll see give me a sec

Answer 16

\(m=2k+1=p_1p_2p_3...p_m\) where p is an odd prime. \(n=2k=2p_1p_2p_3...p_n\) where p is an odd prime. (the primes can be to powers too, but I'm not typing that. you get the point

Answer 17

write m/n as it's prime factorizations.

Answer 18

Actually, the original problem is : Prove that if G be a finite group of odd order, then subgroups of G cannot have even order.
My work is
Let H <G, |G| = 2m +1
suppose |H| = 2n
suppose |H|| |G|, that is 2n |2m+1
then \(2m\equiv 1\) (mod 2n)
but \(2m\equiv 0\) (mod 2n)
hence \(0\equiv 1\) (mod 2n)
contradiction,
Hence \(2n\cancel{|} 2m+1\)
And I got full credit for this argument.
BUT SHAME ON ME. I review for final and don't get how \(2m\equiv 0\) (mod n\)
You see!! I am such a "good" student, hahaha....

Answer 19

wait you don't get how 2m is 0 mod 2n?

Answer 20

I did, but not now, hahaha

Answer 21

oh lol that's a lot simpler than the one I was making ok what is 2mod2=?

Answer 22

0

Answer 23

ok how about 4 mod2?

Answer 24

0

Answer 25

6mod 2?

Answer 26

0

Answer 27

10987654321098765432 mod 2

Answer 28

0
but it is 2n, not 2

Answer 29

hint it ends in a 2 :)

Answer 30

how else can you write 6?

Answer 31

2*3

Answer 32

okhow about 12?

Answer 33

2*6

Answer 34

ok, and how about 26?

Answer 35

2*13

Answer 36

doesn't that look just like 2 times a number? aka 2n

Answer 37

then?

Answer 38

that's it. because it is even, it is 0 mod 2

Answer 39

you should have used just 2 instead of 2n in my opinions, but your teacher didn't mind that

Answer 40

if it is 2, quite simple, I confuse when it is 2n

Answer 41

ok, so you have a generic even number

Answer 42

I actually don't think this works with 2n... O.O

Answer 43

I can explain using prime factorizations, but using mods I would so use 2

Answer 44

not 2n

Answer 45

hahaha... you confuse also....

Answer 46

oh wait

Answer 47

now i get it

Answer 48

ok i always get uber confused on proofs by contra

Answer 49

okkkk so if 2n|2m+1
we know that 2m+1=0mod 2n right?

Answer 50

yes

Answer 51

if it is divisible by 2n we have to have a 0 congruency, but we also know that 2m+1 is an odd number. That means that is has to be the same as 2m+1=1mod(2n)

Answer 52

but when we subtract one from both we get 2m=0mod2n, which since 2m+1=0mod2n, attains our contradiction

Answer 53

Yeah... I think it is back to me, I got it . Thanks friend.

Answer 54

lol np, give me things I can do any day :D