Question

Prove if if a ⊥m and b ⊥ m then ab⊥ m

Answer 1

Remind me the significance of that notation?

Answer 2

Its co prime

Answer 3

coprime
is coprime to
number theory
x ⊥ y means x has no factor greater than 1 in common with y. 34 ⊥ 55.

Answer 4

http://en.wikipedia.org/wiki/List_of_mathematical_symbols

Answer 5

Seems to beg for a proof by contradiction, eh?

Answer 6

@SmoothMath I thought you had sorta of ideas on how to at least solve this problem

Answer 7

Iw as thinking of some thing like If a|m and b|m, then we know a=k*m and b=l*m for some constants k and l.

lcm(a,b) is the least common multiple of a and b.

Multiples of a are 1km, 2km, 3km... etc
Multiples of b are 1lm, 2lm,...

We search the two lists until we find a common multiple. Which will have the form c*m for some constant c.

i.e. lcm(a,b) = cm
Which is clearly divisible by m.

Answer 8

is that correct ?

Answer 9

Hmm I think you're on the wrong track...

Answer 10

why ?

Answer 11

if they are co prime they sould have the same remainder I guess

Answer 12

You're assuming a|m and b|m, but if a is coprime to m, then a does not divide m.

Answer 13

to make them equal then you have to multiply by a number

Answer 14

The proof I was trying to construct was something like:
Assume a comprime to m and b coprime to m, but AB not comprime.
Show a contradiction.

Answer 15

ok I see

Answer 16

Do you know any refrence that talks about the co prime theorems If I have one of them I can figure out how they proof the others I dont think you need to go the contrdaction way

Answer 17

@SmoothMath

Answer 18

Bah. I give up on this. Sorry, bro.

Answer 19

Let be a prime. Then every integer that is not a multiple of is coprime to , so the non-zero integers mod form a group. This group has elements, so by Lagrange's theorem every element of the group has order dividing . Therefore, mod for every integer that is not a multiple of . This is Fermat's little theorem

Answer 20

I found out this if a ^ m-1 = 1 (mod m) b ^ m -1 = 1 mod m