How to Prove the Divisibility Property Theorem?

[annoymage]

Homework Statement

proof the theorem

if a l b and b l a then a=+-b

Homework Equations

The Attempt at a Solution

there exist integer p,q such that ap=b and bq=a, then I've no idea how i can relate it to a=+-b.. clue please T_T

 

[Raskolnikov]

there exist integer p,q such that ap=b and bq=a, then I've no idea how i can relate it to a=+-b.. clue please T_T

$$ap = b = (\frac{a}{q})$$

Multiply by q and divide be a (since b|a <--> a is not 0), giving us:

$$pq = 1$$

Do you follow?

 

[annoymage]

yea yah, i thought that too, but don't know to continue from there too, owhoho, more clue please, ;P

 

[annoymage]

wait, let me think first

 

[Mark44]

Homework Statement

proof the theorem

if a l b and b l a then a=+-b

Homework Equations

The Attempt at a Solution

there exist integer p,q such that ap=b and bq=a, then I've no idea how i can relate it to a=+-b.. clue please T_T

So b = pa and a = qb, for some integers p and q.
Then b = pa = p(qb) = (pq)b.

What can you say about pq?

 

[Raskolnikov]

Remember, p and q are both nonzero integers.

 

[annoymage]

hmm, so pq=1 , hence, p=1 and q=1, hence a=b and b=a, still cannot get +-b, T_T

 

[annoymage]

WAIIITTTT (-1)(-1) also equal 1, wait wait let me think again

 

[annoymage]

so i get

a=b or (a=-b and b=-a)

=> (a=b or a=-b) and (a=b or b=-a)

(a=b or a=-b) is enough to verify it right?

 

[Raskolnikov]

Yep, though technically you're proving, not verifying. "Proving" is a stronger word...makes you sound "smarter/cooler" :P!

 

[annoymage]

owho, thankyou very much