ajprincess Group Title Please help:) Prove that if a|b and b|a, then $$a=\pm b$$ one year ago one year ago

1. sauravshakya Group Title

Let x=a/b then since b divides a, x is a integer other than 0 Now, Since a also divides b b/a is also integer other than 0 Here, b/a =1/(a/b) =1/x So, x can be either +1 or -1

2. sauravshakya Group Title

Now, when x=1 then a/b=1 ===>a=b when x=-1 then a/b=-1 =====>a=-b

3. sauravshakya Group Title

Thus, a=+-b

4. ajprincess Group Title

Thanx a lottt @sauravshakya.:)

5. jiteshmeghwal9 Group Title

@sauravshakya i couldn't understand ur statement  Now, Since a also divides b b/a is also integer other than 0 

6. jiteshmeghwal9 Group Title

@ajprincess can u tell me the statement in easier way ??

7. ajprincess Group Title

a|b is a notation that means a divides b and b/a is an integer. b|a is a notation that means b divides a and a/b is an integer.

8. ajprincess Group Title

a|b can also be expressed in this way b/a=x, x is an integer. If x is o , b must be equal to 0. b|a can also be expressed in this way a/b=y, y is an integer a/0=infinity. infinity is not an ineger that is y we give the condition that x is an integer except 0. getting it @jiteshmeghwal9?

9. ParthKohli Group Title

Hmm, I take it that you meant a/0 is not defined.

10. jiteshmeghwal9 Group Title

ya! I gt it a little but $$\frac{a}{b}$$ doesn't this mean b divides a

11. ajprincess Group Title

ya u can take it that way @parthkohli

12. ajprincess Group Title

a|b and a/b are different.

13. ParthKohli Group Title

$a\ne 0, b \ne 0$because if one of them is zero, then one of $$a|b$$ or $$b|a$$ will be false. =)

14. jiteshmeghwal9 Group Title

ohh ! i thought it is a/b i gt it completely, thanx :)

15. ParthKohli Group Title

$$b|a$$ is the same as $$bx = a :x\in \mathbb{Z}^{+}$$

16. ajprincess Group Title

ya u are absolutely right:) @ParthKohli and welcome:) @jiteshmeghwal9

17. ParthKohli Group Title

Nice proof! @sauravshakya