Question

For a natural number a, let aZ be the set of all integer multiples of a, prove that for all a, b elemelts N, a = b iif aZ=bZ (Z and N are the number sets)

Answer 1

I know its an easy thing to prove concecpt wise, i am more concerned with the notation and how I should write it up.

Answer 2

it should say "b is an element of N"

Answer 3

I think it is important that a,b are natural and not integers otherwise we run into -3Z=3Z in terms of set equality. So maybe there is a place to start in recognizing that we are trying to disprove that there could exist a <> b but aZ=bZ.

Answer 4

Perhaps we go with Let a <> b and WLOG a < b. a*1 is an element of aZ but not an element of bZ because a*1 is positive and the smallest positive element of bZ is b*1

Answer 5

Hence if a <>b, aZ <> bZ, now just do the other way around.

Let aZ <> bZ then WLOG there exists some element of aZ which is not an element of bZ. (we can no longer assume a<b) Let us call that element E. Then E = a*z for some integer z. Also, b*z <> E or else E would be an element of bZ. Therefore a*z <> b*z. Unless z = 0, we can conclude that a <>b. If z is zero though, then E = zero. And we know zero is an element of bZ, so that is also not possible. Therefore since a<>b => aZ<>bZ and aZ<>bZ => a<>b we can conclude that a=b iff aZ=bZ.

Answer 6

I have not read everything yet, but a and b are natural.

Answer 7

what is wlog?

Answer 8

what is <> we dont use this

Answer 9

iif?

Answer 10

If a=b then ax=bx for every integer x then
\[ a\mathbb{Z}=\{ax:x\in\mathbb{Z}\}=\{bx:x\in\mathbb{Z}\}=b\mathbb{Z} \]
this is the trivial part

Answer 11

Conversely if
\[ a\mathbb{Z}=\{am:m\in\mathbb{Z}\}=\{an:n\in\mathbb{Z}\}=b\mathbb{Z} \]
why have to prove a=b