Show that z is a divisor of w iff all multiples of w is a subset of all multiples of z.?

Question

Show that z is a divisor of w iff all multiples of w is a subset of all multiples of z.?

Answer 1

Proof

▬►

Assume z is a divisor of w,

then w is a multiple of z

so a multiple of w is a multiple of a multiple of z

which says, via the associative law, that

a multiple of w is a multiple of z

◄▬

Assume all multiples of w is a subset of

all multiples of z, then clearly w itself is a

multiple of z, hence z divides w

The end

Answer 2

First we show that if the set of all multiples of w is a subset of the set of all multiples of z, then z is a divisor of w.

Let W be the set of all multiples of W.

W = {w, 2w, 3w,... }

Let Z be the set of all multiples of z.

Z = {z, 2z, 3z,... }

If we use proof by contradiction, assume that z is not a divisor of w, that means that w/z is not an integer n.  In other words w is not equal to nz.  But w must be in the set Z, therefore the assumption is incorrect and the first part of the proof is complete.

Now we show that if z is a divisor of w, then the set of all multiples of w is a subset of the set of all multiples of z.

z is a divisor of w, therefore w/z = n (where n is an integer)

or we can write w = nz

Let...

W = the set of all multiples of w = {w, 2w, 3w,... } = {nz, 2nz, 3nz,...}

Z = the set of all multiples of z = {z, 2z, 3z,...}

All of the elements of W must be elements of Z, but not the other way around.  Therefore Z is a subset of W and the proof is complete.

Take care,

David