Thread: Set Theory

Show that for all m,n in N, m<n if and only if m is a proper subset of n.

Here is my reasoning.

Assume m is a proper subset of n. Then m doesn't equal n by definition of a proper subset. Since m is a subset of n, then m must be less than n.

Is this a concrete enough?

Originally Posted by taypez

Show that for all m,n in N, m<n if and only if m is a proper subset of n.

Here is my reasoning.

Assume m is a proper subset of n. Then m doesn't equal n by definition of a proper subset. Since m is a subset of n, then m must be less than n.

Is this a concrete enough?

The reasoning looks ok so far to me, but you have an "if and only if" which means you have to do the reverse. I'd probably try to say what you did a little fancier, like:

Assume m is a proper subset of n. Then there is some x in n that is not in m (of course we'd use the symbols here, you noe the one that looks like a capital e and then the same symbol with a slash through it). Thus |m|<|n| for all m and n.

Are you sure the question didn't ask |m|<|n|? That seems to be what you're proving.