# Basic Set Theory Proof

## Homework Statement

Prove that if A is a subset of B then A/D is a subset of B/D.

## The Attempt at a Solution

Consider element x of A. Since A is a subset of B then for all x element of A, x is an element of B. Consider element x of A/D. If x is an element of D then x is not a member of A and thus it does not matter if x is an element of B. If x is not an element of D and is an element of A than x is also in B because x is an element of A. Thus, A/D is a subset of B.

Not even sure if this much is correct. How do I prove this basic "subtracting a set from both subsets" identity??

## Answers and Replies

For one, you should not say "If x is an element of D then x is not a member of A ...", you should say "If x is an element of D then x is not a member of A/D", likewise for B/D.

But you take a lengthy confusing approach at this point. it would be more concise to say "Consider element x of A/D.." (here comes my part): x is an element of A/D implies x is in A and x is not in D....

Okay, I'll change that. But how do I finish the proof?

Pretty much like you did, I think, just with that shorter way of presenting it. Talk about why it implies x in is in B/D instead of considering B and D separately.

Oh! I think I get it. So the proof is: Consider x element of A/D. This means x is an element of A and not of D. This means x is a member of B, because all members of A are members of B, and since we already know that x is not an element of D we can combine these two facts and say x is an element of B/D.

It seems like you missed the point of considering x is in A/D instead of considering x is D and x is not in D.

You begin well here: "Consider element x of A. Since A is a subset of B then for all x element of A, x is an element of B. Consider element x of A/D." but then you begin taking cases of x in D or not in D. Don take cases. Just go straight into x is in A/D implies... whatever it implies. Leading into something about x is in B/D, right?

Um...I don't get it. :( What does A/D imply other than x is in A and not in D?

Oh, sorry, you're right. You're answer is good, I was thrown off by all the words. (we use all symbols in my class) :) Yay set theory *waves flag*

Perfect! Thanks! :D