Proof: A Subset of B implies A/D Subset of B/D

Click For Summary

Homework Help Overview

The discussion revolves around proving that if set A is a subset of set B, then the set difference A/D is a subset of the set difference B/D. Participants are exploring the implications of set operations within the context of set theory.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of elements belonging to A/D and how this relates to their membership in B/D. There are attempts to clarify the definitions and properties of set differences, as well as to streamline the proof process.

Discussion Status

There is an ongoing exploration of the proof structure, with participants providing feedback on each other's reasoning. Some guidance has been offered regarding how to present the proof more concisely, but there is still uncertainty about the final steps and clarity in the argumentation.

Contextual Notes

Participants express confusion over the wording and structure of their arguments, indicating a need for clearer definitions and a more straightforward approach to the proof. There is also a mention of a preference for symbolic representation in their class discussions.

marinalee
Messages
5
Reaction score
0

Homework Statement


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


Homework Equations





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??
 
Physics news on Phys.org
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
 

Similar threads

Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
34
Views
4K
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K