-3
$\begingroup$

To prove:$ \left( A-B\right) \cup B = A \cup B$
I want it to be done in two ways:1. The algebraic way and 2 Using the method where we say,for example, $x \in A \text{ or }x\notin B $(I dont know what that method is called.)
I have done the algebraic method, somebody help me do it using the second method where we assume that let x be a element that belongs to LHS and then we prove that LHS is a subset of the RHS. Then we do it all again but with y such that it belongs to RHS and finally we prove that RHS is a subset of LHS.

$\endgroup$
  • 1
    $\begingroup$ What are your thoughts on the matter? What have you trid? What sort of proof are you looking for - an algebraic proof like the one you wrote below, or a prose proof emphasizing quantifiers? $\endgroup$ – Carl Mummert Nov 14 '13 at 13:55
  • 1
    $\begingroup$ @CarlMummert The OP already answered the question $\endgroup$ – Amr Nov 14 '13 at 13:56
  • $\begingroup$ @CarlMummert I posted this question earlier here:math.stackexchange.com/questions/566763/… but was asked by lord_farin to make a separate question for it. $\endgroup$ – Shaurya Gupta Nov 14 '13 at 14:00
  • 2
    $\begingroup$ @Amr: yes, but I don't think that is relevant to the issue of the question being written in a way that is not yet suitable for this site. Here is a link to "How to ask a good question": meta.math.stackexchange.com/questions/9959/… $\endgroup$ – Carl Mummert Nov 14 '13 at 14:03
  • $\begingroup$ @CarlMummert I think it's relevant. Some users post questions and answer it, but I believe their questions are interesting/non-trivial. I consider this question is trivial. It is uninteresting to see a lot of trivial questions on the site. I know trivial questions are being asked, but at least the posters don't know the answer, but I don't think its a good idea to tolerate trivial questions being asked by users who know their answer $\endgroup$ – Amr Nov 14 '13 at 14:07
1
$\begingroup$

Let $x\in (A-B)\cup B$, then $x\in A-B$ or $x\in B$. If $x\in A-B$, then $x\in A$ and $x\notin B$ and so $x\in A\cup B$. If $x\in B$, then $x\in A\cup B$. Hence $(A-B)\cup B\subset A\cup B$. Let $x\in A\cup B$, then $x\in A$ or $x\in B$. If $x\in A$, then $x\in A\cup B$. If $x\in B$, then $x\in A\cup B$. Hence $A\cup B \subset (A-B)\cup B$. Thus $(A-B)\cup B=A\cup B$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Which method should be used when? $\endgroup$ – Shaurya Gupta Nov 15 '13 at 13:00
  • $\begingroup$ You need to show that the left-hand side is a subset of the right-hand side and that the right-hand side is a subset of the left-hand side. $\endgroup$ – 1233dfv Nov 15 '13 at 13:59
  • $\begingroup$ What I am asking is which method should i use the algebraic or the one u used? $\endgroup$ – Shaurya Gupta Nov 15 '13 at 15:54
  • $\begingroup$ The standard way to prove these identities is the way I did it. However, if you can prove this in a different way that is fine too. $\endgroup$ – 1233dfv Nov 15 '13 at 17:27
  • $\begingroup$ This is the best way to prove that two sets are the same $\endgroup$ – Stella Biderman Mar 24 '14 at 2:13
1
$\begingroup$

$ \left( A-B\right) \cup B = (A \cap B') \cup B = (B \cup A) \cap (B \cup B') = (A \cup B) \cap X = A \cup B $

Note that $A,B \in X$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.