# DSE 2012 Option A Classic List Threaded 18 messages Open this post in threaded view
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## DSE 2012 Option A

 QUESTION 18. A and B are two non empty sets. A-B = {x belongs to A | x does not belong to B} and A + B = (A - B) U (B- A) Consider the following statements is true or false Statement 1: A + B = B implies A IS SUBSET OF B The doubt regarding this problem is for statement 1 to be true A has to be a empty set. Which contradicts the given condition. Please help regarding this.
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## Re: DSE 2012 Option A

 getting same problem
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## Re: DSE 2012 Option A

 In reply to this post by MI Hi.. :) Given, A+B = B ⇒ A+B ⊆ B and B ⊆ A+B ⇒ (A-B) ∪ (B-A) ⊆ B   [From A+B ⊆ B] ⇒ (A-B) ⊆ B ⇒ A ⊆ B Hence, proved! :)
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## Re: DSE 2012 Option A

 Hi Duck :) Th problem I am facing is as follows A + B = (A-B) U (B-A) This can also be written as (using venn diagram) as A + B = (A U B) - (A ∩ B) ......(I) Lets assume A⊆B => (A U B) = B   ALSO  (A ∩ B) = A    ...........(II) Using (I) AND (II) A+B=A-B WHICH IS TRUE ONLY WHEN B= ∅ Doesn't this contradict with the initial assumption " A and B are two non-empty sets"? Please let me know if there are any logical fallacy in my approach. Thanks :)
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## Re: DSE 2012 Option A

 In reply to this post by duck How did u infer  A ⊆ B from (A-B) ⊆ B (in general, given A and B being non empty) ? ---  "You don't have to believe in God, but you should believe in The Book." -Paul Erdős
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## Re: DSE 2012 Option A

 In reply to this post by duck duck wrote Hi.. :) Given, A+B = B ⇒ A+B ⊆ B and B ⊆ A+B ⇒ (A-B) ∪ (B-A) ⊆ B   [From A+B ⊆ B] ⇒ (A-B) ⊆ B⇒ A ⊆ B Hence, proved! focussing at the bold part of your text: A-B means all those elements of A which are not present in B. ie if we remove all the elements of A which are in B and if still this set is a subset of B implies it still contains some elements (at least 0 elements) of B. But this contradicts our definition of A-B since A-B cant have any element which is present in B. So A has to be an empty set... what am I missing? ---  "You don't have to believe in God, but you should believe in The Book." -Paul Erdős
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## Re: DSE 2012 Option A

 In reply to this post by Sinistral Hey Thanks Sinistral for pointing out that.. :) Sorry for the earlier proof. Please ignore that. Please see the below proof: Given: A+B = B To show : A⊆B Proof: Let, x∈A . If we could show that x∈B then, we'll be done! Lets prove it by contradiction. Supose, x∉B ⇒ x∈ (A-B) ⇒ x∈ A+B ⇒x∈ B which is a contradiction. Therefore, x∈B. Hence, A⊆B :)
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## Re: DSE 2012 Option A

 This post was updated on . In reply to this post by MI Hey MI.. :) What you're showing is "If A⊆B then A+B=B". And the above statement doesnot hold as pointed by you. Because, if A⊆B then, A-B will be an empty set. ⇒A+B = B-A ≠ B ( as A is a non empty set) :)
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## Re: DSE 2012 Option A

 This post was updated on . In reply to this post by duck ok mathematically ur proof is just fine. but somehow its not convincing me. (read below quoted message) duck wrote Hey Thanks Sinistral for pointing out that.. :) Sorry for the earlier proof. Please ignore that. Please see the below proof: Given: A+B = B To show : A⊆B Proof: Let, x∈A . If we could show that x∈B then, we'll be done! Lets prove it by contradiction. Supose, x∉B ⇒ x∈ (A-B) ⇒ x∈ A+B ⇒x∈ B which is a contradiction. Therefore, x∈B. Hence, A⊆B now suppose this proof is good. and  (given) A+B = B ⇒ (to prove) A⊆B lets verify this. I'll go in sync with ur proof. now since its given that A⊆B and u said it will be all fine if for x∈A if we can show that x∈B. And it has been proved also. so now lets verify it:     x∈B ⇒ x∉ A-B     x∈A ⇒ x∉ B-A ⇒ x∉ A+B ⇒ x ∉  B contradiction. why??? or putting it more simply can you give me example of such sets A and B which satisfy this: A+B = B ⇒ A⊆B ---  "You don't have to believe in God, but you should believe in The Book." -Paul Erdős
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## Re: DSE 2012 Option A

 This post was updated on . First of all, i didnt assume A⊆B. It was what i proved. Secondly, i couldn't understand. "so now lets verify it:     x∈B ⇒ x∉ A-B     x∈A ⇒ x∉ B-A ⇒ x∉ A+B ⇒ x ∉  B contradiction. why??? "Its not at all sync with the proof which i gave. I think you haven't read the proof carefully Or you're misinterpreting! Thirdly, A+B = B can never hold for any non empty sets A and B. So, the antecedant is always false. Hence, the statement A+B=B ⇒ A⊆B will hold trivially. :)
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## Re: DSE 2012 Option A

 I went ahead with all these proofs and verification because I dint want to use the line A+B=B ⇒ A⊆B will hold trivially at the very first instance. Now this will hold trivially; doesn't it violate the initial assumption given in the question A and B are two non empty sets??ofcourse statements 1,2,3 are true but only if we assume A and B to be any sets (including non empty sets). whereas statement 2 & 3 are valid for even non trivial cases. ---  "You don't have to believe in God, but you should believe in The Book." -Paul Erdős
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## Re: DSE 2012 Option A

 If the antecendant is false then, P⇒Q holds. Thats what "trivially hold" mean. Now, A+B = B ⇒ A⊆B holds even for non empty sets A and B. As the antecendant (A+B = B) is always false for non empty sets A and B. :)
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## Re: DSE 2012 Option A

 thank you so much. :) In your last reply you said "so the antecedent is always true". this got me confused. I think now its clear. thanx once again :) ---  "You don't have to believe in God, but you should believe in The Book." -Paul Erdős
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## Re: DSE 2012 Option A

 i didn't get d last step . when x belongs to a +b why does it imply that it belongs to b ? doesn't it mean that it can be in a or b ?
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## Re: DSE 2012 Option A

 because its given that A+B=B ---  "You don't have to believe in God, but you should believe in The Book." -Paul Erdős
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## Re: DSE 2012 Option A

 ooh ya thanks :-)