Symmetric Difference Proof for Sets A, B, and C

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SUMMARY

The discussion focuses on proving the symmetric difference identity for sets A, B, and C, specifically that A%B=C if and only if B%C=A and A%C=B. The proof begins by analyzing the elements of the symmetric difference and their relationships to the sets involved. Participants suggest methods to approach the proof, emphasizing the need to manipulate the definitions of symmetric difference and set operations effectively.

PREREQUISITES
  • Understanding of set theory, particularly symmetric difference
  • Familiarity with set operations such as union and intersection
  • Knowledge of logical implications in mathematical proofs
  • Basic skills in manipulating algebraic expressions involving sets
NEXT STEPS
  • Study the properties of symmetric difference in set theory
  • Learn about set identities and their proofs
  • Explore logical equivalences in mathematical reasoning
  • Practice proving set relationships using Venn diagrams
USEFUL FOR

Students of mathematics, particularly those studying set theory, educators teaching proof techniques, and anyone interested in formal logic and mathematical reasoning.

bedi
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Homework Statement



Let % be the symmetric difference.
Prove that for any sets A, B, C;

A%B=C iff B%C=A iff A%C=B

Homework Equations



(I will use forwardslash as I cannot find the backslash on my keyboard.)

The Attempt at a Solution



Take x in A%B. Then x is either in A/B or in B/A and in C. Choose the former. Then x is in A and x is in C which implies x is in AnC. This is what I know.
Now, using these I will prove that B%C=A.
Take x in (B/C)U(C/B). Then x is either in the first or in the second. Actually it cannot be in the first because that would imply that x is not in C, which contradicts that x is in C. So x is in C/B. But I already know that x is in A, so nothing to prove in that direction. Take x in A... I don't know how to proceed.
 
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hi bedi! :smile:
bedi said:
A%B=C iff B%C=A …

(to prove from left to right …)

you need to start "B%C = B%(A%B) = … " :wink:
 

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