Question

Prove that: (A\B) intersects (B\A) = empty set

Answer 1

can you just draw a diagram?

Answer 2

otherwise, suppose (A\B) intersects (B\A) is not the empty set. then there exist x such that x is in (A\B) and x is in (B\A). then: x is in A but not in B and x is in B but not in A. contradiction and absurd hence it must be the case that (A\B) intersects (B\A) = empty set

Answer 3

suppose (A\B) intersects (B\A) is not the empty set. then there exist x such that x is in (A\B) and x is in (B\A). then: (x is in A and x not in B) and (x is in B and x not in A. [so x is in A and x is not in A; absurd. and x is in B and x is not in B; absurd as well]. contradiction and absurd hence it must be the case that (A\B) intersects (B\A) = empty set

does it make sense at all?

Answer 4

ooh i get it! i did something similar to that but the proof of contradiction was a bit confusing to me so i kinda get it :)

Answer 5

cool

Answer 6

basically when using the proof of contradiction, you're just assuming its the opposite of what the question is asking??

Answer 7

yes

Answer 8

aaaahh lol thank you :)

Answer 9

np

Answer 10

btw do you use the proof of contradiction for every proof you're given?

Answer 11

no

Answer 12

contradiction is only good for some types of proof just like induction is good for some other proofs. usually when you want to prove that somethings equals something, use contradiction; not always though.