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August 30, 2017 at 12:16 pm (This post was last modified: August 30, 2017 at 12:17 pm by Longhorn.)
(August 30, 2017 at 10:32 am)Fireball Wrote: Lightweights like us can get ripped on a cup of strong coffee!
Strangely enough, coffee never had that effect on me. Never once felt the slightest buzz, even after a strong cuppa. Maybe I just didn't drink enough of it. But it had bad side effects so I had to stop entirely. Not that I'm denying I'm a lightweight mind you :|
Btw I have stopped buzzing. Genghis Khan seemed a lot taller in illustrations
(August 27, 2017 at 11:04 pm)LadyForCamus Wrote: Feeling awful. Seven days since I ran out of Zoloft. Hopefully it will be in the mail tomorrow. Antidepressant withdrawal is not fun...🙁
August 30, 2017 at 11:22 pm (This post was last modified: August 30, 2017 at 11:31 pm by Kernel Sohcahtoa.)
I just finished proving the following proposition via proof by contradiction: If A and B are sets, then A∩(B– A)=∅.
IMO, this was a pretty cool proof, as it involved a lot of rewriting of statements into equivalent forms in order to generate new forward statements that lead to a contradiction.
Hint
Rewrite the conclusion in the equivalent form "A∩(B– A) ⊆ ∅ and ∅ ⊆ A∩(B– A)." Now, to apply the contradiction method, we take the negation of the re-written conclusion which results in the forward statement, either A∩(B– A) ⊈ ∅ or ∅ ⊈ A∩(B– A) (note we used DeMorgan's laws here). Thus, the keywords, "either/or" suggests proceeding with a proof by cases, where we prove each case separately.
Now, when we proceed with each case, since the keyword "not" is present in each case (⊈ means "is not a subset of"), we can place each statement in parentheses with a negation symbol on the outside of it. For example, A∩(B– A) ⊈ ∅ becomes ~(A∩(B– A) ⊆ ∅). Now, writing the definition of subset for A∩(B– A) ⊆ ∅ in parentheses results in a statement containing the universal quantifier. For example, ~(A∩(B– A) ⊆ ∅) becomes ~(for every x in A∩(B– A), x is in ∅). Distributing the negation symbol through the statement in parentheses gives "there is an x in A∩(B– A) such that x is not in ∅ (note "there is" is the existential quantifier). We then work forward from this statement in order to reach a contradiction (please note that rewriting the statement for the second case works in a very similar manner).
Spoiler alert
In each case, it was interesting to see how moving forward with the proof (in each case) lead us to a statement where the empty set contained an element, which then resulted in a contradiction.
P.S. Please let me know if I've made any errors, as I'm far from perfect and still have lots and lots to learn.
August 31, 2017 at 2:43 am (This post was last modified: August 31, 2017 at 2:46 am by Little lunch.)
What's happened to Isis? Did something happen while I was gone?
S'okay, I just looked up her profile and apparently life's busy and she's away.
Carry on.
I couldn't think of it either.
Bella, used to be a fella.
I didn't agree with all her opinions but I got a soft spot for her.
Good kid. And Scottish, to boot. :-)
