A few more proofs involving sets.

We look at a few more proofs involving sets.

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[JalebJay]

I believe the last part should be (A and B) or (A and C). All of your ands/ors got flipped somehow when writing the final line. What's on the board comes out to be A or (B and C).

[bigjazbo]

At 8:40, I think DeMorgan rule was applied incorrectly. In the result he derived, x could belong to both B and C but not A and still satisfy the final statement. But the original statement requires membership in A.

[schweinmachtbree]

Note that the hypothesis C =/= ∅ is necessary because in the case C = ∅ it is false that "A x ∅ = B x ∅ ⇒ A = B". Michael also didn't point out where the hypothesis was used in the proof, so you check your skills by seeing where it is used. In my opinion this is one of the best results to illustrate the point that one has to be careful with the empty set: if you ask someone to prove "A x C = B x C ⇒ A = B", most of the time they will come up with an incorrect proof that only works when C =/= ∅; often the mistake in their proof is using the phrase "Let c ϵ C", which one can't do when C = ∅! The moral is that whenever you use the phrase "Let x ϵ X", make sure to check that X is non-empty, because if you don't verify this then your proof is wrong!

[joyofmath]

Hi Michael,

As some others have pointed out there is an error.
You can tell at a glance as your last line is not the right side of what you were trying to prove in the first place (the unions and intersects are switched).
The cause of the error happens at 8:39, again the ands & ors are flipped.
Also at 8:05 there is another option, that x is an element of both B and C, i.e. BnC.

Still frustratingly human. You're in good company.

[jhansam]

Really thankyou. I study for my final exam of Algebra and this video help me a lot!! You are awesome!

[schweinmachtbree]

by the way, to verify that you understand the proof of the first proposition in the video, you might like to prove the related result "for all non-empty sets A, B, C, D we have A x B = C x D ⇒ A = C and B = D" as an exercise. what this result says is that, for non-empty sets, the "function" taking (A, B) to A x B is injective.

[tomatrix]

Where I am from the A-B set is written as A/B and read ‘A without B’ but follows the same definition. Addition: we also write A compliment as A’

[marienbad]

Ok, so this must be like the first way but the other way round, right?

b member B, then (b, c) member BxC, and (b, c) member AxC so b member A, hence A must equal B.

Not sure how you do that last bit the other way round though, surely it is the same proof?

[OriginalSuschi]

At the end it should mean „x is element of (A and B) OR (A and C)
This also follows by logic because if
„x is element of A and (B or C), then the only way this can hold is when it‘s A and B or A and C.“ It‘s like the law of distribution in arithmetic lol

[ДометдеВоргес]

Proof by contradiction: suppose A=/=B, taking the cartesian product with C to both sides of the inequality gives the negation of the hipothesis.

[_judge_me_not]

A question :-
If x, y € [1, 10]
3^(sec²x-1) * √(9y²-6y+2) <= 1 is :-
A)2
B)1
C)4
D)∞
Note :- the root part is not an exponent of 3 (no bracket)

[josepazp]

At the Morgan law is and not or. Its little mistake.

[VenSensei]

I've only seen an Exclusive Or (XOR) as ⊕.

e.g. P ⊕ Q