Proof: A'-B' = B-A (Double Inclusion) | Set Theory

We prove A'-B'=B-A. That is, the complement of A minus the complement of B equals B minus A. We prove this using double inclusion, meaning we prove both sets are subsets of each other, which by definition establishes set equality. This will require basic applications of set theory definitions. #SetTheory
What is a Set Complement? • What is a Set Complement? z
What is Set Subtraction? • What is Set Subtraction?
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[@tahiyaorsha2564]

Thanks for the video. I was confused so i searched and now I'm never going to make mistake. ❤️

[@tomkerruish2982]

You could also apply the definition of set difference, along with the commutativity of set intersection and the fact that (X')' = X for any set X (set complement is an involution). Explicitly:
A' - B' = A' int (B')' = A' int B = B int A' = B - A
However, your method of showing mutual set inclusion is more broadly applicable to such problems.
Good video!

[@inmathswetrust5293]

I would taught my students your method but if it was me solving it, I would prefer video's method

[@WrathofMath]

Thanks! I like your method a lot! Those string of equality type set proofs are always satisfying.

[@tomkerruish2982]

@@WrathofMath Thank you! I was likely inspired by Conway's 'one-line' proofs from On Numbers and Games, which are simply strings of equalities, often more than one line long.

[@marleny750]

i found this in a book
A minus B is equal to C times five.
can you explain it to me please?

[@bhikaricharansahu4206]

Thankyou for explain this problem in easy method

[@WrathofMath]

Glad it helped, thanks for watching!

[@Shaurya-Kushwaha]

Sir can I solve it like this ? : -
A' = U-A , B' = U-B
A'-B'= (U-A)-(U-B)
= U-A-U+B
=U-U+B-A = B-A

[@kartikmishra8615]

Thanks

[@Arzoo-wc5uk]

Well explained...

[@WrathofMath]

Thank you!

[@coolATHARV001]

Love from India sir 🙏🙏

[@vinebadboi001]

Wow❤ thnks sir ........Like this video .Upload a lot of video ..Ok sir🎉

[@WrathofMath]

Thank you! Always more on the way!

[@quak1037]

THANKS

[@WrathofMath]

You're welcome!