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!
@@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.
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!
I would taught my students your method but if it was me solving it, I would prefer video's method
Thanks! I like your method a lot! Those string of equality type set proofs are always satisfying.
@@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.
Thanks for the video. I was confused so i searched and now I'm never going to make mistake. ❤️
i found this in a book
A minus B is equal to C times five.
can you explain it to me please?
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
Thankyou for explain this problem in easy method
Glad it helped, thanks for watching!
Wow❤ thnks sir ........Like this video .Upload a lot of video ..Ok sir🎉
Thank you! Always more on the way!
Love from India sir 🙏🙏
Thanks
Well explained...
Thank you!
Gora
THANKS
You're welcome!