I just finished my discrete math final today, thanks so much, I used this channel all semester to understand graph theory concepts. It helped me out a ton!!
@@WrathofMath let (a,b) in Ax(BUC) a inA and b in BUC a in A and b in B or b in C a in A and b in B or a in A and b in C (a,b) in AxB or (a,b)in AxC (a,b) in(AxB)U(AxC). Isn't it two way? I used Union for better symbolism
Yes indeed, the reverse of the first direction is sufficient for the second direction! That's what I said in the video - I just did not emphasize the equivalence aspect connecting the statements, but of course when the implication is in both directions, equivalence is what we're dealing with. In your proof, one typo/mistake you've got is the "(a,b) in AxB and (a,b)in AxC" in the second to last line. I think you meant "(a,b) in AxB or (a,b) in AxC". But good proof otherwise!
And I've got gifts for all the boys and girls! ....as long as they're leaning set theory! What little one doesn't want Charles Pinter's book of set theory in their stocking??
@@WrathofMath Please do help us with concepts of group theory and linear algebra, such as homomorphism, kernel, linear maps... It would be a great Christmas gift! :)
Check out my set theory playlist for more set equality proofs! ua-cam.com/play/PLztBpqftvzxWUF1psif8R7aUph4tsIuNw.html
I just finished my discrete math final today, thanks so much, I used this channel all semester to understand graph theory concepts. It helped me out a ton!!
So glad to hear it, hope you did well on the final! Thanks a lot for watching!
What if it was a union instead of intersection, does the same steps apply then?
If we use equivalence in your part, isn't it ok for the reverse;
Thanks for watching, I am not sure what you mean - could you clarify?
@@WrathofMath let (a,b) in Ax(BUC) a inA and b in BUC a in A and b in B or b in C a in A and b in B or a in A and b in C (a,b) in AxB or (a,b)in AxC (a,b) in(AxB)U(AxC). Isn't it two way?
I used Union for better symbolism
Yes indeed, the reverse of the first direction is sufficient for the second direction! That's what I said in the video - I just did not emphasize the equivalence aspect connecting the statements, but of course when the implication is in both directions, equivalence is what we're dealing with. In your proof, one typo/mistake you've got is the "(a,b) in AxB and (a,b)in AxC" in the second to last line. I think you meant "(a,b) in AxB or (a,b) in AxC". But good proof otherwise!
@@WrathofMath you are right. I fixed it
Santa Math is back!
And I've got gifts for all the boys and girls! ....as long as they're leaning set theory! What little one doesn't want Charles Pinter's book of set theory in their stocking??
@@WrathofMathSo true!
You make it so simple.
That's my job! Thanks for watching!
@@WrathofMath Please do help us with concepts of group theory and linear algebra, such as homomorphism, kernel, linear maps... It would be a great Christmas gift! :)