How to do a PROOF with CARTESIAN PRODUCTS - Discrete Mathematics
ฝัง
- เผยแพร่เมื่อ 25 ก.ค. 2024
- We learn how to do formal proofs in set theory using intersections, unions, complements, differences, and cartesian products (or cross products).
0:00 - [Intro]
0:32 - [Language of Cartesian Products]
2:33 - [Proof #1]
6:29 - [Proof #2]
10:29 - [Proof #3]
#SetTheory #Proofs #CartesianProducts
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Discrete Mathematics 1: • Discrete Math (Sets, L...
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Check out my new course in Set Theory: trevtutor.com/p/master-discrete-mathematics-set-theory
It comes with video lectures, text lectures, practice problems, solutions, and a practice final exam!
Thank you so much for this informative video!
Thanks man it helped a lot
Example three was wrong to see if we were following, pairs (a, d) or (c, b) are not surely in both, thanks to others who pointed it out.
The legend has returned!
thanks!
In example 3, the initial proof that the left side is a subset or equal to the right side makes sense to me. But the reverse proof to show both sides are equal does not. It seems that it should be: (I will use E for element of)
(x,y) E (AuC) x (BuD)
(x E A or x E C) and (y E B or y E D)
This yields four possible combinations,
x E A and y E B, or x E A and y E D, x E C and y E B, or x E c and y E D.
In other words, if for example each set A,B,C, and D contain a single element, none of them the same, then the left side of the equation would yield 2 unique ordered pairs, and the right side would yield 4 unique ordered pairs. The left side would be a subset of the right, but they can not be equal. Right?
Am I missing something?
Also, I’m new to set theory, but it seems that the “or”s in the above case should be inclusive, because in a union an element can be part of either set or both sets in the union. Is that right?
Thank you for these videos btw, it’s making a nice time of learning set theory!!
You are correct. It's not an equality.
thats exactly what i was confused about when trying the question myslef, thanks for pointing it out
Absolutely you are correct. Thank you...
the step where he asks "where can x/y go" is wrong because he added 2 new possible conditions for x and y. The new expression holds true for x in A and y in C likewise for x in C and y in B whereas in the expression before, those conditions don't affect the outcome, so the expression might hold true for x and y even if (x,y) is not in the initial set which contradicts to the definition that (x,y) is a member of that set.
That said, i guess this would count as a valid "subset" proof because he basically added more elements to the set and if i have a set, add more elements to it and it's still a subset of another set then the previos set must also be a subset of that set.
This guy is a Hero
Thank you sir, surprised the views are low ngl since it’s so easy to understand :/
Unfortunately, your example 3 at 10:29 is not actually an equality, only a one-way subset relation.
A concrete counter example is a pair of (x, y) where x∈C, but x∉A, x∉B, x∉D, and where y∈B, but y∉A, y∉C, y∉D. In other words, x is *only* ∈ C, and y is *only* ∈ B. Then nevertheless, still x∈(A∪C) and y∈(B∪D), so:
(x,y) ∈ ((A∪C) × (B∪D))
But since x∉A, then:
(x,y) ∉ (A×B)
And since y∉D, then:
(x,y) ∉ (C×D)
Thus, since (x,y) is neither a member of (A×B) nor of (C×D), then:
(x,y) ∉ ((A×B) ∪ (C×D))
Thus, since it is *_not always_* the case that ((A∪C) × (B∪D)) ⊆ ((A×B) ∪ (C×D)), therefore:
((A∪C) × (B∪D)) ≠ ((A×B) ∪ (C×D))
I think you meant to use intersections instead of unions, as that would indeed have yielded an equality:
((A⋂C) × (B⋂D)) = ((A×B) ⋂ (C×D))
yes!! you are correct!
Was happy to see other people thought the same. I am implementing some set theory and interval operations in a C++ library and had an intuition about how to do it, but I wanted to learn how to prove the operations are correct. My intuition didn’t match the proof for problem 3 and now I can confirm my intuition was correct. Thank you!
What's a good book to read this? My teacher uses set algebra to solve algebraic inequalities, and it's cool. But I tried to do this for the triangle inequality and I didn't know how to do operations between sets with a different variable.
Can you do a video on Disjunctive normal form?
I’ll try to work on one of those!
can you recommend me literature for mathematical logic and set theory? or some course about it?
For sets, you can start with 'Naive Set Theory' by P.R.Halmos, then study the ZFC axiomatic Set Theory.
However, I've watched this series up to this point, supplemented it with CS103 course notes from Stanford which are available for free... And I understand it for know, I feel confident with the knowledge I currently have.
My point is - don't try to overcomplicate your life. Get the knowledge of the basis down and try to do some actual tasks like writing proofs. Then you can expand your knowledge, but start doing the maths on your own and see that it's not that difficult!
Best, Jordan
Trev: I think you can do this one on your own
me: proceeds to try to do it by myself and fails, watches the rest of the video anyway.
But after watching the whole video I did everything from the start by myself, keep up with the good work!
Question: if i have (A × B)^2 how would this look like for example:
A = {1,2}
B = {3,7}
({1,2}×{3,7})^2
What is the complement of a cartesian product???
"and distributes over the or" was a surprise here. were we supposed to know that? I know it's a principle of logic but we hadn't seen it for sets.
i am so pissed i just now found you while studying 2 hours before my exam🤦🏻♂️🤦🏻♂️
Trick question, would you know how to do this?
Prove:
A u B = (A n B) u (A − B) u (B − A)
man the amount of views proves that ppl no longer care about learning and make a better future even if its hard u have to struggle in order to live a nice life
What do you mean
Not everyone is cut out for this kind of thing and that’s ok 👌🏼
@_Anna_Nass_ bro youre replying to my 2 year old comment, the view count was like 50 views
@@YujiItadori102 Okay and? You’re still here obviously.
thanks!