Proofs clicked for me after reading Intro to Analysis by Brtle, I made several weak attempts at logic and sets two years ago, But engaging with the books definitions, then having remembered you said start with definitions, I did it used the definition used the math lingo, and it felt like I casted a spell, It was like wizzardry was guiding my pen and I knew what to write. Sometimes we get really excited when casting our spell and then find out IT WAS TOTALLY WRONNG. But that's okay! Thats part of the fun of being a sorcerer! Sometimes you think you extracted the annals of universe itself, only to find out its an adorable duck! Some say that yellow rubber duckies are all the adorable ducks from all the Math Sorcerers out there, We don't talk about it often because someone profited off our adorable duckies, but that neither nor their. P.S. Discrete Mathematics with Ducks is also a great book (hence my analogy)
Isn't it necessary to prove for the empty sets as well in a seperate case? Since there wont be any elements and we cant show the proof by using an arbitrary element? 🙄
sometimes you can write it differently and just do both directions at once, so really it's double inclusion but you are showing both directions at once. Let's see, here we go, an example of that: th-cam.com/video/0TgVnZCU4_w/w-d-xo.html
Hello, I would like to ask, how does one solve these kind of problems, if the formula is not true? Our professor showed us some (this may be lost in translation) "kontra-example", which disproved the formula.
@@TheMathSorcerer Ah! A counter example! Great, finally I am able to find some results. Google only found rhymes on that word in my language :D Thank you!
How do you know when to prove the proof on both sides? From left to right And from right to left Isnt showing one proof enough? How do we recognize when to do the method of double inclusion?
Whenever you see a set = another set, for example A = B, then you should prove that A is a subset of B, but also B is a subset of A. Remember the definition of a subset: Let A and B be sets. If all elements in set A is in set B, then A is a subset of B. This means that B COULD still have elements that is not in A, ie. A is a proper subset of B. If A is subset of B, then it also means that B could have all the elements that A has, ie. A=B. So that is why you MUST prove that A is a subset of B, AND B is a subset of A. I hope this helps (BTW A and B I used are just placeholders)
this concept is so simple I had trouble proving it
Yeah...it seems too obvious
Now we will have the power to beat the topper
This looks like absorption law. Can't wait to come back here in a few months to add on to this concept the teacher glossed over.
Is it possible to also prove this using the Distributive law and going from there?
Proofs clicked for me after reading Intro to Analysis by Brtle,
I made several weak attempts at logic and sets two years ago,
But engaging with the books definitions, then having remembered you said start with definitions,
I did it used the definition used the math lingo, and it felt like I casted a spell,
It was like wizzardry was guiding my pen and I knew what to write.
Sometimes we get really excited when casting our spell and then find out IT WAS TOTALLY WRONNG.
But that's okay! Thats part of the fun of being a sorcerer!
Sometimes you think you extracted the annals of universe itself, only to find out its an adorable duck!
Some say that yellow rubber duckies are all the adorable ducks from all the Math Sorcerers out there,
We don't talk about it often because someone profited off our adorable duckies, but that neither nor their.
P.S. Discrete Mathematics with Ducks is also a great book (hence my analogy)
Would you mind doing A=B iffy A is a subset of B and B is a subset of A
Life is a journey,lets soldier on till the end
Isn't it necessary to prove for the empty sets as well in a seperate case? Since there wont be any elements and we cant show the proof by using an arbitrary element? 🙄
Life is a journey,let's soldier on
What other methods are there to prove equality of sets?
sometimes you can write it differently and just do both directions at once, so really it's double inclusion but you are showing both directions at once. Let's see, here we go, an example of that: th-cam.com/video/0TgVnZCU4_w/w-d-xo.html
Hello, I would like to ask, how does one solve these kind of problems, if the formula is not true? Our professor showed us some (this may be lost in translation) "kontra-example", which disproved the formula.
Counter example yes, so you find one example of sets where the statement is false
@@TheMathSorcerer Ah! A counter example! Great, finally I am able to find some results. Google only found rhymes on that word in my language :D Thank you!
👍
so nicely shown
How do you know when to prove the proof on both sides?
From left to right
And from right to left
Isnt showing one proof enough?
How do we recognize when to do the method of double inclusion?
Whenever you see a set = another set, for example A = B, then you should prove that A is a subset of B, but also B is a subset of A. Remember the definition of a subset: Let A and B be sets. If all elements in set A is in set B, then A is a subset of B. This means that B COULD still have elements that is not in A, ie. A is a proper subset of B. If A is subset of B, then it also means that B could have all the elements that A has, ie. A=B. So that is why you MUST prove that A is a subset of B, AND B is a subset of A.
I hope this helps
(BTW A and B I used are just placeholders)
@@stefanvorster4410 thanks man, helped a lot
Absorption Law ….. Done
proofs involving sets are gross
Agreed but failing the course would be more gross so I'm learning them lol
@@beri4138 I said this when I was angry honistly they are fine
@@raynebenson9040 You were angry because discrete math is a frustrating course. I'm currently taking it.
@@beri4138 I was angry because proofs involving sets are difficult. I would say the majority of discreet math isn't so difficult.