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!
It's almost like those universities don't really *want* you to learn. It's infuriating how much potential is lost by that. I'm in my 30s now and I always thought I hated math, but this is actually a nice little puzzle and logic game instead of panic and confusion.
Thank you so much for your videos. I went to a school where we were divided into groups dependent on maths ability and once in your group you could never really learn above that level. My lack of maths skills i know is down to the teachers i had and that they didnt make it fun and understandable. Now as a 33 year old im feeling mature enough to try again and try and get a degree in computer science. Your videos make me feel like it might be possible for me to master maths and actually enjoy it.
I came here after realizing there was no playlist for Discrete Math for the Organic Chemisty Tutor. I’m happy to report I’m equally pleased with your videos so far. Thanks, dude!
Great lectures, thank you. I've been a software engineer for 10+ years, but hoping to get a more rigorous understanding of computer science so I'm starting with discrete maths.
at 8:06, isn't (c,d) and (d,c) the same because set doesn't have order, also if then zoom out, the larger set has two same set of (c,d), (d,c), which would conclude BxB = {(c,c),(c,d),(d,d)}?
Great! Thanks Trev. It would be great if the volume of your videos were just a tad bit louder. I had to max system and UA-cam volume just be to able to hear you, and I'm sporting a pair of 6'' studio monitors that regularly get noise complaints...
thank you sir for making this vlog it really helps me for being an freshman student in Bachelor of science and information course. please im looking forward for more
For anyone reading this later on: those elements written in normal parentheses are ordered pairs, so their order still matters and thus they aren't the same. If you wanted to represent those ordered pairs as sets then you'd write {{c}, {d,c}} and they still wouldn't be the same thing.
Hello i have a question on minute 8:32, would BxB be ( (c) , (c,d) , (d) ) since in the last video it was explained that repeated elements are listed once and theres no order in a set meaning that c,d is the same as d,c? just a bit confused 😅 Thanks
The set is unordered, but individual elements in cartesian products are ordered. If you see {, } it's unordered. If you see (, ) or then it's ordered and we must preserve that order.
I'm wondering what the link is between {a, {ab}} and a Cartesian product? None of the Cartesian products have subsets of different lengths so I don't see the connection
Why are Cartesian coordinates represented as {{x}, {x,y}}? Is it just so that you can differentiate between two ordered pairs that have the same x and y, but flipped?
Sets by definition have no predefined order according to the previous video. This is why the coordinates are represented specifically as ordered pairs. The first variable stated (in an ordered pair) dictates the order of the pair.
I have to agree with comments below. I had to take a course at the the university I attended for my BS degree. The text book was "Elementary Functions" that addressed Cartesian products. The class professor and text were a waste. They should have name the course deciphering bullshit. Thanks for making this easy.
I can't see the relation with the cartesian product as you define it and as it is defined in linear algebra, with vectors. Is there any? Actually, in linear algebra AxA would be zero.
Regarding Ø × A, I understood the cardinality argument, that it's size should be zero, so the result is Ø. Taking that detour makes sense. But what threw me off was, if you carried through the process as shown prior to this example, we would get { } × { a, b } = { ( , a), ( , b) }. Is this wrong or do we define the size of ( , a) to be zero? Or is (m, n) only defined if both elements are present? If something like ( , n) is not defined, how is the number zero derived from it? Thanks!!!
( ,a) is in fact not defined, i.e. does not correspond to any object. It is similar to how you can write x∈{}, but x does not correspond to any object. I think another way to see this is as follows. We can say that t∈AxB iff there are some a∈A and b∈B, such that t=(a,b). This means that t∈{} x {0,1} is equivalent to saying that there are some a∈{} and b∈{0,1} such that t=(a,b). However, no such a exists, and thus no such t exists either. This means that no element is in {} x {0,1}.
I'm not sure why do you write a singleton when we have a set for example (1,2) why did you write the singleton {1} then the set {1,2} I hope I'm making sense.
How do i solve something like this? For each of the following pairs on the natural numbers N (in other words, elements from N x N), list the ordered pairs in each set. R = {:2x+y=9} S = {:x+y=7}
Well it appears to me that since natural numbers are positive integers between 1 and infinity, the answer to S should be {(1,6), (6,1), (2,5), (5,2), (4,3), (3,4)} because those are the pairs of natural numbers that add up to 7. For R I would look at it the same way starting with X values. If x is 1 then y would be 7 to equal nine, if x was 2 then y would be 5 to equal 9 so on and so forth. Therefore R = {(1,7), (2, 5), (3,3), (4,1)}. I may be completely off but that is how I logic through it.
Cartesian products, 5 mins in, (0,0) looks like a face. Okay, w/e. Then all the other pairs start to look like faces. I think its time for a nap. Be back soon
if A = { 0, 1} and B = { 2,3,4 } What is A x A x B? If I write out all the possible combinations it's: { {0,1}, {0,1}, {0,2}, {0,3}, {0,4}, {1,2}, {1,3}, {1,4} } = 8 But A x A = 4 and A x B = 6 which is 12. Am I doing something wrong?
Thank you....I have a question. If S is the set of real number and I be set of rational number. Aalpha( the alpha is the symbol, but there is no symbol for alpha on my phone pad here. moreover the alpha is index set of the family. Aalpha = { x € S: x is greater than or equal to alpha} for any alpha is an element of index set. show that UAalpha = S.
On the first video dealing with an intro to sets, I was wondering whether the set "Desk" might not be best understood as belonging to the set "noun" since the desk is not a set that may contain itself as one of the elements. Just as "humans" is the set of specific mammals, but it itself is not a human, or, an element of itself. I am asking because in the example it is difficult to see, that might be the point, any relation between elements and sets. With numbers is slightly less difficult, but with linguistic examples, I thought that words belonging to logical-algebraic categories of language might be better.
Hey, I noticed that you mentioned linguistics. I am currently studying linguistics to better help me understand maths. I’m currently using crash course linguistics to study. Do you recommend any other materials that are still simply enough written but dives much deeper into linguistics?
@@gloriakalengelayi8294 From what I have seen Saussure opened up (also psychoanalysis actually) the field in terms of introducing the phonetic aspect. If one has taken classes in other languages in school, or, is bilingual one realizes how the phonetic aspect (out of it) one can begin to derive the words of other languages. The latter element is crucial to answer the question of he way human language is universal. The only material on ("generative"/Godel is the censored word used. It's actually dialectics) linguistics that I'm aware of that is worth taking a look at is that of Saussure, then all the three branches of formal logic and dialectical logic. It's difficult to find good material on the latter since it has been censored/mediated vía Kurt Godel's work on generativity. The only crucial works on "generative"/grammar (introductions) that I've read are in Spanish. The books were written by someone called Eduardo Vázquez. One of the crucial ones can be found on pdf on the net. The book is called "Los Puntos Fundamentales de la Filosofía de Hegel". He has others and can be found on the net. The stuff presented by this called called the "Einstein of Linguistics" doesn't look like anything. After being crowned with this title he was asked to at least (Einstein apparently had to) present his general theory. He was not able. After this individual retired he was asked again about the general theory that would acct for him being designated someone with a general theory of human language. Once again he claimed that he did not have it. That it was too difficult etc.
@@gloriakalengelayi8294 In a set of questions posed to this individual that goes by the name Chomsky he claimed that he had studied the phonetic (as well as other [differential]) element introduced by structural linguistics, but that he could not see the relevance to it regarding questions of "generativity" and the universality of language. The lecture and the set of questions can be found in UA-cam.
@@gloriakalengelayi8294 I would recommend presentations of Bertrand Russell's paradox in UA-cam. The looping logic is crucial to grasp the self-reflexivity of dialectics. Along with Vazquez's books I would recommend a book presented by Rebecca Goldstein and another by Todd McGowan. Goldstein's book is called "Incompleteness" and introduces the looping/generative/dialectical logic presented by Godel as his own. McGowans book is called "Emancipation After Hegel". The crucial thing is to concentrate upon formal and dialectical logic since that's what German Idealists were working with. McGowans propositions about contradiction begin with how they are found in language and then in the logic of people's practices etc.
Thank you Sir; but I have a question; I answered the question ∅×A as ( {a} , {a,b} ) but got it wrong . I did this because i thought that the null set doesn't count hence we treat set A alone as the ordered pair. Please what am I missing?
An empty set has a cardinality (size) of zero. When we Multiply two sets, the formula is the product of the cardinality of the 2 sets. A and B both has 2 a cardinality of two. 2 x 2 is 4, so you get 4 ordered pairs. In the case of an empty set, it becomes 0 x 2 which is 0, thus you get no ordered pair. Therefore, an empty set. Hope that answered your question.
What would be the result to the following A^13 * A^ 21 = ? Do we follow the exponent "product rule"? If so does this sound right to you A^13 * A^ 21 = A^34 !! please someone explain to me
Yes. Because as he showed in the video, |A| = m, |B| = n, and |A×B| = m ⋅ n. Since |A²| is just another way of writing |A×A|, |A²| = |A| ⋅ |A|, in other words, |A²| = |A|². For |A×A×A|, if we look at it as |(A×A)×A|, we get |A×A×A| = |A×A| ⋅ |A| = |A| ⋅ |A| ⋅ |A| = |A|³, so by the exponent properties it follows that |Aˣ| = |A|ˣ, so at that point we can just use them to show that |A¹³×A²¹| = |A¹³| ⋅ |A²¹| = |A|¹³ ⋅ |A|²¹ = |A|³⁴.
Halfway through the video, you mention "Ordered Triplets.' Can you explain what that is, in relation to an ordered pair? Like, how do you represent and ordered triplet as a set?
Exactly what it sounds like. An example of an ordered pair you're most familiar with would be (X,Y) or (X,Y(X)), where the first element comes from your set 'X' and the second element is obtained from your output set 'Y' or 'Y(X)'. You can imagine an ordered triplet as just adding another set, in which case you'd now have three elements in every "ordered pair" like (X,Y,Z)
Why are the empty set crosses equal to the empty set when you can obviously pair with the second set? The cross product would just result in the ordered pairs with only the second set's elements, which is obviously not an empty set.
Yati Kasera I think it is because a nul set is an empty set/equal to zero. When you multiply anything with zero your answer will always be zero, or empty in this case
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
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!
It's almost like those universities don't really *want* you to learn. It's infuriating how much potential is lost by that. I'm in my 30s now and I always thought I hated math, but this is actually a nice little puzzle and logic game instead of panic and confusion.
You are such a great human being. Thank you
Да, уроки у него супер) Отлично объясняет)
I agree!
I need this guy as my therapist. Actually, with his videos I no longer need a therapist.
Nice, I love that
Thank you so much for your videos. I went to a school where we were divided into groups dependent on maths ability and once in your group you could never really learn above that level. My lack of maths skills i know is down to the teachers i had and that they didnt make it fun and understandable. Now as a 33 year old im feeling mature enough to try again and try and get a degree in computer science. Your videos make me feel like it might be possible for me to master maths and actually enjoy it.
You can do it! I have faith!
Literally have an exam in 2 days and you have taught me more than my lecturer did the whole semester
I came here after realizing there was no playlist for Discrete Math for the Organic Chemisty Tutor. I’m happy to report I’m equally pleased with your videos so far. Thanks, dude!
Hello... I really appreciate how you've made this content publicly available.It's helping me go over fundamentals of discrete mathematics.
And this is what happens when a linguist explains mathematics .... Simply brilliance in every which way.... Thank you!
Your style of teaching is kind of similar to that of "the organic chemistry tutor", and I like it
Haha, this hits home. Now that you mention it, Trevor reminds me a lot of a friend of mine who studied organic chemistry.
I love Organic Chemistry he's my man in Math
ive been watching trev for discrete and org chem tutor for computing logic. both complete lifesavers
@@CL41R3xI’ve been doing that same 😂
You have a really, really nice handwriting. :) Keep up the good work!
Probably a smooth function in the software he uses and his own neat handwriting.
Yes, very nice presentation - a pleasure to watch.
Great lectures, thank you. I've been a software engineer for 10+ years, but hoping to get a more rigorous understanding of computer science so I'm starting with discrete maths.
Thank you very much for this video, literally couldn't find any other that included the actual definition of cartesian product
I don't have right words to express my gratitude. But Thank you! a million times
Thank you for this wonderful video. Such a clear and easy to understand explanation.
Thanks for your videos man. I havent taken this math yet so I am learning all I can before I have to. Ur appreciated
How did it go? did you understand everything while taking the class?
You explain everything so well, THANK YOU!! Literally saving my grade
these videos are so good that they are recommended by our uni professor. Now thats epic!
The colors you use are great, they aren't hard on the eyes yet they're vibrant enough to catch attention
That's because the background is black. Feels nice.
your saving my grade for this semester thank youuuuuuu
I'm so happy I found this channel
Your teaching skill is so amazing. So clear simple and easy. Thanks a lot
at 8:06, isn't (c,d) and (d,c) the same because set doesn't have order, also if then zoom out, the larger set has two same set of (c,d), (d,c), which would conclude BxB = {(c,c),(c,d),(d,d)}?
no because (c,d) and (d,c) arent the same, (c,d)={{c},{c,d}} and (d,c)={{d},{d,c}={{d},{c,d}}. As you can see with one its a c the other a d
Great! Thanks Trev. It would be great if the volume of your videos were just a tad bit louder. I had to max system and UA-cam volume just be to able to hear you, and I'm sporting a pair of 6'' studio monitors that regularly get noise complaints...
thank you sir for making this vlog it really helps me for being an freshman student in Bachelor of science and information course. please im looking forward for more
very helpful video ,YOU explain better than my teacher
I'm just paying my uni to learn from youtube
you will be the reason I pass my Discrete midterm
mines in 2 days prey 4 me
mines in 2 days prey 4 me
i have a final after 3 days and i start with this playlist, wish me a good luck 😉
5 days passed.. how does it go?
Francis Lee well, i think i did well. but the result after one week. i’m really thankful to this channel
Francis Lee how about you do you have final exams ?
@@Pages_Perfected Just had it this morning. Well careless got the better of me again
you the goat my guy, saving me
So useful for SQL. Thanks for making these videos - I have to take a discrete math class for comp sci and I’m afraid to go into it without any primer
awesome videos i understood it more clearly for my exam tomorrow
3:02 wow thats a real nice owl you just drew, but what about (0,0)?
If I don't pass this DM exam next week, I don't know anymore. You explain this way better...
did you pass your exam
@@georgian1421 I think he has gone mad
bro i had a mental breakdown learning this stuff in English(i don't speak English) but thank you for making this easy even though it been 4 years
This is really great content, thank you! Very helpful.
You really help me with your videos to do my maths. Thank you soo muuch !
Question.
At 8:21 when doing the BxB wouldn't the sets (c, d) and (d, c) be the same set, therefore making them only one set?
Someone, please answer
You cannot change its order because it was given at the question such that B = {c,d}. You should just multiply BxB.
For anyone reading this later on: those elements written in normal parentheses are ordered pairs, so their order still matters and thus they aren't the same. If you wanted to represent those ordered pairs as sets then you'd write {{c}, {d,c}} and they still wouldn't be the same thing.
you are far the best teacher bro
Thank you ver much sir.🚀
♥from Namibia 🇳🇦.
Come and teach at my university lol.
Hello i have a question on minute 8:32, would BxB be ( (c) , (c,d) , (d) ) since in the last video it was explained that repeated elements are listed once and theres no order in a set meaning that c,d is the same as d,c? just a bit confused 😅 Thanks
The set is unordered, but individual elements in cartesian products are ordered. If you see {, } it's unordered. If you see (, ) or then it's ordered and we must preserve that order.
@@Trevtutor ah now it makes sense, thanks ❤️
I'm wondering what the link is between {a, {ab}} and a Cartesian product? None of the Cartesian products have subsets of different lengths so I don't see the connection
Yes me tok
Why are Cartesian coordinates represented as {{x}, {x,y}}? Is it just so that you can differentiate between two ordered pairs that have the same x and y, but flipped?
Sets by definition have no predefined order according to the previous video. This is why the coordinates are represented specifically as ordered pairs. The first variable stated (in an ordered pair) dictates the order of the pair.
I have to agree with comments below. I had to take a course at the the university I attended for my BS degree. The text book was "Elementary Functions" that addressed Cartesian products. The class professor and text were a waste. They should have name the course deciphering bullshit. Thanks for making this easy.
Omg IOVE THE WAY U TEACH MY MIND THANK U SO MUCH.
Little bit long video which makes the video little boring but it's really useful and made me understand the lesson very easily Ty💙
watch @ 1.25x speed it's perfect.
You rock! Using this for my DB design course for learning Relational Algebra.
How are n-tuples represented as sets if an ordered pair is a 2-tuple?
What is Cartesian product good for? Combinatorics come to mind, but there must be other uses. What are those?
Question: could you explain ø x A = ø in terms of set notation? I arrive at { ø, {a},{b}} ø , assuming your example with A= {a,b}
As empty set contains nothing in it..
Where we can get the element for pairing..
(a, ?) So pairing is no possible..
So..
That's why it's an empty set
I can't see the relation with the cartesian product as you define it and as it is defined in linear algebra, with vectors. Is there any? Actually, in linear algebra AxA would be zero.
We are just thinking of these elements as arbitrary numbers not vectors.
Hi there. Yet another great explanation of Mathematics. May God Bless You.
what a great series thanks man keep up the good work
thanks a lot, helped to prepare for a test
What drawing tablet and software do you use to record these videos?
Thank you very much , you've cleared all of ma confusion
Thankyou, I understand finally.
10:15 Cartesian product is not associative. Does BxA^2 equal BxAxA or Bx(AxA)?
BxAxA.
Left to right order of composition. BxAxA= Bx(AxA) !=(BxA)xA
If a power set multiply the set where it came from, does it mean 2 to power of n element(for power set) multiply n element of set?
Regarding Ø × A, I understood the cardinality argument, that it's size should be zero, so the result is Ø. Taking that detour makes sense. But what threw me off was, if you carried through the process as shown prior to this example, we would get { } × { a, b } = { ( , a), ( , b) }. Is this wrong or do we define the size of ( , a) to be zero? Or is (m, n) only defined if both elements are present? If something like ( , n) is not defined, how is the number zero derived from it? Thanks!!!
( ,a) is in fact not defined, i.e. does not correspond to any object.
It is similar to how you can write x∈{}, but x does not correspond to any object.
I think another way to see this is as follows.
We can say that t∈AxB iff there are some a∈A and b∈B, such that t=(a,b).
This means that t∈{} x {0,1} is equivalent to saying that there are some a∈{} and b∈{0,1} such that t=(a,b).
However, no such a exists, and thus no such t exists either.
This means that no element is in {} x {0,1}.
I'm not sure why do you write a singleton when we have a set for example (1,2) why did you write the singleton {1} then the set {1,2} I hope I'm making sense.
great explanation! Keep doing good stuff) thanks
Thanks for the video, indeed it has been helpful
How do i solve something like this?
For each of the following pairs on the natural numbers N (in other words,
elements from N x N), list the ordered pairs in each set.
R = {:2x+y=9}
S = {:x+y=7}
Well it appears to me that since natural numbers are positive integers between 1 and infinity, the answer to S should be {(1,6), (6,1), (2,5), (5,2), (4,3), (3,4)} because those are the pairs of natural numbers that add up to 7. For R I would look at it the same way starting with X values. If x is 1 then y would be 7 to equal nine, if x was 2 then y would be 5 to equal 9 so on and so forth. Therefore R = {(1,7), (2, 5), (3,3), (4,1)}. I may be completely off but that is how I logic through it.
Cartesian products, 5 mins in, (0,0) looks like a face. Okay, w/e. Then all the other pairs start to look like faces. I think its time for a nap. Be back soon
what board do you use ? Is it electronic ?
If set A={a,b.c} then what is the cardinality of B where B={1,2,3,{a,b,c}}. I will greatly appreciate your response.
should be 4 since u have 4 elements in set b
Awesome!
08:03 Roll for Initiative!
Actually made me laugh out loud thank you
What is the vertical line in between (a,b) and a is an element of A
“Such that”
@@Trevtutor Thanks!
What software do you use for your videos (to write)? You inspire me.
if A = { 0, 1} and B = { 2,3,4 }
What is A x A x B?
If I write out all the possible combinations it's:
{ {0,1}, {0,1}, {0,2}, {0,3}, {0,4}, {1,2}, {1,3}, {1,4} } = 8
But A x A = 4 and A x B = 6 which is 12. Am I doing something wrong?
Dude thank you so much❤
Great video 🙏but I didn't get the ordered pair and its sets relation can someone clear it out for mr please
Thank you....I have a question. If S is the set of real number and I be set of rational number. Aalpha( the alpha is the symbol, but there is no symbol for alpha on my phone pad here. moreover the alpha is index set of the family. Aalpha = { x € S: x is greater than or equal to alpha} for any alpha is an element of index set. show that UAalpha = S.
On the first video dealing with an intro to sets, I was wondering whether the set "Desk" might not be best understood as belonging to the set "noun" since the desk is not a set that may contain itself as one of the elements. Just as "humans" is the set of specific mammals, but it itself is not a human, or, an element of itself. I am asking because in the example it is difficult to see, that might be the point, any relation between elements and sets. With numbers is slightly less difficult, but with linguistic examples, I thought that words belonging to logical-algebraic categories of language might be better.
Hey, I noticed that you mentioned linguistics. I am currently studying linguistics to better help me understand maths. I’m currently using crash course linguistics to study. Do you recommend any other materials that are still simply enough written but dives much deeper into linguistics?
@@gloriakalengelayi8294 From what I have seen Saussure opened up (also psychoanalysis actually) the field in terms of introducing the phonetic aspect.
If one has taken classes in other languages in school, or, is bilingual one realizes how the phonetic aspect (out of it) one can begin to derive the words of other languages. The latter element is crucial to answer the question of he way human language is universal.
The only material on ("generative"/Godel is the censored word used. It's actually dialectics) linguistics that I'm aware of that is worth taking a look at is that of Saussure, then all the three branches of formal logic and dialectical logic. It's difficult to find good material on the latter since it has been censored/mediated vía Kurt Godel's work on generativity. The only crucial works on "generative"/grammar (introductions) that I've read are in Spanish.
The books were written by someone called Eduardo Vázquez. One of the crucial ones can be found on pdf on the net. The book is called "Los Puntos Fundamentales de la Filosofía de Hegel". He has others and can be found on the net.
The stuff presented by this called called the "Einstein of Linguistics" doesn't look like anything. After being crowned with this title he was asked to at least (Einstein apparently had to) present his general theory. He was not able. After this individual retired he was asked again about the general theory that would acct for him being designated someone with a general theory of human language. Once again he claimed that he did not have it. That it was too difficult etc.
@@gloriakalengelayi8294 In a set of questions posed to this individual that goes by the name Chomsky he claimed that he had studied the phonetic (as well as other [differential]) element introduced by structural linguistics, but that he could not see the relevance to it regarding questions of "generativity" and the universality of language. The lecture and the set of questions can be found in UA-cam.
@@gloriakalengelayi8294 I would recommend presentations of Bertrand Russell's paradox in UA-cam. The looping logic is crucial to grasp the self-reflexivity of dialectics.
Along with Vazquez's books I would recommend a book presented by Rebecca Goldstein and another by Todd McGowan. Goldstein's book is called "Incompleteness" and introduces the looping/generative/dialectical logic presented by Godel as his own. McGowans book is called "Emancipation After Hegel". The crucial thing is to concentrate upon formal and dialectical logic since that's what German Idealists were working with. McGowans propositions about contradiction begin with how they are found in language and then in the logic of people's practices etc.
@@gonzogil123 thanks for the reply. I’ll check out Saussure and hopefully I can find good material on the other 2 you mentioned.
I am starting to like discrete mathematics.
Good tutor. Thanks!
Thank you Sir; but I have a question;
I answered the question ∅×A as
( {a} , {a,b} ) but got it wrong .
I did this because i thought that the null set doesn't count hence we treat set A alone as the ordered pair.
Please what am I missing?
An empty set has a cardinality (size) of zero. When we Multiply two sets, the formula is the product of the cardinality of the 2 sets. A and B both has 2 a cardinality of two. 2 x 2 is 4, so you get 4 ordered pairs. In the case of an empty set, it becomes 0 x 2 which is 0, thus you get no ordered pair. Therefore, an empty set. Hope that answered your question.
@@muddtheboss415 Thanks a lot.makes a whole lot of sense now.
Please how will u writ the cartesian product of a power set
what does he mean in 1:37?
What would be the result to the following A^13 * A^ 21 = ? Do we follow the exponent "product rule"? If so does this sound right to you A^13 * A^ 21 = A^34 !! please someone explain to me
Yes. Because as he showed in the video, |A| = m, |B| = n, and |A×B| = m ⋅ n.
Since |A²| is just another way of writing |A×A|, |A²| = |A| ⋅ |A|, in other words, |A²| = |A|².
For |A×A×A|, if we look at it as |(A×A)×A|, we get |A×A×A| = |A×A| ⋅ |A| = |A| ⋅ |A| ⋅ |A| = |A|³, so by the exponent properties it follows that |Aˣ| = |A|ˣ, so at that point we can just use them to show that |A¹³×A²¹| = |A¹³| ⋅ |A²¹| = |A|¹³ ⋅ |A|²¹ = |A|³⁴.
How would you do A^3 if A={0,1,3}
A^3 = A . A . A = {(0,0), (0,1), (0,3), (1,0), (1,1), (1,3), (3,0), (3,1), (3,3)}
can you please solve the problems from Discrete Mathematics and Its Applications written by Kenneth H. Rosen, especially those of mathematical proofs.
i was your 1000th like fam
Always check your audio recording levels, folks. Always.
Brilliant!
i find your voice soothing .. im not gay
:/
"im not gay" means you are gay
Halfway through the video, you mention "Ordered Triplets.' Can you explain what that is, in relation to an ordered pair? Like, how do you represent and ordered triplet as a set?
Exactly what it sounds like. An example of an ordered pair you're most familiar with would be (X,Y) or (X,Y(X)), where the first element comes from your set 'X' and the second element is obtained from your output set 'Y' or 'Y(X)'.
You can imagine an ordered triplet as just adding another set, in which case you'd now have three elements in every "ordered pair" like (X,Y,Z)
Can anyone help me with proving
A x B=B x A if A=B? I understand that it is but I cannot go on about proving it.
you are a godsend
I'm assuming that the cartesian product is not that same as a cross product of two vectors because that's not how a vector cross product works
Why are the empty set crosses equal to the empty set when you can obviously pair with the second set? The cross product would just result in the ordered pairs with only the second set's elements, which is obviously not an empty set.
Ok, the distributive law applies. The laws of arithmetics for the Cartesian product.
Good videos, you just gotta speak up or turn up your mic
Audio for this set is really low and I'm not sure why. New videos after this discrete set should be normal, hopefully.
Buy new speakers, get loud speakers. Mine are clear.
"Buy new speakers" he says, even though TrevTutor replied with "Audio for this set is really low" hurr durr
why the
nul set * A= nul set?
could u please tell me
Yati Kasera I think it is because a nul set is an empty set/equal to zero. When you multiply anything with zero your answer will always be zero, or empty in this case
hello, do you use a mouse as you write this? im just curious :D
He uses a pen on a tablet or a phone while he records his voice
|2^3.3^2|=72 is it correct??
Much appreciated
Thank you very much