CARTESIAN PRODUCTS and ORDERED PAIRS - DISCRETE MATHEMATICS
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- čas přidán 20. 01. 2018
- We introduce ordered pairs and cartesian products. We also look at the definition of n-tuples and the cardinatliy of cartesian products.
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Hello, welcome to TheTrevTutor. I'm here to help you learn your college courses in an easy, efficient manner. If you like what you see, feel free to subscribe and follow me for updates. If you have any questions, leave them below. I try to answer as many questions as possible. If something isn't quite clear or needs more explanation, I can easily make additional videos to satisfy your need for knowledge and understanding.
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Haha, this hits home. Now that you mention it, Trevor reminds me a lot of a friend of mine who studied organic chemistry.
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@@CL41R3xI’ve been doing that same 😂
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!
Thank you very much for this video, literally couldn't find any other that included the actual definition of cartesian product
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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.
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
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.
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Come and teach at my university lol.
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How are n-tuples represented as sets if an ordered pair is a 2-tuple?
what board do you use ? Is it electronic ?
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
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.
What is Cartesian product good for? Combinatorics come to mind, but there must be other uses. What are those?
nice video !
Thank you
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?
3:02 wow thats a real nice owl you just drew, but what about (0,0)?
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
tnks
Thanks sir
well done
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
I'm just paying my uni to learn from youtube
thanks
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
tks
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08:03 Roll for Initiative!
Actually made me laugh out loud thank you
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.
Please how will u writ the cartesian product of a power set
I am starting to like discrete mathematics.
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.
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.
Always check your audio recording levels, folks. Always.
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.
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?
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.
Great video 🙏but I didn't get the ordered pair and its sets relation can someone clear it out for mr please
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}.
What is the vertical line in between (a,b) and a is an element of A
“Such that”
@@Trevtutor Thanks!
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.
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.
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
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
what does he mean in 1:37?
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
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
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.
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.
|2^3.3^2|=72 is it correct??
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)
Love 💘
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.
Is the relation f ⊆ Z × Z defined by f(x) = 2x a mapping? can you explain this kind of problems?
please don't you have any lecture on de morgan's law