Question

How is T o (S o R) = (T o S) o R?

Let
R be a relation from A to B
S is a relation from B to C
T is a relation from C to D.

I basically wrote down the definition for each.
T o (S o R) = {(a,d)∈AxD | ∃c∈C [ ∃b∈B ((a,b)∈R and (b,c)∈S) and (c,d)∈T] }

(T o S) o R = {(a,d)∈AxD | ∃b∈B [(a,b)∈R and ∃c∈C ((b,c)∈S and (c,d)∈T] }

Answer 1

besides the fact that both have (a,d)∈AxD, I'm having a bit difficulty interpreting the right side of each

Answer 2

I was hesitating about removing some unnecessary parentheses because I'm afraid that might change the meaning.

Answer 3

humm r u sure u need to prove ?
or ur not sure if the statment is true ?

Answer 4

well, yes, i'm basically trying to prove the theorem. Sadly it wasn't proved in the book and left as an exercise -.- Typical in a text book haha

Answer 5

humm
what i really see is giving counter example !
cuz i dnt thnk its true sentence :O
what do u think ?

Answer 6

Composition of functions is always associative. But defining them in terms of relations has always been a bit confusing for me. And I've never seen a proof of this fact using that specific notation.

Answer 7

it is a theorem (as stated in the book). So T o (S o R) = (T o S) o R is true. I just can't to make sense of the definition I wrote

Answer 8

intresting ...
but i cant focus with music :D
https://www.youtube.com/watch?v=rYEDA3JcQqw

Answer 9

haha I'm watching people playing video games while proving the theorem XD

Answer 10

Looking at the definitions you've written down though, I'm relatively positive that you can get rid of some of the parentheses and re-order some of the statements that don't depend on one another. For example,
"∃b∈B [(a,b)∈R and ∃c∈C ((b,c)∈S and (c,d)∈T]"
could be rewritten as
"∃b∈B and ∃c∈C [(a,b)∈R ,(b,c)∈S and (c,d)∈T]"

Answer 11

@KingGeorge yes, I thought that removing some parenthesis and changing the order of each statement within the set should not change the meaning. But then again, when I have doubts, I don't wanna to make an invalid conclusion

Answer 12

And from there you can switch the order of "∃b∈B and ∃c∈C" to "∃c∈C and ∃b∈B" and do similar rearranging of "(a,b)∈R ,(b,c)∈S and (c,d)∈T."

You can make these rearrangements because the only things that might cause trouble, are moving quantifiers (∃ symbols) around statements such as "(a,b)∈R." The only time I did this, was moving a quantifier only involving the element "c∈C" around a statement not involving that same element. So in that case, we can switch the order.

Answer 13

So doing this, you can rewrite the set-theoretic definition of (T o S) o R to be exactly the same as the set-theoretic definition of T o (S o R). Since the two sets are then equal, the theorem is proved.

Answer 14

As far as I know, the order of the qualifies, if all of them are the same, does not matter. However, if there mix then it really changes the meaning.

For example: ∃x∀y is completely different from ∀y∃x

Answer 15

That's also true. But we only have one type of quantifier here, so we don't have to worry about that.

Answer 16

I might be thinking too hard. But as you know, in mathematics, sometimes things that we thought (for sure) that they're true can turn out to be false and counter intuitive after proving them. I guess I'll just take their words (the mathematicians') and believe that the theorem is true hehe

Answer 17

I think my favorite example of how important these quantifiers are, is the following question. Is this statement true or not?\[\forall w\in\mathbb{R},\exists x\in\mathbb{R},\forall y\in\mathbb{R},\exists z\in\mathbb{R}(wxyz=w+x+y+z).\]What about\[\exists w\in\mathbb{R},\forall x\in\mathbb{R},\exists y\in\mathbb{R},\forall z\in\mathbb{R}(wxyz=w+x+y+z)?\]

Answer 18

O.O torture!!!!! XD

Answer 19

I can't even begin to think XD

Answer 20

I'm not exactly here yet >.> but wouldn't wxyx = w * x*y*x?

Answer 21

This is why sometimes I hate mathematics. it's only when you really understand something 100% can you be confident that it's correct. My motto is that, if there is even a slightest doubt, it means I still don't understand it completely.

Answer 22

well , sourwing i love math for that reson only :D
understanding concepts is much more imp than giving a solution :D

Answer 23

While that may be true, there comes a point when you just have to move on and accept that some things you just don't understand completely, and there's nothing wrong with that.

Answer 24

hehe, which is why I'm about to turn on my PS3 and starting trolling with some noobs in Call of duty ahaha!! XD

Answer 25

anyways, thank you for your time @KingGeorge @tHe_FiZiCx99 @ikram002p

Answer 26

CoD >.>

Answer 27

You're welcome. And just btw, the first statement I put up there is true, while the second is not.

Answer 28

no lol i dint help , i enjoyed with the problem :D

Answer 29

Pardon my ignorance but how is w + x + y + z = wxyz?

Answer 30

In general, you're correct. But by choosing specific values for \(x\) and \(z\), I can make it true.

Answer 31

I thought about that as well, but it seemed a bit more work. I don't know how to explain it though :/ I'm horrible with words.

Answer 32

All you have to do, is choose \(x=0\), and \(z=-w-y\). Then both sides are 0. And since \(w\) and \(y\) have already been chosen, \(-w-y\) is a specific value.

Answer 33

check this intrested vedio. sourwing hope u dnt mind posting this in ur thread
https://www.youtube.com/watch?v=w-I6XTVZXww

Answer 34

In the second statement though, we can't do that process anymore.

Answer 35

Seems interesting. ^_^