Question about proof from a guy with a highschool education

[micromass]

To contrast (15), take a look at

\{6x + 2y~\vert~x,y\in \mathbb{Z}\}

This will not equal $\mathbb{Z}$. Do you see which set it will equal instead?

Questions of these kind are solved in abstract algebra. In particuler, the relevant result here is Bezout's theorem which states when

\mathbb{Z}~=~\{ax+by~\vert~x,y\in \mathbb{Z}\}

and what exactly the correct set is if the equality doesn't hold.

 

[reenmachine]

To contrast (15), take a look at

\{6x + 2y~\vert~x\in \mathbb{Z}\}

This will not equal $\mathbb{Z}$. Do you see which set it will equal instead?

ℝ? Because y could be an irrational number?

Questions of these kind are solved in abstract algebra. In particuler, the relevant result here is Bezout's theorem which states when

\mathbb{Z}~=~\{ax+b~\vert~x,y\in \mathbb{Z}\}

and what exactly the correct set is if the equality doesn't hold.

Not sure I understand this part , what is the y doing there?

 

[micromass]

Sorry, typos corrected now.

 

[reenmachine]

\mathbb{Z}~=~\{ax+by~\vert~x,y\in \mathbb{Z}\}

Still not sure I understand.

If a and b are ${\sqrt{2}}$ and x and y are 2 , then it gives us 2.828... + 2.828... = 5.657... which would mean that the set would be ℝ no?

 

[reenmachine]

15.

As I said before, it's the set of all 5a+2b such that a and b are integers. It's pretty obvious that this is the set of all integers, $\mathbb Z$, as you have now concluded. (Edit: Ohh...after reading micromass' reply below, I see it's not as obvious as I thought. It's only obvious that this set will be a subset of $\mathbb Z$). If you want to prove that this "guess" is correct, you need to rely on the axiom that says that two sets are equal if and only if they have the same elements. Define $Z=\{5a+2b\,|\,a,b\in\mathbb Z\}$. We want to prove that $Z=\mathbb Z$.

Let $x\in Z$ be arbitrary. The definition of Z tells us that there exist $a,b\in\mathbb Z$ such that $5a+2b=x$. Since $\mathbb Z$ is closed under addition and multiplication, this implies that $x\in\mathbb Z$.

Let $x\in\mathbb Z$ be arbitrary. Define $a,b\in\mathbb Z$ by $a=x$ and $b=-2x$. We have $a,b\in\mathbb Z$ and $5a+2b=5x+2(-2x)=5x-4x=x$. This implies that $x\in Z$.

wow this is great! Can you elaborate a little bit on what you mean with '' Since $\mathbb Z$ is closed under addition and multiplication, this implies that $x\in\mathbb Z$ ''.

21. Note that your result can be simplified to $\left\{n^2\,|\,n\in\mathbb N\right\}$. The proof of that is similar to the proof of 15. (Most people use n and m for integers rather than x and y, but it's not necessary to do this. n is a dummy variable here, so you can use x instead).
.

Is it important to simplify it?

thanks Fredrik!

Congratulations on becoming a mentor , well deserved promotion!

 

[Fredrik]

Still not sure I understand.

I think micromass is talking about the following issue: For what integer values of a,b is $\left\{ax+by\,|\,x,y\in\mathbb Z\right\}=\mathbb Z$?

Edit: Problem 15 says that this equality holds when a=5 and b=2. micromass says that it doesn't hold when a=6 and b=2. (Note that when we write the set this way, the dummy variables corresponding to a and b in problem 15 are x and y, not a and b).

Can you elaborate a little bit on what you mean with '' Since $\mathbb Z$ is closed under addition and multiplication, this implies that $x\in\mathbb Z$ ''.

Suppose that X is a set on which an addition operation is defined. A subset $S\subseteq X$ is said to be closed under addition if for all $x,y\in S$, we have $x+y\in S$.

"Closed under multiplication" is defined similarly.

Is it important to simplify it?

Not really. But it's important to know that it can be simplified, because you may encounter the other way of writing the set, and when you do, you need to be able to recognize it as the same set.

Congratulations on becoming a mentor , well deserved promotion!

Thank you.

 

[CompuChip]

As an example, the set of even numbers is closed under addition (because the sum of two even numbers is even), but the set of odd numbers is not - any two numbers will provide a counterexample.

The set of all integers except 0 is closed under multiplication, but the set of all integers except 1 is not (do you see why?).

 

[CompuChip]

thanks Fredrik!

Congratulations on becoming a mentor , well deserved promotion!

Congratulations! Do you get the extended Christmas pack now? ;)

 

[Fredrik]

The set of all integers except 0 is closed under multiplication, but the set of all integers except 1 is not (do you see why?).

Edit: What I said here is wrong. See the two posts below this one.

I think you must have meant to say something other than you did in that last example. For all $n,m\in\mathbb Z-\{1\}$, we have $nm\in\mathbb Z$ and it's impossible that nm=1, because this equality is false if one of n and m is 0, and implies n=1/m or m=1/n otherwise.

So $\mathbb Z-\{1\}$ is closed under multiplication (←Wrong.), but $\mathbb Q-\{1\}$ and $\mathbb R-\{1\}$ are not. Also, $\mathbb Z-\{1\}$ is not closed under addition. Maybe that's what you had in mind.

Congratulations! Do you get the extended Christmas pack now? ;)

Thanks. It doesn't have a lot of benefits I'm afraid.

 

[micromass]

I think you must have meant to say something other than you did in that last example. For all $n,m\in\mathbb Z-\{1\}$, we have $nm\in\mathbb Z$ and it's impossible that nm=1, because this equality is false if one of n and m is 0, and implies n=1/m or m=1/n otherwise.

What about $n=m=-1$?

(Feel free to delete this post if you'll delete your comment)

 

[Fredrik]

What about $n=m=-1$?

(Feel free to delete this post if you'll delete your comment)

Lol. No, I'm not going to delete a post just to cover up how dumb I can be sometimes.

(But I have added a comment to the post where I made that silly mistake).

 

[reenmachine]

To contrast (15), take a look at

\{6x + 2y~\vert~x,y\in \mathbb{Z}\}

This will not equal $\mathbb{Z}$. Do you see which set it will equal instead?

Questions of these kind are solved in abstract algebra. In particuler, the relevant result here is Bezout's theorem which states when

\mathbb{Z}~=~\{ax+by~\vert~x,y\in \mathbb{Z}\}

and what exactly the correct set is if the equality doesn't hold.

I just woke up and I think I know why now , \{6x + 2y~\vert~x,y\in \mathbb{Z}\} can't be the set Z because both multiplications will always result in even numbers , which means the addition will always be an even number as well.

So this would be the set of all even integers.

Let's try it out with all possibilities.

6(3)+2(-2) = 14 odd/even
6(2)+2(5) = 22 even/odd
6(-2)+2(4) = -4 even/even
6(5)+2(-7) = 16 odd/odd

The result is always even.

For the set to be Z , we would need an odd number to be able to multiply it by another odd number like for example 7(7) = 49 + 2(2) = 53

 

[CompuChip]

Correct!

Just to test your intuition, what do you think
\{ 6x + 9z \mid x, y \in \mathbb{Z} \}
will be?

 

[reenmachine]

Correct!

Just to test your intuition, what do you think
\{ 6x + 9z \mid x, y \in \mathbb{Z} \}
will be?

as I mentionned at the end of the post , this will be Z because there's an even and odd number.If you replaced the 9 with a 8 , it would be the set of all even integers.

If both would be odd , it would still be Z.

 

[micromass]

as I mentionned at the end of the post , this will be Z because there's an even and odd number.

But the number is always divisible by 3! (! is not a factorial here)

 

[reenmachine]

But the number is always divisible by 3! (! is not a factorial here)

what does it change? what role does division play in that situation?

 

[Number Nine]

as I mentionned at the end of the post , this will be Z because there's an even and odd number.If you replaced the 9 with a 8 , it would be the set of all even integers.

If both would be odd , it would still be Z.

Would a and b not have to be relatively prime? It seems to me that any linear combination of 6 and 9 would get you a multiple of 3.

 

[reenmachine]

okay you got me , give me a chance I just woke up from a pretty rough night

the set would be ( w in Z : there's exist a v in Z such that w = v/3 ) ?

not sure that even works for all , have to verify

 

[Number Nine]

Any pair of relatively prime integers a and b would do it. We know that there is a linear combination

ax + by = 1

But then we have

n(ax + by) = (nx)a + (ny)b = n

So we can express any integer as a linear combination of a and b.

 

[CompuChip]

Yep. You probably don't know what "relatively prime" means (yet), but that's the official term.
The trick is, if you look at the earlier example with a = 5, b = 2, that you can write 1 = 5 - 2 * 2.
So you can write any number n as $n(5 - 2 \cdot 2) = 5n + 2(n - 2)$, i.e. as 5x + 2y with x = n and y = n - 2.
It doesn't work for 4 and 2 or for 6 and 9, because you can't write 1 as 4x + 2y or 6x + 9y, so you can only get multiples of 2 and 3, respectively.

This is what Number Nine tries to tell you in more general terms.

 

[reenmachine]

Another try:

set = {x ∈ Z : ∃y ∈ Z x = 3y} ?

I have a feeling this is correct.

 

[Number Nine]

Another try:

set = {x ∈ Z : ∃y ∈ Z x = 3y} ?

I have a feeling this is correct.

This would give you the set of all multiples of 3.

 

[reenmachine]

This would give you the set of all multiples of 3.

Isn't all 6x + 9y sums (if x and y are members of Z) a multiple of 3?

EDIT: woah , I just noticed that the z isn't an element of Z , that's just a brain cramp there

 

[reenmachine]

But in that case isn't it the set R? If x=0 and z=1/9 then 6(0)+9(1/9)=1 and you can generate any number in theory no?

 

[Number Nine]

But in that case isn't it the set R? If x=0 and z=1/9 then 6(0)+9(1/9)=1 and you can generate any number in theory no?

You specified that both x and y are integers, in which case you would only have multiples of 3.

 

[reenmachine]

You specified that both x and y are integers, in which case you would only have multiples of 3.

The exemple was \{ 6x + 9z \mid x, y \in \mathbb{Z} \} in which I confused the z with the y.

If z was replaced by y on the left side (like below), I think every elements of that set would be multiples of 3.Am I wrong?

\{ 6x + 9y \mid x, y \in \mathbb{Z} \}

 

[reenmachine]

Suppose that X is a set on which an addition operation is defined. A subset $S\subseteq X$ is said to be closed under addition if for all $x,y\in S$, we have $x+y\in S$.

"Closed under multiplication" is defined similarly.

Very clear thank you!

 

[reenmachine]

I need to be reminded of something , looking at page 24-25 in the book of proof : http://www.people.vcu.edu/~rhammack/BookOfProof/Sets.pdf

why does i=1 below the notations?

 

[Fredrik]

I need to be reminded of something , looking at page 24-25 in the book of proof : http://www.people.vcu.edu/~rhammack/BookOfProof/Sets.pdf

why does i=1 below the notations?

It's just the conventional notation. For example, the sum of all the real numbers in the set $\{x_1,x_2,\dots,x_n\}\subseteq\mathbb R$ can be written as
$$\sum_{i=1}^n x_i.$$ The alternative is to define $I=\left\{i\in\mathbb Z\,|\,1\leq i\leq n\right\}$ and write
$$\sum_{i\in I} x_i.$$

 

[Fredrik]

By the way, the sigma notation for sums can be defined recursively.

For all $n\in\mathbb Z$, we define
$$\sum_{i=n}^n x_i=x_n.$$ For all $n,m\in\mathbb Z$ such that $m>n$, we define
$$\sum_{i=n}^m x_i =\left(\sum_{i=n}^{m-1}x_i\right)+x_m.$$ It's not essential that you understand this now, but you will have to get used to recursive definitions at some point.

Another example (of recursion): For each $n\in\mathbb N$ (where this $\mathbb N$ is defined to include 0), we define $n!$ by
$$n!=
\begin{cases}1 &\text {if }n=0\\
n(n-1)! &\text{if }n>0.
\end{cases}
$$ This definition implies e.g. that
$$5!=5(4!)=5(4(3!))=\cdots =5(4(3(2(1(0!))))) =5\cdot 4\cdot 3\cdot 2\cdot 1.$$

 

[CompuChip]

Another try:

set = {x ∈ Z : ∃y ∈ Z x = 3y} ?

I have a feeling this is correct.

Isn't all 6x + 9y sums (if x and y are members of Z) a multiple of 3?

EDIT: woah , I just noticed that the z isn't an element of Z , that's just a brain cramp there

Yep, you are correct in both instances (I just made your Z boldface, which is alternative notation for the set of integers \mathbb{Z}, rather than some unspecified set Z).

A common shortcut is just to write 3\mathbb{Z} by the way, so if you see that written the above set is meant.

 

[reenmachine]

Yep, you are correct in both instances (I just made your Z boldface, which is alternative notation for the set of integers \mathbb{Z}, rather than some unspecified set Z).

A common shortcut is just to write 3\mathbb{Z} by the way, so if you see that written the above set is meant.

I might have been correct , but in your exemple \{ 6x + 9z \mid x, y \in \mathbb{Z} \} , the small z isn't an element of Z , so isn't it incorrect?

 

[CompuChip]

Ah, yes, that is a typo. The z should have been a y. My apologies for the confusion that may have caused.

 

[reenmachine]

It's just the conventional notation. For example, the sum of all the real numbers in the set $\{x_1,x_2,\dots,x_n\}\subseteq\mathbb R$ can be written as
$$\sum_{i=1}^n x_i.$$ The alternative is to define $I=\left\{i\in\mathbb Z\,|\,1\leq i\leq n\right\}$ and write
$$\sum_{i\in I} x_i.$$

okay so i isn't really 1 , it's just a notation for a dummy variable.

when they write x1 , x2 and so on until xn , does it represent real numbers?

thanks

 

[micromass]

when they write x1 , x2 and so on until xn , does it represent real numbers?

In Fredrik's example: yes, since he detailed that $\{x_1,...,x_n\}\subseteq \mathbb{R}$. But in principle, it could be something entirely different.

For example, we could have $x_1 = 1$, $x_2 = 2$ and $x_3 = 13$, then

\sum_{i=1}^3 x_i = 1 + 2 + 13 = 16

 

[reenmachine]

Little bit of a rewind here , from the textbook I used at the beginning:

Some terminology.
Functions from A to B in the general case are said to be into B.If the range of the function equals B, then the function is onto B(or surjection). A function F: A → B is called one-to-one function (or injection) just in case no member of B is assigned to more than one member of A (so if a ≠ b, then F(a) ≠ F(b)). A function which is both one-to-one and onto is called a one-to-one correspondence (or bijection). It is easy to see that if a function F is one-to-one correspondence, then the relation F–1 is a function and one-to-one correspondence

Just to use an example , suppose A= {1 , 2 , 3} and B= {4 , 5 , 6}.

If F: A → B , then R ⊆ A × B right? (R being the relation)

If F(1)= 4 , F(2)= 5 and F(3)= 6 , then this is a surjection is that right? (and a bijection).This would also be a one-to-one function correct? Is this a one-to-one correspondance?

But if F(1)= 5 , F(2)= 5 and F(3)= 5 , then this is an injection?

About F-1 , is that the same concept as with relations -1? So in the last case above , F-1 would be what? It can't be F(5) = 1 , 2 , 3 can it?

thanks!

 

[reenmachine]

In Fredrik's example: yes, since he detailed that $\{x_1,...,x_n\}\subseteq \mathbb{R}$. But in principle, it could be something entirely different.

For example, we could have $x_1 = 1$, $x_2 = 2$ and $x_3 = 13$, then

\sum_{i=1}^3 x_i = 1 + 2 + 13 = 16

Very clear thank you!

 

[Fredrik]

okay so i isn't really 1 , it's just a notation for a dummy variable.

It's the start of the range of values for the dummy variable i. In the sum $x_1+\cdots+x_n$, the i has a different value in each term. The first is written below the sigma, the last above it. $$\sum_{i=1}^n x_i$$.

 

[Fredrik]

Just to use an example , suppose A= {1 , 2 , 3} and B= {4 , 5 , 6}.

If F: A → B , then R ⊆ A × B right? (R being the relation)

If F: A → B , then F ⊆ A × B. (This is if we use the definition of "function" from the pdf, which is equivalent to my definition 1). There's no need to introduce another symbol for "the relation", since functions are relations.

If F(1)= 4 , F(2)= 5 and F(3)= 6 , then this is a surjection is that right? (and a bijection).This would also be a one-to-one function correct? Is this a one-to-one correspondance?

Yes, to all questions.

But if F(1)= 5 , F(2)= 5 and F(3)= 5 , then this is an injection?

No, because here F(x)=F(y) doesn't imply x=y. F takes two elements of A to the same element of B.

About F-1 , is that the same concept as with relations -1? So in the last case above , F-1 would be what? It can't be F(5) = 1 , 2 , 3 can it?

F^-1 is only defined when F is bijective. So if we use your example bijection above, then $F^{-1}(4)=1$ and so on.

However, the set $F^{-1}(C)$ (called the preimage of C under f) is defined for all sets C, even if F is not surjective or injective. It's defined by $F^{-1}(C)=\{x\in A\,|\,F(x)\in C\}$.

 

[reenmachine]

It's the start of the range of values for the dummy variable i. In the sum $x_1+\cdots+x_n$, the i has a different value in each term. The first is written below the sigma, the last above it. $$\sum_{i=1}^n x_i$$.

very clear thank you !

 

[reenmachine]

If F: A → B , then F ⊆ A × B. (This is if we use the definition of "function" from the pdf, which is equivalent to my definition 1). There's no need to introduce another symbol for "the relation", since functions are relations.

good , so a function is always a relation but a relation is not always a function.

Yes, to all questions.

No, because here F(x)=F(y) doesn't imply x=y. F takes two elements of A to the same element of B.

In that case , F(1)=4 , F(2)=5 and F(3)=6 would be an injection?

F^-1 is only defined when F is bijective. So if we use your example bijection above, then $F^{-1}(4)=1$ and so on.

However, the set $F^{-1}(C)$ (called the preimage of C under f) is defined for all sets C, even if F is not surjective or injective. It's defined by $F^{-1}(C)=\{x\in A\,|\,F(x)\in C\}$.

Not sure I understand the last paragraph.First of all , in your last set notation , wouldn't x be an element of B instead of A since 4 , to use your exemple above , is an element of B? About set C , isn't C like the range since with F-1 we'll always have the inverse of a function and therefore will have elements of B inside the function?

thank you very much , your feedbacks are always greatly appreciated!

 

[Fredrik]

good , so a function is always a relation but a relation is not always a function.

Yes.

In that case , F(1)=4 , F(2)=5 and F(3)=6 would be an injection?

The F defined by these three equalities is an injection, yes. It's also a surjection. Since it's both, it's a bijection.

Not sure I understand the last paragraph.First of all , in your last set notation , wouldn't x be an element of B instead of A since 4 , to use your exemple above , is an element of B?

I don't see any typos in my notation. $F^{-1}(C)$ is a subset of A, for all sets C. I don't see what 4 has to do with it.

 

[reenmachine]

I don't see any typos in my notation. $F^{-1}(C)$ is a subset of A, for all sets C. I don't see what 4 has to do with it.

I'm having a hard time understanding this for some reasons.

$F^{-1}(C)=\{x\in A\,|\,F(x)\in C\}$

What happened to the $F^{-1}$ in the definition? If we are in the universe A×B , so the function is from A to B , suppose F(1)=4 and $F^{-1}$ (4)=1 , then since 4 is in B , why should $F^{-1}$ (C)=m = element of A? That's what I was wondering.

edit: I think I understand now , since x is in A , then function x = x in B , so this is what F-1(C) is made of?

 

[Fredrik]

It's just a definition, so it doesn't really require explanation. However, if $C\subseteq B$, and $F:A\to B$ is bijective onto B, so that $F^{-1}:B\to A$ is defined, then the preimage of C under F (the left-hand side below) is equal to the image of C under $F^{-1}$ (the right-hand side below). So it makes sense to use the notation $F^{-1}(C)$ for both.
$$\left\{x\in A\,|\,F(x)\in C\right\} =\left\{F^{-1}(x)\in A\,|\,x\in C\right\}.$$

 

[reenmachine]

It's just a definition, so it doesn't really require explanation. However, if $C\subseteq B$, and $F:A\to B$ is bijective onto B, so that $F^{-1}:B\to A$ is defined, then the preimage of C under F (the left-hand side below) is equal to the image of C under $F^{-1}$ (the right-hand side below). So it makes sense to use the notation $F^{-1}(C)$ for both.
$$\left\{x\in A\,|\,F(x)\in C\right\} =\left\{F^{-1}(x)\in A\,|\,x\in C\right\}.$$

Ok I get it now

thanks a lot for the patience!

 

[reenmachine]

Pretty tired today so just going to ask a few quick questions:

http://people.umass.edu/partee/NZ_2006/Set Theory Basics.pdf

The function F: A→A such that F= {<x,x> : x ∈ A} is called the identity function on A, written idA (or 1A).

Given a function F: A→B that is a one-to-one correspondence, we have the following equations:

F–1 °F= idA,
F° F–1 = ?

Is the answer to ? = idA-1 ?

Given two functions F: A→B and G: B→C, we may form a new function from A to C, called the composition of F and G, written G°F.

Function composition is defined as G°F = def{<x,z> for some y, <x,y> ∈ F and <y,z> ∈ G}

I think I understand the concept (except for why they decided to switch the letters in the notation but I guess that's not really important) but there's one thing I would like to be sure about.

Since it's G°F in the first place , why is it ''<x,z> for some y(?) , <x,y> ∈ F and <y,z> ∈ G'' , why isn't it for all y?

thanks a lot!

Now time to attack the book of proof with more consistancy.Will probably re-read the entire thread in the following week , I think it would be a good idea.

cheers!

 

[Office_Shredder]

Is the answer to ? = idA-1 ?

Nope. Think about what set the values you plug in and get out are going to be in.

Since it's G°F in the first place , why is it ''<x,z> for some y(?) , <x,y> ∈ F and <y,z> ∈ G'' , why isn't it for all y?

thanks a lot!

What this is saying is that G(F(x)) = z if F(x) = y and G(y) = z. y here is representing the specific value that F(x) takes - the statement that <x,y> ∈ F and <y,z> ∈ G isn't going to be true for EVERY y because there can only be one value of y with <x,y> ∈ F to begin with!

For a concrete example, let F and G be functions on the real numbers. F(x) = x+1 and G(x) = x².

Then G(F(x)) = (x+1)². Claim: G(F(2)) = 9. Reading off the definition, the reason why this is true is because there exists some real number y, in particular y=3, with F(2) = y and G(y) = 9. Obviously it's not going to be true that F(2) = y and G(y) = 9 for every real number

 

[Fredrik]

I see that Office_Shredder has already written a good answer, but I wrote half of this before I saw it, so I might as well post it.

Is the answer to ? = idA-1 ?

If you mean what I think you mean, then no, it's not. (Could you at least use sub/sup tags?)

why is it ''<x,z> for some y(?) , <x,y> ∈ F and <y,z> ∈ G'' , why isn't it for all y?

I would define $G\circ F:A\to C$ by saying that
$$(G\circ F)(x)=G(F(x))$$ for all $x\in A$. I have never thought about other ways to define it until now, but let's see. $G\circ F$ is the set of all the ordered pairs $(x,G(F(x)))$ with $x\in A$. So
$$G\circ F=\left\{(x,G(F(x)))|\, x\in A\right\} =\left\{(x,z)\,|\, x\in A\text{ and } z=G(F(x))\right\}.$$ Now consider the equality z=G(F(x)). If we define y by y=F(x), then we can write z=G(y). So $G\circ F$ is the set of all (x,z) such that $x\in A$ and $z=G(y)$, where y is an alternative notation for F(x).

You know that the equalities $y=F(x)$ and $z=G(y)$ can be written as $(x,y)\in F$ and $(y,z)\in G$ respectively. Suppose that we say that there exists a $y\in B$ such that $(x,y)\in F$. This statement is true if and only if x is in the domain of F. So it's a way of saying that $x\in A$, and at the same time reserve the symbol y for F(x).

This means that $G\circ F$ is the set of all (x,z) such that there exists a y in B such that (x,y) is in F and (y,z) is in G.
$$G\circ F=\left\{(x,z)\,|\,\exists y\in B~~ \left((x,y)\in F~\land~(y,z)\in G\right)\right\}.$$

 

[reenmachine]

If you mean what I think you mean, then no, it's not. (Could you at least use sub/sup tags?)

Very sorry for this , I pushed my luck long enough , I'll try to have some kind of ''LaTeX session'' this week where I will try to write different equations or formulas in LaTeX without worrying about the meaning just to get used to it.

I would define $G\circ F:A\to C$ by saying that
$$(G\circ F)(x)=G(F(x))$$ for all $x\in A$. I have never thought about other ways to define it until now, but let's see. $G\circ F$ is the set of all the ordered pairs $(x,G(F(x)))$ with $x\in A$. So
$$G\circ F=\left\{(x,G(F(x)))|\, x\in A\right\} =\left\{(x,z)\,|\, x\in A\text{ and } z=G(F(x))\right\}.$$ Now consider the equality z=G(F(x)). If we define y by y=F(x), then we can write z=G(y). So $G\circ F$ is the set of all (x,z) such that $x\in A$ and $z=G(y)$, where y is an alternative notation for F(x).

You know that the equalities $y=F(x)$ and $z=G(y)$ can be written as $(x,y)\in F$ and $(y,z)\in G$ respectively.

Very clear! Feel much better today and it shows in my understanding.

Suppose that we say that there exists a $y\in B$ such that $(x,y)\in F$. This statement is true if and only if x is in the domain of F. So it's a way of saying that $x\in A$, and at the same time reserve the symbol y for F(x).

Very clear again! Though it's pretty obvious that x is in the domain of F from where we come from with this.

This means that $G\circ F$ is the set of all (x,z) such that there exists a y in B such that (x,y) is in F and (y,z) is in G.
$$G\circ F=\left\{(x,z)\,|\,\exists y\in B~~ \left((x,y)\in F~\land~(y,z)\in G\right)\right\}.$$

Crystal clear! Loving it!

Thank you very much Fredrik , once again!

edit: Just for the sake of ultimate clearness , and I know we already talked about this , but in the last set notation , ''for each'' (x,z) is implied on the left side correct? Don't need to write the symbol?

edit2:We can see with your post that there's many many ways to write a notation for the same set.

 

[reenmachine]

Nope. Think about what set the values you plug in and get out are going to be in.




What this is saying is that G(F(x)) = z if F(x) = y and G(y) = z. y here is representing the specific value that F(x) takes - the statement that <x,y> ∈ F and <y,z> ∈ G isn't going to be true for EVERY y because there can only be one value of y with <x,y> ∈ F to begin with!

For a concrete example, let F and G be functions on the real numbers. F(x) = x+1 and G(x) = x².

Then G(F(x)) = (x+1)². Claim: G(F(2)) = 9. Reading off the definition, the reason why this is true is because there exists some real number y, in particular y=3, with F(2) = y and G(y) = 9. Obviously it's not going to be true that F(2) = y and G(y) = 9 for every real number

Also very clear! Thanks guys this really made me understood the concept perfectly!