Question about proof from a guy with a highschool education

[Fredrik]

If I were to say that a is in R and b is in [0,1] , would that make sense or would it be irrelevant because a and b are dummy variables?

I would interpret your statement as "there exists a,b such that z=(a,b)", so both variables are the target of a "there exists", and strictly speaking, that makes them dummy variables, and a reference to a or b in the next sentence doesn't make sense. However, since of the abuse of language I described earlier is so common, I would have interpreted a reference to a or b in the next sentence as if you had made two separate statements: "There exists c,d such that z=(c,d)" and "Let a and b be such that z=(a,b)".

It makes sense to change "This implies that z is an element of the form (a,b)" to "This implies that z is an element of the form (a,b) with $a\in\mathbb R$ and $b\in[0,1]$", and I consider this an improvement. But what's really missing from your statement is something that makes it clear that you are assigning values to the variables a and b. (So that it makes sense to refer to them later).

You could e.g. say "define $a\in\mathbb R$ and $b\in[0,1]$ by $z=(a,b)$" or "let a,b be the unique real numbers such that z=(a,b)"

I'm not sure I understand this , doesn't it say that z=(x,y) and that those x and y are the x in R and y in [0,1] I was talking about?

You said that for all real numbers x,y with y in [0,1], we have z=(x,y). This implies that all of the following equalities and infinitely many more are all true: $(1,0)=z,\ (-1/12,0)=z,\ (\pi,1/2)=z$...and they all contradict each other.

Let $z \in R × [0,1]$ be arbitrary.This implies that there's an $x \in R$ and a $y \in [0,1]$ such that $z \in (x,y)$.

The last $\in$ should be an equality, but I assume that was just a typo (hmm...repeated at the end of the proof). The statement is fine apart from that. You could say that there's a unique $x\in\mathbb R$ and a unique $y\in[0,1]$ such that z=(x,y) if you want to emphasize that the values of x and y are fully determined by z, but it's not necessary to do that in this proof.

This implies that $x \in \bigcup_{a\in R}\{a\}$ and that $y \in [0,1]$ such that $(x,y) \in \bigcup_{a \in R}\{a\} × [0,1]$.

The "such that" makes the sentence weird. You can end the sentence after $y\in[0,1]$, and then say "this implies that...". But you don't actually have to mention that $y\in[0,1]$, since you did that earlier.

It looks like what you're trying to do here is what I would say like this: Since $\mathbb R=\bigcup_{a\in R}\{a\}$, this implies that we have $z=(x,y)\in\left(\bigcup_{a\in R}\{a\}\right)\times[0,1]$.

But this right-hand side isn't the one we're interested in. We want to prove that $z\in \bigcup_{a\in R}\left(\{a\}\times[0,1]\right)$.

 

[reenmachine]

I meant z = (x,y) , both times.Sorry about that.Also not sure if you saw the edited version.

 

[reenmachine]

It looks like what you're trying to do here is what I would say like this: Since $\mathbb R=\bigcup_{a\in R}\{a\}$, this implies that we have $z=(x,y)\in\left(\bigcup_{a\in R}\{a\}\right)\times[0,1]$.

But this right-hand side isn't the one we're interested in. We want to prove that $z\in \bigcup_{a\in R}\left(\{a\}\times[0,1]\right)$.

Yes this is pretty much what I tried to do.Seems I confused everything again.

 

[Fredrik]

I think the comment "I'm done editing" was there when I started typing the reply.

 

[reenmachine]

In that case , it should be easier to prove in the sense you only have to prove that $z=(x,y)$ and that $(x,y) \in {a} × [0,1]$

Let $z \in R × [0,1]$ be arbitrary.This implies that there's an $x \in R$ and a $y \in [0,1]$ such that $z = (x,y)$.This implies that $(x,y) \in \bigcup_{a \in R}\{a\} × [0,1]$ which implies that $z \in \bigcup_{a \in R}\{a\} × [0,1]$.

?

Is this sufficient? The truth is it looks like much of the same of what I previously attempted.I'm hitting a wall.

 

[Fredrik]

OK, I'll tell you.

Let $z \in R × [0,1]$ be arbitrary.This implies that there's an $x \in R$ and a $y \in [0,1]$ such that $z = (x,y)$.

All you need to say after that is that
$$z=(x,y)\in\{x\}\times[0,1]\subseteq\bigcup_{a\in\mathbb R}\{a\}\times[0,1].$$

 

[reenmachine]

OK, I'll tell you.All you need to say after that is that
$$z=(x,y)\in\{x\}\times[0,1]\subseteq\bigcup_{a\in\mathbb R}\{a\}\in[0,1].$$

$$z=(x,y)\in\{x\}\times[0,1]\subseteq\bigcup_{a\in\mathbb R}\{a\} ×[0,1]?$$

× instead of $\in$ at the end no?

Okay so I had to go through the $(x,y) \in \{x\} × [0,1] \subseteq\bigcup_{a\in\mathbb R}\{a\} ×[0,1] ...$.

But why is it necessary to say that $(x,y) \in \{x\} × [0,1]$ , isn't implied in $(x,y) \in R × [0,1]$ when it was already said that $z=(x,y)$ and that $z \in R × [0,1]$ ?

 

[reenmachine]

I guess my next move is to sleep on it and see what happens tomorrow

This has been one of my roughest day in this thread.

 

[Fredrik]

× instead of $\in$ at the end no?

Yes. (I have edited that now).

Okay so I had to go through the $(x,y) \in \{x\} × [0,1] \subseteq\bigcup_{a\in\mathbb R}\{a\} ×[0,1] ...$.

But why is it necessary to say that $(x,y) \in \{x\} × [0,1]$ , isn't implied in $(x,y) \in R × [0,1]$ when it was already said that $z=(x,y)$ and that $z \in R × [0,1]$ ?

Actually, what you said in #560 is fine. I just think it's clearer with the extra step. I mean, it follows immediately from the definition of $\times$ that $(x,y)\in\{x\}\times[0,1]$, and now the definition of "union" implies that $(x,y)\in\bigcup_{a\in\mathbb R}\left(\{a\}\times[0,1]\right)$.

 

[reenmachine]

Yes. (I have edited that now).Actually, what you said in #560 is fine. I just think it's clearer with the extra step. I mean, it follows immediately from the definition of $\times$ that $(x,y)\in\{x\}\times[0,1]$, and now the definition of "union" implies that $(x,y)\in\bigcup_{a\in\mathbb R}\left(\{a\}\times[0,1]\right)$.

This was the problem , I wasn't sure what was missing.I would lie if I told you I'm 100% confidant in my understanding of everything we've discussed today.But I'll get there eventually.

Not sure that it was the easiest exemple to try either.

thank you infinitely for the patience!

 

[reenmachine]

By the way , it's been a month now since this thread has been created!

I am extremely satisfied with the help I've received.This has been an incredibly positive experience for me to the point where I deeply regret not doing it years ago.

Learning some completely unknown mathematics has been tough and remains tough , especially with my poor background , but I'm impressed by the quality of the helpers helping me getting through these concepts despite my lack of knowledge.

A special thank to you Fredrik , who has been the most active helper all along , you truly deserve to be called a mentor! I think I should also name micromass who has been extremely helpful to me.

thank you very much to everybody who helped me in the thread!

 

[reenmachine]

When we said $(x,y)\in\{x\}\times[0,1]$ yesterday , what does $\{x\}$ truly mean here? I know it's suppose to be any real number , but what difference does it make to put an $x$ out there instead of $R$ or $a$ like in the union notation? Is it because we want to pinpoint that $x$ is there as $\{x\}$ confirming both the position of $x$ and the position of $y$ in the formula ? Isn't $(x,y)$ enough? I'm trying to understand the importance of this statement.Is it because we want to pass through $\{x\} × [0,1]$ to reach $\subseteq\bigcup_{a\in\mathbb R}\{a\} ×[0,1]$? If so , why is that crucial? Why not $\{a\}$ instead of $\{x\}$?

 

[reenmachine]

I would interpret the notation on the right here as
$$\bigcup_{a\in \mathbb{R}} \left(\{a\}\times [0,1]\right),$$ not as
$$\left(\bigcup_{a\in \mathbb{R}} \{a\}\right)\times [0,1].$$ The notation is kind of ambiguous though. I don't know if one of these interpretations are "standard". I'm also not sure how you're interpreting it.

I think this was the most problematic aspect of how I handled the whole thing yesterday.

I think I even confused which notations I was trying to prove in the middle of my whole thought process.Here we want to prove the first one correct?

So by saying that $(x,y)\in\{x\}\times[0,1]$ , we're saying that $z \in (G)$ and that $z \in \bigcup (G)$ if $z = (x,y)$ and $(G)=\{\{x\}×[0,1]\}$.Is that a correct way of viewing things? Because in my mind , if that's correct , it's much clearer now than yesterday.

 

[reenmachine]

A very simple thing seems to confuse me at this very moment , what's the difference between a set $A$ and $\bigcup A$?

Suppose $A=\{ (0,1) , (1,2)\}$.

The reason I'm asking is I'm not sure of the difference between

$$\bigcup_{a\in \mathbb{R}} \left(\{a\}\times [0,1]\right),$$
and

$\{a\} × [0,1]$ with $a \in R$

 

[Fredrik]

When we said $(x,y)\in\{x\}\times[0,1]$ yesterday , what does $\{x\}$ truly mean here? I know it's suppose to be any real number , but what difference does it make to put an $x$ out there instead of $R$ or $a$ like in the union notation? Is it because we want to pinpoint that $x$ is there as $\{x\}$ confirming both the position of $x$ and the position of $y$ in the formula ? Isn't $(x,y)$ enough? I'm trying to understand the importance of this statement.Is it because we want to pass through $\{x\} × [0,1]$ to reach $\subseteq\bigcup_{a\in\mathbb R}\{a\} ×[0,1]$? If so , why is that crucial? Why not $\{a\}$ instead of $\{x\}$?

x and y are the unique real numbers such that z=(x,y). {x} is the singleton set with x as its only element. I'm not sure where you would like to put an $a$ instead of $x$.

We want to prove that every element of $\mathbb R\times [0,1]$ is an element of $\bigcup_{a\in\mathbb R}\left(\{a\}\times[0,1]\right)$. So we start by saying "Let $z\in\mathbb R\times[0,1]$ be arbitrary", and set out to prove that z is an element of the set on the right-hand side.

We define x and y by z=(x,y). Since $z\in\mathbb R\times[0,1]$, this ensures that $x\in\mathbb R$ and $y\in[0,1]$. The three statements $z=(x,y)$, $x\in\mathbb R$ and $y\in[0,1]$, together imply that $z\in\{x\}\times[0,1]$. Since $x\in\mathbb R$ and $y\in[0,1]$, this result implies that $z\in\bigcup_{a\in\mathbb R}\left(\{a\}\times[0,1]\right)$.

Another way of saying that last bit: Since x is a real number and z is an element of $\{x\}\times[0,1]$, z is an element of one of the infinitely many sets $\{a\}\times[0,1]$ with $a\in\mathbb R$, and by definition of "union", that implies that z is an element of the union of those sets.

 

[Fredrik]

A very simple thing seems to confuse me at this very moment , what's the difference between a set $A$ and $\bigcup A$?

Suppose $A=\{ (0,1) , (1,2)\}$.

You're making it complicated by choosing a set with ordered pairs as elements, but maybe you meant { when you wrote (? I will consider the example with { instead.

If $A=\{\{0,1\},\{1,2\}\}$, then $\bigcup A=\{0,1\}\cup\{1,2\}=\{0,1,2\}$.

 

[reenmachine]

You're making it complicated by choosing a set with ordered pairs as elements, but maybe you meant { when you wrote (? I will consider the example with { instead.

If $A=\{\{0,1\},\{1,2\}\}$, then $\bigcup A=\{0,1\}\cup\{1,2\}=\{0,1,2\}$.

Okay , but what if $A = \{1,2\}$? Is it the same set?

 

[micromass]

A very simple thing seems to confuse me at this very moment , what's the difference between a set $A$ and $\bigcup A$?

Suppose $A=\{ (0,1) , (1,2)\}$.

Everything in mathematics is a set, so $(0,1)$ and $(1,2)$ are sets. So $(0,1)\cup (1,2)$ makes sense. So the answer is

\bigcup A = (0,1)\cup (1,2)

However, while it does make sense, it is something you never really want to do. If you ever encounter a situation where you would have to evaluate $(0,1)\cup (1,2)$, then you made a mistake somewhere.

That said, I absolutely hate the notation $\bigcup$. I think it's a very bad notation. But it's used quite a lot.

 

[micromass]

Okay , but what if $A = \{1,2\}$? Is it the same set?

In that case, you have

\bigcup A = 1\cup 2

Again, this makes perfect sense since $1$ and $2$ are sets. But it is not something you ever want to do.

 

[reenmachine]

In that case, you have

$$\bigcup A = 1\cup 2$$

Again, this makes perfect sense since $1$ and $2$ are sets. But it is not something you ever want to do.

So if $1=\{u,y\}$ and $2=\{z,x\}$ , then

$$\bigcup A = 1\cup 2 = \{u,y,z,x\} ?$$

Basically it's just not good to reach the level where you can't split sets into elements.

 

[micromass]

So if $1=\{u,y\}$ and $2=\{z,x\}$ , then

$$\bigcup A = 1\cup 2 = \{u,y,z,x\} ?$$

Yes.

 

[reenmachine]

x and y are the unique real numbers such that z=(x,y). {x} is the singleton set with x as its only element. I'm not sure where you would like to put an $a$ instead of $x$.

We want to prove that every element of $\mathbb R\times [0,1]$ is an element of $\bigcup_{a\in\mathbb R}\left(\{a\}\times[0,1]\right)$. So we start by saying "Let $z\in\mathbb R\times[0,1]$ be arbitrary", and set out to prove that z is an element of the set on the right-hand side.

We define x and y by z=(x,y). Since $z\in\mathbb R\times[0,1]$, this ensures that $x\in\mathbb R$ and $y\in[0,1]$. The three statements $z=(x,y)$, $x\in\mathbb R$ and $y\in[0,1]$, together imply that $z\in\{x\}\times[0,1]$. Since $x\in\mathbb R$ and $y\in[0,1]$, this result implies that $z\in\bigcup_{a\in\mathbb R}\left(\{a\}\times[0,1]\right)$.

Another way of saying that last bit: Since x is a real number and z is an element of $\{x\}\times[0,1]$, z is an element of one of the infinitely many sets $\{a\}\times[0,1]$ with $a\in\mathbb R$, and by definition of "union", that implies that z is an element of the union of those sets.

Clear enough thank you!

Just a last thing I'm wondering about , when we say $R × [0,1]$ it's definitely not ALL elements of $R$ as the $\{x\}$ here right? $R$ means any elements of $R$ but not all of them at the same time correct?

This would gives us an ordered pair like $(allrealnumbers , 1)$.

 

[Fredrik]

So if $1=\{u,y\}$ and $2=\{z,x\}$ , then

$$\bigcup A = 1\cup 2 = \{u,y,z,x\} ?$$

Yes, but you could use a more common definition of 1 and 2:

0=∅
1={0}
2={0,1}

Edit: I edited the following string of equalities after reenmachine spotted a silly mistake and asked about it below:
$$\bigcup \{1,2\}=\{0\}\cup\{0,1\}=\{0,1\}=2.$$ Of course, this isn't anything you'd ever want to do either.

Basically it's just not good to reach the level where you can't split sets into elements.

They always can be. That's what we mean by "everything in mathematics is a set".

But sometimes, you will not want to consider their elements, and in those cases, it's rather pointless to consider the union of the elements.

 

[Fredrik]

Just a last thing I'm wondering about , when we say $R × [0,1]$ it's definitely not ALL elements of $R$ as the $\{x\}$ here right? $R$ means any elements of $R$ but not all of them at the same time correct?

This would gives us an ordered pair like $(allrealnumbers , 1)$.

I don't understand what you're saying, but since we know that the cartesian product of any two sets is a set, we can write $$\mathbb R\times[0,1]=\{(x,y) : x\in\mathbb R,\ y\in[0,1]\}.$$ If you read the right-hand side the way you've been taught, you should understand what the members of $\mathbb R\times[0,1]$ are.

I'm heading out to get some food. Gone for an hour at least.

 

[reenmachine]

Yes, but you could use a more common definition of 1 and 2:

0=∅
1={0}
2={0,1}
$$\bigcup U\{1,2\}=\{0\}\cup\{0,1\}=\{0\}=1.$$ Of course, this isn't anything you'd ever want to do either.

I don't get it , why does $\{0\}\cup\{0,1\}=\{0\}$?

 

[Fredrik]

Lol, now I confused unions with intersections. Obviously, $\{0\}\cup\{0,1\}=\{0,1\}$.

 

[reenmachine]

I don't understand what you're saying, but since we know that the cartesian product of any two sets is a set, we can write $$\mathbb R\times[0,1]=\{(x,y) : x\in\mathbb R,\ y\in[0,1]\}.$$ If you read the right-hand side the way you've been taught, you should understand what the members of $\mathbb R\times[0,1]$ are.

I'm heading out to get some food. Gone for an hour at least.

Yep I do.Thanks and bon appetit!

 

[micromass]

Yes, but you could use a more common definition of 1 and 2:

0=∅
1={0}
2={0,1}
$$\bigcup U\{1,2\}=\{0\}\cup\{0,1\}=\{0\}=1.$$ Of course, this isn't anything you'd ever want to do either.

Just as an aside. IF $A\subseteq \mathbb{N}$, then under these definitions, we have

\bigcup A = \max A

Which already gives an indication that the union and the maximum/supremum are closely related. This observation is very useful in axiomatic set theory and when dealing with ordinals and cardinals.

Sorry, I couldn't resist

 

[reenmachine]

Lol, now I confused unions with intersections. Obviously, $\{0\}\cup\{0,1\}=\{0,1\}$.

and here I thought I was in for another long day on this very beautiful and sunny friday.

 

[reenmachine]

$$\bigcup \{1,2\}=\{0\}\cup\{0,1\}=\{0,1\}=2.$$

I understand the mechanism , but the result $=2$ at the end while the set was the union of $\{1,2\}$ is a little bit confusing.

I understand why it's $2$ based on the operations , but the result compared to the previous unionized set is bizarre.

$$\bigcup \{1,2\}=2$$

 

[Fredrik]

It just looks weird because of the definitions of 1 and 2. What we're doing is just a special case of the following: For all sets x and y, we have
$$\bigcup\{\{x\},\{x,y\}\}=\{x\}\cup\{x,y\}=\{x,y\}.$$ And this result doesn't look weird at all. But we can make it look weird by choosing to introduce a notation like $a=\{x\},\ b=\{x,y\}$, which turns the above into
$$\bigcup\{a,b\}=a\cup b=b.$$ This last equality holds for all sets a and b such that $a\subseteq b$, not just when there exist sets x,y such that $a=\{x\},\ b=\{x,y\}$.

 

[reenmachine]

It just looks weird because of the definitions of 1 and 2. What we're doing is just a special case of the following: For all sets x and y, we have
$$\bigcup\{\{x\},\{x,y\}\}=\{x\}\cup\{x,y\}=\{x,y\}.$$ And this result doesn't look weird at all. But we can make it look weird by choosing to introduce a notation like $a=\{x\},\ b=\{x,y\}$, which turns the above into
$$\bigcup\{a,b\}=a\cup b=b.$$ This last equality holds for all sets a and b such that $a\subseteq b$, not just when there exist sets x,y such that $a=\{x\},\ b=\{x,y\}$.

very clear thank you!

So $\bigcup\{a,b\}=b \leftrightarrow a \subseteq b$ ?

 

[Fredrik]

So $\bigcup\{a,b\}=b \leftrightarrow a \subseteq b$ ?

Yes. Note however that the equality is almost always written as $a\cup b=b$. The notation $\bigcup S$ for the union of the elements of some set S is only useful when we don't want to think of a notation for the elements of S.

 

[reenmachine]

Yes. Note however that the equality is almost always written as $a\cup b=b$. The notation $\bigcup S$ for the union of the elements of some set S is only useful when we don't want to think of a notation for the elements of S.

Good! Very clear!

Thanks!

 

[reenmachine]

(Just using this thread to post a proof that was directed at a specific person just in case I would want to go back at it for some reasons)

We want to prove that to give a unique room number (in an hotel with an infinite number of rooms) to an infinite number of passengers coming from an infinite number of bus , we can use the technic of:

5th passenger on 2nd bus = 5 + ( 00000 ) + 2 + ( 00 ) = 500000200 = $RN$ = Room Number and that everytime we use this technic , no matter the passenger or bus number , it will generate a unique room number for the new guests.

I would first like to introduce some notations I had to use to get my point across.Apologize if these notations exist in other form I am unaware of.I want to split the room number in four parts , the passenger number , the zeroes added after it , the bus number and the zeroes added after it.

$K$ = {Passenger number}
$V$ = {Bus number}
$a$ = {$K$ amount of zeroes added after $K$ in $RN$ }
$b$ = {$V$ amount of zeroes added after $V$ in $RN$ }

Let me explain quickly what I meant.

If $K=2$ and $V=4$
Then $a=00$ and $b=0000$
By definition , $a=00$ is placed after $2$ and $0000$ is placed after $4$ in $20040000 = RN$

This means that $K_1=2$ then $a_1=00$ and if $V_3=4$ then $b_3=0000$ and finally:
$K_1 \ a_1 \ V_3 \ b_3 = 20040000 = RN$

(Take note that $K_1$ could be any passenger number , not necessarily $K = 1$ , but that $a$ is dependant on $K$ and $b$ is dependant on $V$)

I have to specify that we accept that if a room number has a different amount of digits than another room number , then $RN_1 ≠ RN_2$.

For example , 13 = 2 and 3455 = 4.It's impossible than a number with 2 digits is the same as a number with 4 digits.

We will use $x$ as $\{amount \ of \ digits \ in \ K\}$ and $c$ as $\{amount \ of \ digits \ in \ V \}$.

We want to do this to simplify our job to find the amount of digits that will be in the room number based on $K$ and $V$.

If $K=2$ and $V=4$ , then to find the amount of digits in the room number , then we can use $\{K + x + V + c\}$.

Amount of digits in $K$ is 1 since 2 is only one digit.So $x=1$.Same with $V$ , since 4 is a one digit number than $c=1$.

To get back to our little formula $\{K + x + V +c\}$ = number of digits in $RN$ , if $K=2$ and $V=4$ , then to amount of digit of the room will be $\{2+1+4+1=8\}$.

20040000 = 8 digits

The reason I'm using this is to prove that $\{K_1+x_1\} + \{V_1 + c_1\} ≠ \{K_1 + x_1\} + \{V_z + c_z\}$ if $z ≠ 1$ and therefore that
every $RN$ will be different for any $K_1$ if you change $V$ (or vice versa) since the amount of digits of $RN$ will be different.

To prove this , we can take $K_1 = 5$.

$\{K_1+x_1\}$ will always equal to $\{5+1\}=6$.

If we keep the same $K$ but change the $V$ , the amount of total digits in $RN$ will always change and therefore the room number will always be different.

6 + (2+1) =9
...
6 + (7+1) =14
6 + (8+1) =15
6 + (9+1) =16
6 + (10+2) =18

(9 ... 14 , 15 , 16 , 18 ...)

I want to say that I must intuitively accept that this series always goes up and remains constant.The series will always be +1 ... +1 +2 +1 ... +1 +2 +1 with the ''...'' being the amount of +1s between the +2 which will happen once everytime you add a digit to $V$.That being said , the series will always go +1 or +2 and nothing else , therefore the if you have the $K$ and change $V$ or vice versa , you will always have a different $RN$ because two such $RN$ won't have the same amount of digits.

This implies that anytime to add $1$ to $K$ or $V$ , you either add $1$ or $2$ digits to $RN$ (if the other $K$ or $V$ doesn't change at the same time).

1 = 10 (2) +1
2 = 200 (3) +1
3 = 3000 (4) +1
...
8 = 800000000 (9) +1
9 = 9000000000 (10) +2
10 = 10 00000 00000 (12)

That proves that if $RN_3 = K_1 \ a_1 \ V_2 \ b_2$ , then $RN_3 ≠ K_1 \ a_1 \ V_u \ b_u$ if $u ≠ 2$ and $RN_3 ≠ K_n \ a_n \ V_2 \ b_2$ if $n ≠ 1$.

In clearer terms , this proves that all passengers from the same bus won't ever have the same room number (since they have a different $K$ (passenger number)) and all passengers with the same passenger number won't ever have the same room number (since they are from a different $V$ (bus)).

Now we want to prove that $K_1 \ a_1 \ V_2 \ b_2 ≠ K_3 \ a_3 \ V_4 \ b_4$ to complete the proof.

As I already demonstrated , if you take out the $K$ part of all this problem (or the $V$ part) , each different $V$ (or $K$) will simply add $1$ or $2$ digit to $RN$.

Every $RN$ will always end with one 0 or more based on $V$.Each different $V$ will give a different ending to $RN$ based on the unique amount of $0$ the $RN$ will end with.

(which was proved with the: 10 - 200 - 300 -... 9000000000 - 10 0000000000) If 1,2,3,9,10 are some $V$ , then all their $RN$ won't finish the same way which proves that their $RN$ won't be the same.

This means that independantly of the $K$ part , any different $V$ will generate a unique $RN$.This proves that $K_1 \ a_1 \ V_2 \ b_2 ≠ K_3 \ a_3 \ V_4 \ b_4$.

Combines this with the earlier proof that if $RN_3 = K_1 \ a_1 \ V_2 \ b_2$ , then $RN_3 ≠ K_1 \ a_1 \ V_u \ b_u$ if $u ≠ 2$ and $RN_3 ≠ K_n \ a_n \ V_2 \ b_2$ if $n ≠ 1$.

This should complete the proof that the technic generate a new and unique room number based on the passenger and bus numbers.

Basically , to resume , either you take the same $V$ and changes the $K$ which will generate a unique $RN$ for each situation , or either you take two different $V$ and this also implies that both $RN$ will be different.

 

[reenmachine]

Allright I've been re-reading some of the earlier pages in the thread in order to refresh my mind and there's two things I was wondering about in this following post:

Definition 1

A set $f\subseteq X\times Y$ is said to be a function from X into Y, if
(a) For all $x\in X$, there's a $y\in Y$ such that $(x,y)\in f$.
(b) For all $x,x' \in X$ and all $y\in Y$, if $(x,y)\in f$ and $(x',y)\in f$, then $x=x'$.
X is said to be the domain of f. Y is said to be a codomain of f. f is also called the graph of f. So the function and its graph is the same thing.

Definition 2

A triple $f=(X,Y,G)$ such that $G\subseteq X\times Y$ is said to be a function from X into Y, if
(a) For all $x\in X$, there's a $y\in Y$ such that $(x,y)\in G$.
(b) For all $x,x' \in X$ and all $y\in Y$, if $(x,y)\in G$ and $(x',y)\in G$, then $x=x'$.
X is said to be the domain of f. Y is said to be the codomain of f. G is said to be the graph of f.

I don't recall what $x'$ means in the (b) part of the first definition.

Why is the $G$ placed after $X,Y$ in $f=(X,Y,G)$ ?

thanks!

 

[CompuChip]

x' is just a variable, just like x and y. We could also have called it u, or a, or $\lambda$. However, in the definition we use two pairs, with the same entry y in the second slot, and by convention for one of them the first entry is called x. For symmetry / aesthetic reasons we prefer x' - after all it is very much "like" x.
For example, the following are all equivalent statements:
(1) "For all x, y, u and v, if (x, y) = (u, v) then x = u and y = v"
(2) "For all x, y, $\xi$ and $\eta$, if (x, y) = $(\xi, \eta)$ then $x = \xi$ and $y = \eta$"
(3) "For all x, y, x' and y', if (x, y) = (x', y') then x = x' and y = y'"
(4) "For all x₁, y₁, x₂ and y₂, if (x₁, y₁) = (x₂, y₂) then x₁ = x₂ and y₁ = y₂"

Most people prefer (3) or (4) to indicate that the variables have a similar function in the statement (and to keep other letters free for other purposes). (3) is mostly used if you have two or three "similar" variables - I could introduce another variable x'' - whereas (4) is more common for many, an unspecified number - as in x₁, x₂, ..., x_n - or an infinite number -as in x₁, x₂, ...

For your section question: it's just a matter of definition. You could put them in any order, although this one is the one that (intuitively, to me) makes the most sense.

 

[Fredrik]

I don't recall what $x'$ means in the (b) part of the first definition.

It's just another variable. I chose to use x and x' for elements of X, and y for elements of Y.

When I write "for all $x,x'\in X$,...", it means "for all x in X, and for all x' in X,..."

Why is the $G$ placed after $X,Y$ in $f=(X,Y,G)$ ?

No reason. We could write down a "definition 3" that says $f=(G,X,Y)$ instead of $f=(X,Y,G)$. This definition would be yet another way to make the idea of "a rule that associates exactly one element of Y with each element of X" mathematically precise. This "definition 3", would be technically different from both definitions 1 and 2, but it would be for all practical purposes equivalent to definition 2, since it (unlike definition 1) ensures that each function has a unique codomain.

We also don't need to include X in the triple to accomplish that, because if we know G, we know X too. So we could write down a "definition 4" that says $f=(Y,G)$ and a "definition 5" that says $f=(G,Y)$.

It's not uncommon that there are lots of ways to make the same idea mathematically precise.

 

[reenmachine]

x' is just a variable, just like x and y. We could also have called it u, or a, or $\lambda$. However, in the definition we use two pairs, with the same entry y in the second slot, and by convention for one of them the first entry is called x. For symmetry / aesthetic reasons we prefer x' - after all it is very much "like" x.
For example, the following are all equivalent statements:
(1) "For all x, y, u and v, if (x, y) = (u, v) then x = u and y = v"
(2) "For all x, y, $\xi$ and $\eta$, if (x, y) = $(\xi, \eta)$ then $x = \xi$ and $y = \eta$"
(3) "For all x, y, x' and y', if (x, y) = (x', y') then x = x' and y = y'"
(4) "For all x₁, y₁, x₂ and y₂, if (x₁, y₁) = (x₂, y₂) then x₁ = x₂ and y₁ = y₂"

Most people prefer (3) or (4) to indicate that the variables have a similar function in the statement (and to keep other letters free for other purposes). (3) is mostly used if you have two or three "similar" variables - I could introduce another variable x'' - whereas (4) is more common for many, an unspecified number - as in x₁, x₂, ..., x_n - or an infinite number -as in x₁, x₂, ...

For your section question: it's just a matter of definition. You could put them in any order, although this one is the one that (intuitively, to me) makes the most sense.

very clear thank you!

 

[reenmachine]

It's just another variable. I chose to use x and x' for elements of X, and y for elements of Y.

When I write "for all $x,x'\in X$,...", it means "for all x in X, and for all x' in X,..."

I see , the reason I got confused was probably that I was reading about relations and sets at the same time and sometimes you have $R'$ or $A'$ so the $'$ had a specific meaning in those cases.

No reason. We could write down a "definition 3" that says $f=(G,X,Y)$ instead of $f=(X,Y,G)$. This definition would be yet another way to make the idea of "a rule that associates exactly one element of Y with each element of X" mathematically precise. This "definition 3", would be technically different from both definitions 1 and 2, but it would be for all practical purposes equivalent to definition 2, since it (unlike definition 1) ensures that each function has a unique codomain.

We also don't need to include X in the triple to accomplish that, because if we know G, we know X too. So we could write down a "definition 4" that says $f=(Y,G)$ and a "definition 5" that says $f=(G,Y)$.

It's not uncommon that there are lots of ways to make the same idea mathematically precise.

Perhaps I'm a little bit more confused than I thought.

$G$ is the graph of f? Don't think I ever got introduced to this concept before.Not sure why I didn't ask about it earlier this month.

Let's try to make this all clearer.

Is $f = (X,Y,G)$ something like function from $X$ to $Y$ to $G$ ?

BTW , just as a refresher again , in $X × Y$ , are both $X$ and $Y$ subsets of $X × Y$? Another stupid question that I should know by now , but there's a difference between using the symbol $×$ to describe a cartesian product and a function is that right?

Sorry lol , going back at it makes a lot of questions pop in my mind.Thank you man!

 

[Fredrik]

I see , the reason I got confused was probably that I was reading about relations and sets at the same time and sometimes you have $R'$ or $A'$ so the $'$ had a specific meaning in those cases.

I forgot that you used ' for complement. I use $^c$ myself, i.e. $A^c$ denotes the complement of $A$.

Is $f = (X,Y,G)$ something like function from $X$ to $Y$ to $G$ ?

I don't know what that would mean, but G is what definition 1 calls a "function from X into Y". By definition 2, it's the triple (X,Y,G) that's called a "function from X into Y". G is then called "the graph" of that function, i.e. the graph of (X,Y,G).

BTW , just as a refresher again , in $X × Y$ , are both $X$ and $Y$ subsets of $X × Y$?

Neither of them is. The elements of $X\times Y$ are ordered pairs (x,y) with $x\in X$ and $y\in Y$.

Edit: You need to pick up the habit of trying to answer these questions yourself when they come up. You know that for X to be a subset of X×Y, every element of X must be an element of X×Y. So take an arbitrary element of X and ask yourself, is it an element of X×Y? If you just use the definition of "cartesian product" here, you will see that it's not.

One of the best reasons to learn how to prove these things is that you won't ever remember all the simple results, but once you have reached a certain level, you can prove them in less time than it will take to look them up. Then you will never have to wonder if (for example) $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$. You can just take 2 minutes off to find out.

...there's a difference between using the symbol $×$ to describe a cartesian product and a function is that right?

I don't follow you here. The symbol $\times$ denotes the cartesian product of two sets. It's also used to denote the product of real numbers, or the "cross product" of elements of $\mathbb R^3$. (Never mind if you don't know what that is).

 

[reenmachine]

I forgot that you used ' for complement. I use $^c$ myself, i.e. $A^c$ denotes the complement of $A$.

Then I use $^c$ from now on.It's a better notation anyway.

I don't know what that would mean, but G is what definition 1 calls a "function from X into Y". By definition 2, it's the triple (X,Y,G) that's called a "function from X into Y". G is then called "the graph" of that function, i.e. the graph of (X,Y,G).

Is there any exemple suppose using $X=\{1\}$ and $Y=\{2\}$?

What is $G$? Is it $G: X \rightarrow Y$ or $G(1) = 2$?

Neither of them is. The elements of $X\times Y$ are ordered pairs (x,y) with $x\in X$ and $y\in Y$.

Yeah I figured it out but you posted before I could edit.

Edit: You need to pick up the habit of trying to answer these questions yourself when they come up. You know that for X to be a subset of X×Y, every element of X must be an element of X×Y. So take an arbitrary element of X and ask yourself, is it an element of X×Y? If you just use the definition of "cartesian product" here, you will see that it's not.

You are completely right.I planned to do a massive amount of exercises yesterday but got hooked on something else all day.I will do it this week to try to really control the knowledge I gained here this last month.Once I'm done with the exercises , I will post the interesting ones in the homework forum.

One of the best reasons to learn how to prove these things is that you won't ever remember all the simple results, but once you have reached a certain level, you can prove them in less time than it will take to look them up. Then you will never have to wonder if (for example) $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$. You can just take 2 minutes off to find out.

Yes you are right it's better that way to be sure.

I don't follow you here. The symbol $\times$ denotes the cartesian product of two sets. It's also used to denote the product of real numbers, or the "cross product" of elements of $\mathbb R^3$. (Never mind if you don't know what that is).

Yes sorry got confused for a moment.

Thanks man!

 

[Fredrik]

Is there any exemple suppose using $X=\{1\}$ and $Y=\{2\}$?

What is $G$? Is it $G: X \rightarrow Y$ or $G(1) = 2$?

If you had picked larger sets, the question wouldn't have made sense, since you didn't specify a function. But since you picked singleton sets, there's exactly one function from X into Y. Let's denote it by f. We have f(1)=2, and therefore $G=\{(1,f(1))\}=\{(1,2)\}$. This happens to be equal to $X\times Y$ because of your choice of X and Y.

The general formula is $G=\left\{(x,f(x)):x\in X\right\}$. This is a subset of $X\times Y$.

 

[reenmachine]

If you had picked larger sets, the question wouldn't have made sense, since you didn't specify a function. But since you picked singleton sets, there's exactly one function from X into Y. Let's denote it by f. We have f(1)=2, and therefore $G=\{(1,f(1))\}=\{(1,2)\}$. This happens to be equal to $X\times Y$ because of your choice of X and Y.

The general formula is $G=\left\{(x,f(x)):x\in X\right\}$. This is a subset of $X\times Y$.

Very clear thank you!

 

[Fredrik]

Note that in the case of a function from ℝ into ℝ, the graph can be visualized as a curve. In the case of of a function from ℝ² into ℝ, it can be visualized as a surface. See http://en.wikipedia.org/wiki/Graph_of_a_function for a few pictures.

Maybe it was a picture like these that gave you the idea that X and Y are subsets of X×Y. It certainly looks like they are in the picture. The entire plane can be viewed as a visual representation of X×Y, and the "x axis" (the horizontal line at the center) can be viewed as a visual representation of X. However, if we view the plane as a representation of X×Y, the x-axis is really a representation of the set $\{(x,0):x\in X\}\subseteq X\times Y$. Because of the obvious correspondence between the elements of X and the elements of this set, it can also be thought of as a representation of X.

You should make sure that you understand that when $X\subseteq\mathbb R$ and $Y=\mathbb R$ the two conditions in my definition can be interpreted like this:

1. For all x in X, a vertical line through (x,0) intersects the graph at least once.
2. For all x in X, a vertical line through (x,0) intersects the graph at most once.

 

[reenmachine]

I'm back!

I want something to be crystal clear in my head:

$A=\{1,2\}$
$B=\{2,3\}$
$A \cup B = \{1,2,3\}$
$A \cap B = \{2\}$
$A × B = \{(1,2) , (1,3) , (2,2) , (2,3)\}$

Then what exactly is $$\bigcup(A × B)$$ ?

Is it $\{1,2,3\}$?

Basically , when you have to unionize a cartesian product set , you take all the elements inside the ordered pairs to make the set?

So any notation like $$\bigcup(A × B)$$ with a cartesian product will be equal to $A \cup B$?

 

[ArcanaNoir]

The union over AxB is the same as what you wrote for AxB.
Better example:
\cup_{n\in \mathbb{N}} [0,\frac{1}{n}]\text{x} [0,1] = [0,1]\text{x}[0,1]

 

[micromass]

The union over AxB is the same as what you wrote for AxB.

Not true. The union here is

\bigcup A\times B = (1,2)\cup (1,3)\cup (2,2)\cup (2,3)

The right-hand side is well-defined since everything is a set. But it is not something you will encounter much in mathematics. It's well-defined, but not something you really want to do.

 

[reenmachine]

Not true. The union here is

\bigcup A\times B = (1,2)\cup (1,3)\cup (2,2)\cup (2,3)

The right-hand side is well-defined since everything is a set. But it is not something you will encounter much in mathematics. It's well-defined, but not something you really want to do.

So does $(1,2)\cup (1,3)\cup (2,2)\cup (2,3) = \{1,2,3\}$ ?

Or would this happened only if you unionize the set once again on a higher level?

For example $$\bigcup(\bigcup A\times B)$$

Or does that don't even make sense at all?

 

[micromass]

So does $(1,2)\cup (1,3)\cup (2,2)\cup (2,3) = \{1,2,3\}$ ?

What is the definition of $(1,2)$ and the other couples? What if you take the union?