# Thread: Constructing Z from N (integers from naturals)

1. ## Constructing Z from N (integers from naturals)

Having some trouble understanding this...

This method uses an equivalence relation defined on NxN such that: (a,b)~(c,d) iff a+d=c+b

Note that we can subtract any k in N from both a and b to get an equivalent pair, providied k < min(a,b) (makes sense)

It follows that every class [a,b] contains a special pair, in which atleast one of the components is equal to 1. Moreover there is only one such pair in each class:

(1,b)=(1,d) => b=d
(a,1)=(b,1) => a=b
(a,1)=(1,b) => a=b=1

What's going on here? Shouldn't the = signs be ~ signs?

The special pairs thus form a transversal,

T={(1,n+1): n in N} union {(1,1)} union {(n+1,1): n in N}

Not sure what's going on here either???

2. Can you possibility give us a reference on this?
Is there a textbook or lecture notes?
Frankly, I do not follow the construction.

PS
I have given thought to your posting. I had fallen to the sin of familiarity.
I know a similar approach: (a,b)~(c,d) iff (a-b)=(c-d).
But of course that is equivalent to the relation you posted.
Have you shown that that is an equivalence relation on NxN?

The equivalence class for [3,1] contains (4,2),(5,3),(9,7) etc.
The equivalence class for [1,3] contains (2,4),(3,5),(7,9) etc.
The equivalence class for [0,0] contains (2,2),(3,3),(7,7) etc.
Thus we can identify n with [a,b] iff a-b=n.
Thus we can identify -n with [b,a] iff a-b=n.
Thus we can identify $0_z$ with [0,0].

We define $+_z$ as $(a,b)+_z (c,d)=(a+c,b+d)$.

Can you carry on from here?

3. Originally Posted by Plato
Can you possibility give us a reference on this?
Is there a textbook or lecture notes?
Frankly, I do not follow the construction.

PS
I have given thought to your posting. I had fallen to the sin of familiarity.
I know a similar approach: (a,b)~(c,d) iff (a-b)=(c-d).
But of course that is equivalent to the relation you posted.
Have you shown that that is an equivalence relation on NxN?

The equivalence class for [3,1] contains (4,2),(5,3),(9,7) etc.
The equivalence class for [1,3] contains (2,4),(3,5),(7,9) etc.
The equivalence class for [0,0] contains (2,2),(3,3),(7,7) etc.
Thus we can identify n with [a,b] iff a-b=n.
Thus we can identify -n with [b,a] iff a-b=n.
Thus we can identify $0_z$ with [0,0].

We define $+_z$ as $(a,b)+_z (c,d)=(a+c,b+d)$.

Can you carry on from here?
Apologies, I think i'm at fault here. I only posted the very beginning of the construction. I kind of assumed it was a standard approach and other people would be familiar with it. I'll post it in its entirety now since I don't really follow the last 3 lines of your explanation. The construction I have here seems a lot more thorough but i wonder if it is essentially the same as yours?
By the way the textbook i'm quoting from is 'Elements of Logic via Numbers and Sets' by D L Johnson. I could link you to some handwritten lecture notes online but i don't think they'd be very illuminating.

This method uses an equivalence relation defined on NxN such that: (a,b)~(c,d) iff a+d=c+b

Note that we can subtract any k in N from both a and b to get an equivalent pair, providied k < min(a,b)

It follows that every class [a,b] contains a special pair, in which atleast one of the components is equal to 1. Moreover there is only one such pair in each class:

(1,b)=(1,d) => b=d
(a,1)=(b,1) => a=b
(a,1)=(1,b) => a=b=1

The special pairs thus form a transversal,

T={(1,n+1): n in N} union {(1,1)} union {(n+1,1): n in N}

Next, attempt to define the addition of classes in a natural way:

[a,b] + [c,d] = [a+c,b+d]

We need to check for independence of representatives in order that this operation be well defined. The effect of this operation on the members of T works out as follows:

for all (a,b) in NxN, [a,b] + [1,1] = [a+1,b+1] = [a,b]

so that [1,1] is an identity for this +. Next,

[m+1,1] + [n+1,1] = [m+n+2,2] = [(m+n)+1,1]
[1,m+1] + [1,n+1] = [2,m+n+2] = [1,(m+n)+1]

so that the first and last sets of T both look like copies on N

Finally,

[n+1,1] + [1,n+1] = [n+2,n+2] = [1,1]

so that [1, n+1] is the additive inverse of [n+1,1]

Hope this makes sense. It seems to have similarities to yours. I understand the intention of the construction and what it arrives at seems to make sense, I think it's mainly just the initial steps i'm confused about.

4. Originally Posted by Obstacle1
It seems to have similarities to yours. I understand the intention of the construction and what it arrives at seems to make sense, I think it's mainly just the initial steps i'm confused about.
In fact, the approaches are identical. I think that your textbook uses a slightly different definition for the set of natural numbers. That is, I am accustomed to using $N=\{0,1,2,3,\cdots\}$ whereas your textbook probability does not include 0 in $N$. So if you have a particular question, I may be able to answer it.

5. Originally Posted by Plato
In fact, the approaches are identical. I think that your textbook uses a slightly different definition for the set of natural numbers. That is, I am accustomed to using $N=\{0,1,2,3,\cdots\}$ whereas your textbook probability does not include 0 in $N$. So if you have a particular question, I may be able to answer it.

I think i've resolved all my problems now. One thing, what does a copy of N mean exactly?

6. Originally Posted by Obstacle1
One thing, what does a copy of N mean exactly?
You do know that we must show that addition is well defined?
Also we must show that there is a well defined linear ordering.
As well as a well defined muptiplicative operation.

That is, each idea is a copy of the natural numbers.

8. Originally Posted by ThePerfectHacker
Actually that reference is not very helpful.
In fact it may be misleading.

The material that I have used with undergraduates is found in this text: An Outline of Set Theory by James M Henle.

9. Originally Posted by Plato
You do know that we must show that addition is well defined?
Also we must show that there is a well defined linear ordering.
As well as a well defined muptiplicative operation.

That is, each idea is a copy of the natural numbers.
Got it. Thanks