# 2 very basic questions

1. Aug 13, 2013

### bobby2k

Hi

I am trying to learn how the number systems was created, and there are two very basic thing I don't get.

first question:

My book describes and proves that addition is well-defined for integers Z.

that is if, z2=z3, then
z1+z2 = z1+z3

It also does the same for rational number, it gives a proof that

if q2 = q3, then
q1+q2=q1+q3

However one thing that puzzles me is that I can not find a proof that it holds for naturlar numbers.

That is if:

n2= n3

then:
n1+n2 = n1+n3

I know this is probably very basic, can I assume it is correct, or should it also be proved?

They define the natural numbers as cardinal numbers of sets. And proves many laws like m+n=n+m etc. for natural numbers, but not the one I asked above.

second question:
This question is probably very stupid, but since it seems like everything should be proved at this basic level, why can I assume that if a = b, then b = a, is this how = is defined?

2. Aug 13, 2013

### Staff: Mentor

All natural numbers are integers (and all integers are rational numbers).
If something is true for all integers, it is true for all natural numbers.

It is a part of its definition.

3. Aug 13, 2013

### phinds

Well, natural numbers are a subset of integers so wouldn't it follow that if it's true for integers it HAS to be true for natural numbers?

EDIT: I see mfb beat me to it

Last edited: Aug 13, 2013
4. Aug 13, 2013

### bobby2k

The problem with what you are saying with, N beeing a subset of Z, therefore if addition is well defined in Z then it is in N, is that for me it seemed like they used that addition in N was well defined when they proved it in Z.

the proof for Z is like this:
[m,n], [p,q] and [r,s] are integers. Where m,n,p,q,r,s is in N.
then we have that [p,q] = [r,s], so p+s = q+r

then they prove that [m+p,n+q] = [m+r,n+s]

they do it like this:

(m+p) + (n+s) = m+n+p+s = m+n+q+r = (n+q) + (m+r),
so by the definition of Z [m+p,n+q] = [m+r,n+s]

but when we write :
m+n+p+s = m+n+q+r
Aren't we using that addition is well defined for N?

EDIT:

Because if m+n = a, p+s=b, and q+r = c, then we have since p+s=q+r, then b=c
so we use that a+b = a+c, if b=c

Last edited: Aug 13, 2013
5. Aug 13, 2013

### SteamKing

Staff Emeritus
'derfor'?

6. Aug 13, 2013

### phinds

That's Upper Slobovian for "therefore"

7. Aug 13, 2013

### bobby2k

Sorry for using a word in a foreign language, it is fixed now.

8. Aug 13, 2013

### bobby2k

I tried making a proof, is this proof valid:

First my book defines addition in W(whole numbers) as. If, m,n$\in$ W, and A and B are sets suck that m = #(A) and n = #(B) and A $\cap$ B = $\phi$, then m+n = #(A$\cup$B)
#is the cardinal number

then I want to prove that if b=c, then a+b=a+c
To prove this I assume that
a = #(A) for a set A, and b = #(B) for a set B, I also assume that A$\cap$B = $\phi$
I also assume that c = #(C) for a set C

Now since A and B are disjoint I get directly from the definition that a+b = #(A$\cup$B)
I also have that c=#(C)=#(B)=b

Now instead of saying that c=#(C), I can just use that c = #(B), and since this it all it takes to use the definition of addition(?, can I just choose to use the other set), then I have that since A and B are disjoint
a+c = #(A$\cup$B)
and because of this
a+b = #(A$\cup$B) = a+c