Proving a+b=b+a in Axiomatic Construction of Naturals

[sutupidmath]

Prove that a+b=b+a?

Hi all,

I am learning the axiomatic construction of Naturals, from the Peano axiom. Hence proving every property of naturals based solely on these axioms. I have come to some point where i am not very sure that what i am doing is okay, since there is not an explicit proof of this, it is left as an exercise to the reader.

I am trying to prove that for every a,b from naturals, a+b=b+a.

Here it is how i am going about it

S=\{b\in N | a+b=b+a\} where a is any fixed natural numeber.

My attempt is to show that S is an inductive set.
We first need to prove that

1\in S(?)

Here it is how i go about it, this is the part that i am not sure,

Using another theorem, where it says that a+1=a', where a' is the successor of a...this in fact is the theorem where they define the operation '+'. But to prove this, the author defines this: 1+b=b' for every b in N.------(1)

Now on my proof above, i am not using the definition of '+' but rather this other definition 1+b=b' for every b in N. ... so my question is am i allowed to do this, or there is another way of going about it.

Here is how i go about it
a+1=a' from the definition 0f +
=1+a from definition (1)
this actually means that 1 is an element of S. the rest i can easily prove if this is correct.

So?

thnx

 

[HallsofIvy]

I did much the same thing, proving all the properties of the natural numbers from Peano's axioms, the going on to integers, rational numbers, and real numbers.

The definition of addition should have two parts: a+ 1 is defined as a' and, if b is NOT 1, so that b= c' for some c, a+ b is defined as (a+ c)'.

Also, I used a separate induction to prove that a+ 1= 1+ a: let U(a) be the set of all a such that a+ 1= 1+ a. It is clear that 1 is in that set. Assume a is in that set. Then a'+ 1 is, by definition (a')'. 1+ a', by the second part of the definition of addition, is (1+ a)' and the rest follows from the fact that a+ 1= 1+ a.

 

[bkelly]

Knowing that you mathematicians must have a mathematical proof for everything, let's try a layman’s test of a + b = b + a.

If I put 3 buttons in a box, then put 4 more buttons in the box, I have 7 buttons. If I take them out and put 4 buttons in, followed up by 3 buttons, I again have 7.

Understanding that one example, or even Z examples, does not make a proof, can we derive a layman’s proof from this?

Maybe: If we put some portion ( X ) of N buttons in a box, then put in the remainder of N buttons, ( Y ), then we have N buttons in the box. If we do this in reverse, we always have N buttons in the box.

I could add some more verbiage to ensure preciseness, but the concept is clear.

 

[D H]

First off, a+b is not always equal to b+a. Mathematicians have invented several number systems in which addition is not commutative.

You are talking about the positive integers, which are described by the Peano postulates. Addition is commutative in the positive integers. One way to prove this is by using induction in conjunction with the associativity of addition.

First, prove that a+1=1+a for all positive integers a

1. Assume the hypothesis is true for some a.
2. (a+1)+1 = (1+a)+1 by assumption.
3. (1+a)+1 = 1+(a+1) by associativity of addition.
4. (a+1)+1 = 1+(a+1) by transitivity of equality.
5. a+1=1+a for a=1 by reflexivity of equality (all natural numbers are equal to themselves).
6. i]a[/i]+1=1+a for all positive integers a by induction.

Now prove a+b=b+a:

1. Assume the hypothesis is true for some positive integers a, b.
2. (b+1)+a = (1+b)+a by 1+k=k+1.
3. (1+b)+a = 1+(b)+a) by associativity of addition.
4. 1+(b)+a) = (b)+a)+1 by 1+k=k+1.
5. (b)+a)+1 = (a)+b)+1 by assumption.
6. (a)+b)+1 = a+(b+1) by associativity of addition.
7. (b+1)+a = a+(b+1) by transitivity of equality.
8. b+a = a+b for b=1 by 1+k=k+1.
9. b+a = a+b for all positive integers a, b by induction.

 

[Werg22]

Knowing that you mathematicians must have a mathematical proof for everything, let's try a layman’s test of a + b = b + a.

If I put 3 buttons in a box, then put 4 more buttons in the box, I have 7 buttons. If I take them out and put 4 buttons in, followed up by 3 buttons, I again have 7.

Understanding that one example, or even Z examples, does not make a proof, can we derive a layman’s proof from this?

Maybe: If we put some portion ( X ) of N buttons in a box, then put in the remainder of N buttons, ( Y ), then we have N buttons in the box. If we do this in reverse, we always have N buttons in the box.

I could add some more verbiage to ensure preciseness, but the concept is clear.

It's interesting that you bring this up. Arithmetic is axiomized from the everyday experience with quantities of whole objects. The commutative law is obtained by inductive reasoning; then we proceed to list it as an axiom for the deductive system we wish to build (arithmetic). Because a result obtained by inductive reasoning can never be certain, it may be that the commutative law sometimes does not hold in nature (as much as this is against our intuition). If this is the case, arithmetic is not a theory which fits the behavior of whole objects.

 

[sutupidmath]

I did much the same thing, proving all the properties of the natural numbers from Peano's axioms, the going on to integers, rational numbers, and real numbers.

The definition of addition should have two parts: a+ 1 is defined as a' and, if b is NOT 1, so that b= c' for some c, a+ b is defined as (a+ c)'.

Also, I used a separate induction to prove that a+ 1= 1+ a: let U(a) be the set of all a such that a+ 1= 1+ a. It is clear that 1 is in that set. Assume a is in that set. Then a'+ 1 is, by definition (a')'. 1+ a', by the second part of the definition of addition, is (1+ a)' and the rest follows from the fact that a+ 1= 1+ a.

Well, i guess this is what i was looking for. I did not look at it carefuly, i am in a rush right now, but i am going to come to it later, so if i have anything unclear, i will get back to u guys.

Because i was only looking how to prove that a+1=1+a, because i can do the rest.

THank you guys.