# 3 probably simple questions

• Oct 6th 2009, 09:27 PM
ChrisBickle
3 probably simple questions
Im doing a section on arithmetic and ordering in natural numbers and i got 3 questions i cant figure out how to get started.

1. 0 does not equal 1 (duh?)

2. for a,b element of natural numbers if ab = 1 a = 1 b = 1.
Can we use the a1=a somehow?

3. a >= 0

• Oct 6th 2009, 10:41 PM
Matt Westwood
How you approach this problem depends completely on what approach was made in the definition of what natural numbers are.

How were they defined in your source work? From the Peano postulates, as a naturally ordered semigroup, or from the Zermelo-Fraenkel axioms?

I expect you already have that $\mathbb{N}$ is a well-ordered set such that 0 is the "smallest" member, which gives you the answer to 3 straight away.
• Oct 7th 2009, 07:00 PM
ChrisBickle
First ill give you what we got to work with our teacher wasnt very specific but a list of properties is on the top of the page so i figure thats what we get to use. We have that a and b are closed under multiplication and addition, a = b implies a+c=b+c, a +1 not equal 0, if a not 0 then a = d +1, then we also have associativity, commutivity, distributive law, identities in both addition and multiplication, and additive cancellation.

I was working with someone today and I think we came up with a solution but i dont know if it is correct.

Assume 0 = 1 then there exists b such that 0 + b = 1 + b by cancellation 0=1 contradiction
I dont imagine the solution to this is very hard, but for some reason that doesnt look legal to me