# Thread: Proof that 1 is in the Natural Numbers

1. ## Proof that 1 is in the Natural Numbers

1 is an element of the Natural Numbers.

Hey all, I would really appreciate some help with the above proof. Thanks a bunch!

2. The answer to this question, and others you posted in this forum recently, depends on the context. If you need a secondary-school proof, then this is the wrong forum because Number Theory is an advance university subject. Even pre-university forums deal mostly with high-school math, whereas the fact that 1 is a natural number is known in elementary school.

On the other hand, if you need a formal proof, then you need to specify the definitions and axioms you are using, such as the definition of natural numbers in set theory, ring axioms, Peano axioms, etc.

3. Originally Posted by jstarks44444
1 is an element of the Natural Numbers.

Hey all, I would really appreciate some help with the above proof. Thanks a bunch!
Tell us what definition of the Naturals you are using.

In Peano arithmetic if we start with 0, and the successor operation s and define n to be a natural if it is either 0 or the successor of a natural number, then as "1" is the name of s0 it is trivially a natural number.

If we start Peano arithmetic with 1 and the successor operation then "1" is by definition a natural number.

Simular arguments apply to the set theoretic construction on the Naturals.

CB

4. The Naturals in this context is defined by the following:

There exists a subset N in Z (integers) with the following properties:

(i) If m,n are elements of N then m+n is an element of N
(ii) If m,n are elements of N then mn is an element of N
(iii) 0 is not an element of N
(iv) For every m that is an element of Z, we have m is an element of N or m=0 or -m is an element of N

This proof needs to be proved by contradiction by the way.

5. Originally Posted by jstarks44444
There exists a subset N in Z (integers) with the following properties:
(i) If m,n are elements of N then m+n is an element of N
(ii) If m,n are elements of N then mn is an element of N
(iii) 0 is not an element of N
(iv) For every m that is an element of Z, we have m is an element of N or m=0 or -m is an element of N
This proof needs to be proved by contradiction by the way.
Suppose that $\displaystyle 1\notin \mathbb{N}$.
We know that $\displaystyle 1\in \mathbb{Z}$ so then either $\displaystyle 1=0$ or $\displaystyle -1\in \mathbb{N}$ (iv).
From the integer properties you should have that $\displaystyle 1\not= 0$.
That leaves $\displaystyle -1\in \mathbb{N}$.
Can you finish? [hint: (ii)]

6. (-1)(-1) leads to a contradiction! Thanks a lot for the help!