# Proof that 1 is in the Natural Numbers

• Feb 15th 2011, 12:10 AM
jstarks44444
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!
• Feb 15th 2011, 01:17 AM
emakarov
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.
• Feb 15th 2011, 05:32 AM
CaptainBlack
Quote:

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
• Feb 16th 2011, 10:55 AM
jstarks44444
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.
• Feb 16th 2011, 11:04 AM
Plato
Quote:

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 $1\notin \mathbb{N}$.
We know that $1\in \mathbb{Z}$ so then either $1=0$ or $-1\in \mathbb{N}$ (iv).
From the integer properties you should have that $1\not= 0$.
That leaves $-1\in \mathbb{N}$.
Can you finish? [hint: (ii)]
• Feb 16th 2011, 11:12 AM
jstarks44444