1 is an element of the Natural Numbers.

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

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- Feb 15th 2011, 12:10 AMjstarks44444Proof 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 AMemakarov
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 AMCaptainBlack
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 AMjstarks44444
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 AMPlato
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)] - Feb 16th 2011, 11:12 AMjstarks44444
(-1)(-1) leads to a contradiction! Thanks a lot for the help!