# Prove 1 is the least element in the positive integers

• Nov 3rd 2008, 12:27 PM
superevilcube
Prove 1 is the least element in the positive integers
I did a proof, it sounds right to me, but I want to make sure.

Prove:
1 is the smallest element in Z+ (the positive integers)

Suppose not, then there exists some integer $a\in Z+$ such that $a<1$

Let $S=[x\in Z+|x<1]$. By our assumption that $a<1$, $a\in S$. Thus S is not the empty set and by the Well Ordering Principle there exists a smallest element, call it $b$. Since $b<1$ it follows that $b-1<0$.

Thus, $b(b-1)<0$ which implies that $b^2. It follows that $b^2<1$. Thus $b^2\in S$. But $b^2, a contradiction since $b$ is the smallest element in S. Thus 1 must be the smallest element in the positive integers.

Is this right?
• Nov 4th 2008, 05:16 AM
poutsos.B
Quote:

Originally Posted by superevilcube
I did a proof, it sounds right to me, but I want to make sure.

Prove: 1 is the smallest element in Z+ (the positive integers)

Suppose not, then there exists some integer $a\in Z+$ such that $a<1$

Let $S=[x\in Z+|x<1]$. By our assumption that $a<1$, $a\in S$. Thus S is not the empty set and by the Well Ordering Principle there exists a smallest element, call it $b$. Since $b<1$ it follows that $b-1<0$.

Thus, $b(b-1)<0$ which implies that $b^2. It follows that $b^2<1$. Thus $b^2\in S$. But $b^2, a contradiction since $b$ is the smallest element in S. Thus 1 must be the smallest element in the positive integers.

Is this right?

Using the well ordering principal you can prove what you want to prove in the following way:

The principal says that:

For all sets S, S $\subseteq$ N and S=/= Φ ( S being non empty) S has a least element.

Since the above holds for all non empty subsets of N and since N is non empty and a subset of itself then N has a least member.

Alternatively we can prove the above by induction in the following way:

for n=1 we have .............1>=1 and the theorem is true for n=1

suppose now the theorem true for n=k and k>=1 .And we need to prove that : k+1>= 1.

But k>=1 ====> k+1>= 1+1 >1 +0, since 1>0 and so

................k+1>1 =====> k+1>=1 and the theorem is proved
• Nov 5th 2008, 03:04 AM
I think what poutsos is trying to say is that the well ordering principle is a consequence of 1 being the smallest element of $\mathbb{Z}^+$. He may or might not be right, depending on what axioms you are using. In the construction I learnt, 1 was defined as the smallest positive integer.