# Thread: Prove 1 is the least element in the positive integers

1. ## 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 $\displaystyle a\in Z+$ such that $\displaystyle a<1$

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

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

Is this right?

2. 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 $\displaystyle a\in Z+$ such that $\displaystyle a<1$

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

Thus, $\displaystyle b(b-1)<0$ which implies that $\displaystyle b^2<b$. It follows that $\displaystyle b^2<1$. Thus $\displaystyle b^2\in S$. But $\displaystyle b^2<b$, a contradiction since $\displaystyle b$ 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 $\displaystyle \subseteq$ 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

3. I think what poutsos is trying to say is that the well ordering principle is a consequence of 1 being the smallest element of $\displaystyle \mathbb{Z}^+$. 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.

Your proof is entirely sound if you are allowed the well ordering principle.