(A simple

number theory proof.)

The **intuitive explanation** of the proof is as follows:

One of the fundamental properties of the positive integers (also known as the natural numbers) is that any nonempty set of them has a least element. Compare this to the set of all integers--The negative numbers, a subset of all the integers, has no least element. This means that, unlike all the integers, which go to negative infinity, the positive integers "stop" at some point, at least in the direction of small numbers.

So what is that least element of the positive integers? Assume it is `z`, an element smaller than 1. But if we square `z` to get `z`^{2}, another positive integer smaller than 1 and, indeed smaller than `z`. (Another property of the positive integers is that any two, multiplied together, produce another positive integer.) But this goes against the assumption that `z` is the least element of the positive integers. So `z` cannot exist and 1 is the smallest positive integer.

See, if there were a positive integer smaller than 1, it would have powers that telescope infinitely into the space between 0 and 1. But this would not fit our intuitive notion of the positive integers and, indeed, violates the Peano postulates that are used to construct the positive integers.

**Formal Proof:**

Assume that we have a nonempty set `S` of integers between 0 and 1.

By the well ordering principle, `S` must have a least element. (The well ordering principle is a property of the positive integers.) Call this element `z`. `z` < 1, by our assumption.

By our

lemma,

`z` < 1 ⇒

`z`^{2} <

`z`. But, by

closure,

`z`^{2} is a positive integer. Moreover,

`z`^{2} ∈

`S` (by definition, as

`S` contains all positive integers between 0 and 1). Clearly,

`z` is not the smallest element of

`S`.

⇒ ⇐

Thus, `S` is empty and there are no integers between between 0 and 1.