# Thread: Prove that there is no positive integer between 0 and 1

1. ## Prove that there is no positive integer between 0 and 1

Theorem : Prove that there is no positive integer between 0 and 1

<proof>
Suppose there is a positive integer 'a' between 0 and 1. Let S = {n E Z+ | o < n <1}. Since 0 < a < 1, a E S, so S in nonempty.
</proof>

<me> So far fine from side</me>

<proof>
Therefore, by well ordering principle, S has a least element 'l', where o < l < 1
</proof>

<me> fine from my side</me>

<proof>
Then 0 < l^2 < l, so l^2 E S.
</proof>

<me>
This is not fine. First of all from the set S is supposed to hold only +ve integers, given by the author as
S = {n E Z+ | 0 < n 1}. And the square of a no is less than the original no, only if the no is less than 1. But when the author has assume that 'S' contains only +ve integers Z+, he can't assume that 'l' is a non-integer which he has done in this case by saying 0 < l^2 < l.

Here is the rest of the proof anyways. If I am wrong I would greatly appreciate your help in understanding why I am wrong.
</me>

<proof>
But l^2 < l, which contradicts our assumption that 'l' is a least element of S.Thus there are +ve integers
</proof>

2. Originally Posted by scw_0611
Therefore, by well ordering principle, S has a least element 'l', where o < l < 1
Then 0 < l^2 < l, so l^2 E S.

This is not fine. First of all from the set S is supposed to hold only +integers, given by the author as S = {n E Z+ | 0 < n 1}. And the square of a no is less than the original no, only if the no is less than 1.
Given any $\displaystyle x \in (0,1)$ we have $\displaystyle 0 < x^2 < x < 1$. To see that simply multiply the inequality $\displaystyle 0 < x < 1$ through by $\displaystyle x$.

3. Got it. I just went on to think that the inequality can't be true because an inequality a^2 < a only if 0 < a < 1, but I forgot that we are proving by contradiction.

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# prove that there is no any integer between 0 and 1?

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