well-ordering property

logistic_guy said:
Use the well-ordering property to show that \(\displaystyle \sqrt{3}\) is irrational.
Click to expand...
This is one of the beautiful results in advanced calculus. But one needs to master rather advanced concepts to understand the proof.
Here is an outline for one. The floor of real number is ls the largest integer not exceeding that number.
Examples: [imath]\lfloor 4,95\rfloor]=4\,\,\,[/imath], [imath]\lfloor \pi\rfloor=3\,\,\,[/imath], [imath]\lfloor -1.9\rfloor=-2[/imath].

Lemma: If [imath]x-y>1[/imath] then [imath](n\in\mathbb{Z})[/imath] such that [imath]x<n<y[/imath].
Proof: [imath]\lfloor y\rfloor\le y[/imath] add [imath]1[/imath] to get [imath]y \le\lfloor y\rfloor+1\le y+1<x[/imath]
Thus [imath]y<\lfloor y\rfloor+1<x[/imath] and [imath]\lfloor y\rfloor+1[/imath] is an integer between [imath]y\,\&\, x[/imath]

Suppose that [imath]p[/imath] is a non-square positive real number, We proceed to show that [imath]\sqrt{\,p\,}[/imath] is irrational.
That will answer the question, HOW? We that [imath]\sqrt{\,p\,}=\frac{a}{b}[/imath] where [imath]\{a,b\}\subset\Z^+[/imath] i.e. is rational.
Now moreover If that means that [imath]b\cdot\sqrt{\,p\,}=a[/imath], Is that correct?
So say [imath]\mathcal{T}=\{n\in Z^+:n\cdot\sqrt{\,p\,}\in\Z^+[/imath], We know that [imath]b\in\mathcal{T}[/imath], HOW & WHY?
Thus [imath]\mathcal{T}\ne\emptyset[/imath] So being bounded below the set [imath]\mathcal{T}[/imath] by completeness has a first term call it [imath]\mathbb{K}[/imath].
Now let [imath]\mathbb{H}=\lfloor\sqrt{\,p\,}\rfloor[/imath], We know that because [imath]p[/imath] is a non-square positive integer [imath]\sqrt{p}>1[/imath]
Because [imath]\mathbb{H}[/imath] is the greatest integer NOT exceeding [imath]\sqrt{\,p\,}[/imath], we know that [imath]\sqrt{\,p\,}< \mathbb{H}+1[/imath]. Which gives us [imath]\sqrt{\,p\,}-\mathbb{H}<1[/imath]
Now here comes the contradiction this gives [imath]\sqrt{\,p\,}\cdot\mathbb{K}-\mathbb{H}\mathbb{K}<\mathbb{K}[/imath]
[imath]\sqrt{\,p\,}\cdot\mathbb{K}-\mathbb{H}\mathbb{K}[/imath] is an integer but also [imath]\sqrt{\,p\,\cdot(\mathbb{K}-\mathbb{H}\mathbb{K}))[/imath]
 
pka said:
This is one of the beautiful results in advanced calculus. But one needs to master rather advanced concepts to understand the proof.
Here is an outline for one. The floor of real number is ls the largest integer not exceeding that number.
Examples: [imath]\lfloor 4,95\rfloor]=4\,\,\,[/imath], [imath]\lfloor \pi\rfloor=3\,\,\,[/imath], [imath]\lfloor -1.9\rfloor=-2[/imath].

Lemma: If [imath]x-y>1[/imath] then [imath](n\in\mathbb{Z})[/imath] such that [imath]x<n<y[/imath].
Proof: [imath]\lfloor y\rfloor\le y[/imath] add [imath]1[/imath] to get [imath]y \le\lfloor y\rfloor+1\le y+1<x[/imath]
Thus [imath]y<\lfloor y\rfloor+1<x[/imath] and [imath]\lfloor y\rfloor+1[/imath] is an integer between [imath]y\,\&\, x[/imath]

Suppose that [imath]p[/imath] is a non-square positive real number, We proceed to show that [imath]\sqrt{\,p\,}[/imath] is irrational.
That will answer the question, HOW? We that [imath]\sqrt{\,p\,}=\frac{a}{b}[/imath] where [imath]\{a,b\}\subset\Z^+[/imath] i.e. is rational.
Now moreover If that means that [imath]b\cdot\sqrt{\,p\,}=a[/imath], Is that correct?
So say [imath]\mathcal{T}=\{n\in Z^+:n\cdot\sqrt{\,p\,}\in\Z^+[/imath], We know that [imath]b\in\mathcal{T}[/imath], HOW & WHY?
Thus [imath]\mathcal{T}\ne\emptyset[/imath] So being bounded below the set [imath]\mathcal{T}[/imath] by completeness has a first term call it [imath]\mathbb{K}[/imath].
Now let [imath]\mathbb{H}=\lfloor\sqrt{\,p\,}\rfloor[/imath], We know that because [imath]p[/imath] is a non-square positive integer [imath]\sqrt{p}>1[/imath]
Because [imath]\mathbb{H}[/imath] is the greatest integer NOT exceeding [imath]\sqrt{\,p\,}[/imath], we know that [imath]\sqrt{\,p\,}< \mathbb{H}+1[/imath]. Which gives us [imath]\sqrt{\,p\,}-\mathbb{H}<1[/imath]
Now here comes the contradiction this gives [imath]\sqrt{\,p\,}\cdot\mathbb{K}-\mathbb{H}\mathbb{K}<\mathbb{K}[/imath]
[imath]\sqrt{\,p\,}\cdot\mathbb{K}-\mathbb{H}\mathbb{K}[/imath] is an integer but also [imath]\sqrt{\,p\,\cdot(\mathbb{K}-\mathbb{H}\mathbb{K}))[/imath]
Click to expand...
Beautiful, professor pka. I don't have advanced skills in proofs, but I think that I understand your proof.

You're saying\(\displaystyle \cdots\)

First, you assumed \(\displaystyle \sqrt{p}\) is rational, so that \(\displaystyle \sqrt{p} = \frac{a}{b}\)

you have the set \(\displaystyle \mathcal{T}\) which contains positive integers. It is not empty because \(\displaystyle b \in \mathcal{T}\). It has a least element \(\displaystyle \mathbb{K}\).

You introduced \(\displaystyle \mathbb{H}\) as the floor function of \(\displaystyle \sqrt{p}\). So, \(\displaystyle \mathbb{H} < \sqrt{p}\).

Which means

\(\displaystyle \sqrt{p} < \mathbb{H} + 1\)
\(\displaystyle \sqrt{p} - \mathbb{H} < 1\)
\(\displaystyle \sqrt{p} \ \mathbb{K} - \mathbb{H}\mathbb{K} < \mathbb{K}\)

This shows that there's an integer \(\displaystyle \sqrt{p} \ \mathbb{K} - \mathbb{H}\mathbb{K} = \mathbb{K}(\sqrt{p} - \mathbb{H})\) less than \(\displaystyle \mathbb{K}\)

Which means there's an integer \(\displaystyle (\sqrt{p} - \mathbb{H}) < 1\)

This can only be true when \(\displaystyle \sqrt{p} - \mathbb{H} = 0\).

Which means \(\displaystyle \sqrt{p} = \mathbb{H}\).

This is a contradiction to our assumption that \(\displaystyle \mathbb{H} = \lfloor\sqrt{p}\rfloor\). So, \(\displaystyle \sqrt{p}\) must be irrational. In other words, \(\displaystyle \sqrt{3}\) is irrational.

My Proof

I will assume \(\displaystyle \sqrt{3}\) is rational, so \(\displaystyle \sqrt{3} = \frac{a}{b}, \ \ a,b \in \Z^+\) and \(\displaystyle b \neq 0\).

Let \(\displaystyle M = \{n\sqrt{3} \ | \ n \ \text{and} \ n\sqrt{3} \in \Z^{+} \}\)

\(\displaystyle M\) is not empty because \(\displaystyle b\sqrt{3} \in M\).

By using the well-ordering property, \(\displaystyle M\) has a least element \(\displaystyle m = k\sqrt{3}\).

If we choose the expression, \(\displaystyle m\sqrt{3} - m\), we can say that \(\displaystyle m\sqrt{3} - m\) is an integer because \(\displaystyle m\sqrt{3}\) and \(\displaystyle m\) are integers.

It is a positive integer because \(\displaystyle m\sqrt{3} - m = m(\sqrt{3} - 1)\) and \(\displaystyle \sqrt{3} > 1\).

We know that \(\displaystyle m = k\sqrt{3}\), so \(\displaystyle m(\sqrt{3} - 1) = k\sqrt{3}(\sqrt{3} - 1) = k(3 - \sqrt{3}) < m\).

This is a contradiction that \(\displaystyle m\) is the least integer. So, \(\displaystyle \sqrt{3}\) must be irrational.
 
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