Proving √3 is irrational

[JeffM]

Because you are using Hammack's work you surely can follow this proof.
Suppose that \(\displaystyle P\in\mathbb{Z}^+\) and \(\displaystyle (\forall n\in\mathbb{Z}[P\ne n^2)\)(i.e. P is not a square).
You should know about the floor function: thus \(\displaystyle 0 < \sqrt P - \left\lfloor {\sqrt P } \right\rfloor < 1\) (**)
O.K. here goes: Suppose that \(\displaystyle \sqrt P\) is a rational number.
Then there are two relatively prime positive integers \(\displaystyle a~\&~b\) such that \(\displaystyle \sqrt{P}=\dfrac{a}{b}\) or \(\displaystyle a=b\sqrt{P}\)
Define a set \(\displaystyle T=\{n\in\mathbb{Z}^+: n\sqrt{P}\in\mathbb{Z}^+\}\). \(\displaystyle T\) is not empty because \(\displaystyle b\in T\)
So \(\displaystyle T\) is a non-empty subset of the positive integers as such \(\displaystyle T\) contains a first term. Call it \(\displaystyle k\).
From (**) we get \(\displaystyle 0<k\sqrt P-kH<k\). BUT \(\displaystyle (k\sqrt P-kH)\in\mathbb{Z}^+\). Why??
Moreover, \(\displaystyle (k\sqrt P-kH)<k\) and \(\displaystyle (k\sqrt P-kH)(\sqrt P)\in\mathbb{Z}^+\).
That violates the minimality of \(\displaystyle k\),. Its a contradiction to supposing \(\displaystyle \sqrt P\) is rational.

To restate: If \(\displaystyle P\) is a positive integer that is not a square then \(\displaystyle \sqrt P\) is irrational.

I am presuming that by "k is first" you mean k is the smallest integer in T and that

[MATH]H = \lfloor \sqrt{P} \rfloor \implies H \in \mathbb Z^+.[/MATH]
If those guesses are wrong, I do not get the proof.

If they are correct, it's a very nice proof.

 

[pka]

I am presuming that by "k is first" you mean k is the smallest integer in T and that
[MATH]H = \lfloor \sqrt{P} \rfloor \implies H \in \mathbb Z^+.[/MATH]If those guesses are wrong, I do not get the proof.
If they are correct, it's a very nice proof.

Jeff, I have taught set theory and logic from at least ten different textbook not to mention the books I have written reviews for.
I am conformable saying that to say "k is the first term in a set of positive integers" means that k is the minimal term in the set.
Frankly I am surprised that you would even question that.

 

[JeffM]

Jeff, I have taught set theory and logic from at least ten different textbook not to mention the books I have written reviews for.
I am conformable saying that to say "k is the first term in a set of positive integers" means that k is the minimal term in the set.
Frankly I am surprised that you would even question that.

I always understood that no ordering was implied by a set.

[MATH]\{1,\ 2,\ 3\} = \{3,\ 1,\ 2\}.[/MATH]
Now I know I have been wrong. Thanks.

 

[pka]

I always understood that no ordering was implied by a set.
[MATH]\{1,\ 2,\ 3\} = \{3,\ 1,\ 2\}.[/MATH]Now I know I have been wrong. Thanks.

Suppose that \(\displaystyle S=\{6,19,3,18, 22,4,30, 2, 17\}\) what would you say is the first positive integer in \(\displaystyle S~?\) Would you say it is \(\displaystyle 6~?\).
If so, do you agree that \(\displaystyle \{2,3,4,6,17,18,19,22,30\}\) is the same set \(\displaystyle S~?\)
This time what do you say is the first positive integer in the set? Remember they are the same!
If someone ask you “what is the first positive integer” what say you?
If someone ask you “what is the first positive integer the square of which is exceeds 100? ” what say you?

 

[Steven G]

Because you are using Hammack's work you surely can follow this proof.
Suppose that \(\displaystyle P\in\mathbb{Z}^+\) and \(\displaystyle (\forall n\in\mathbb{Z}[P\ne n^2)\)(i.e. P is not a square).
You should know about the floor function: thus \(\displaystyle 0 < \sqrt P - \left\lfloor {\sqrt P } \right\rfloor < 1\) (**)
O.K. here goes: Suppose that \(\displaystyle \sqrt P\) is a rational number.
Then there are two relatively prime positive integers \(\displaystyle a~\&~b\) such that \(\displaystyle \sqrt{P}=\dfrac{a}{b}\) or \(\displaystyle a=b\sqrt{P}\)
Define a set \(\displaystyle T=\{n\in\mathbb{Z}^+: n\sqrt{P}\in\mathbb{Z}^+\}\). \(\displaystyle T\) is not empty because \(\displaystyle b\in T\)
So \(\displaystyle T\) is a non-empty subset of the positive integers as such \(\displaystyle T\) contains a first term. Call it \(\displaystyle k\).
From (**) we get \(\displaystyle 0<k\sqrt P-kH<k\). BUT \(\displaystyle (k\sqrt P-kH)\in\mathbb{Z}^+\). Why??
Moreover, \(\displaystyle (k\sqrt P-kH)<k\) and \(\displaystyle (k\sqrt P-kH)(\sqrt P)\in\mathbb{Z}^+\).
That violates the minimality of \(\displaystyle k\),. Its a contradiction to supposing \(\displaystyle \sqrt P\) is rational.

To restate: If \(\displaystyle P\) is a positive integer that is not a square then \(\displaystyle \sqrt P\) is irrational.

As an undergraduate student I tried to prove what you did but could never get it. I am so glad to see this proof as now I have settled this in my mind. It can be proved and now I know how to prove it! This site is amazing! Thanks so much for such a beautiful clean simple proof!

 

[JeffM]

Suppose that \(\displaystyle S=\{6,19,3,18, 22,4,30, 2, 17\}\) what would you say is the first positive integer in \(\displaystyle S~?\) Would you say it is \(\displaystyle 6~?\).
If so, do you agree that \(\displaystyle \{2,3,4,6,17,18,19,22,30\}\) is the same set \(\displaystyle S~?\)
This time what do you say is the first positive integer in the set? Remember they are the same!
If someone ask you “what is the first positive integer” what say you?
If someone ask you “what is the first positive integer the square of which is exceeds 100? ” what say you?

I would say that they are meaningless questions if sets do not have order. Sort of like "Do unicorns like oats?"

What is the first in the set {blue, red, and green,} Obviously it is blue. What is the first in the set {green, blue, red}. Obviously it is green. Therefore, if those questions have meaning, the two sets cannot the same because their firsts are not the same.