Various Proofs for Irrationality of sqrt2

[Gib Z]

We all know the standard proof that the square root of two is irrational, and it's easily extended to all integers that are not perfect squares, but It just striked me yesterday that I have only seen one proof (which really is enough, but still =]).

One of the lecturers at the University of Sydney (I went there for work experience) showed me a new proof, which is from the perspective of set theory (it uses the fact a subset of the positive integers must have a smallest element), rather than the standard number theory results of divisibility. I thought some of you would find it interesting, I certainly did =]
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Proof by Contradiction:

The statement that the square root of 2 is irrational is equivalent to the statement that a subset of the natural numbers S= \{ n \in \mathbb{Z}^{+} | \sqrt{2}n \in \mathbb{Z}^{+} \} is not empty.

This set must have a least member, let it be u.
Let w=(\sqrt{2}-1)u
Then \sqrt{2}w= 2u - \sqrt{2} u.

Since u is a member of S, sqrt2*u is a positive integer, so sqrt2*w is also a positive integer. Therefore w is also a member of S.

However, sqrt2 minus 1 is between 0 and 1, so w < u. Yet u is the least member of S.

Contradiction.
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Anyone else with alternative proofs please post here =]

 

[HallsofIvy]

We all know the standard proof that the square root of two is irrational, and it's easily extended to all integers that are not perfect squares, but It just striked me yesterday that I have only seen one proof (which really is enough, but still =]).

One of the lecturers at the University of Sydney (I went there for work experience) showed me a new proof, which is from the perspective of set theory (it uses the fact a subset of the positive integers must have a smallest element), rather than the standard number theory results of divisibility. I thought some of you would find it interesting, I certainly did =]
--------------------------
Proof by Contradiction:

The statement that the square root of 2 is irrational is equivalent to the statement that a subset of the natural numbers S= \left{ n \in \mathbb{Z}^{+} | \sqrt{2}n \in \mathbb{Z}^{+} \right} is not empty. (For some reason the braces won't appear, damn).

Are you quite sure you don't mean "The statement that the square root of 2 is rational??
The braces don't appear because braces are part of the "control" symbols for LaTex. Use "\{" and "\}" instead.

This set must have a least member, let it be u.
Let w=(\sqrt{2}-1)u
Then \sqrt{2}w= 2u - \sqrt{2} u.

Since u is a member of S, sqrt2*u is a positive integer, so sqrt2*w is also a positive integer. Therefore w is also a member of S.

However, sqrt2 minus 1 is between 0 and 1, so w < u. Yet u is the least member of S.

Contradiction.
-----------------------

Anyone else with alternative proofs please post here =]

 

[neutrino]

http://www.cut-the-knot.org/proofs/sq_root.shtml

 

[morphism]

Eisenstein: x^2 - 2 is irreducible over Q.

 

[Gib Z]

Are you quite sure you don't mean "The statement that the square root of 2 is rational??
The braces don't appear because braces are part of the "control" symbols for LaTex. Use "\{" and "\}" instead.

Yup I meant rational :( and ty for the top.

Thanks for the link neutrino =]

morphism - Yes i know that, I was asking for proofs :(

 

[morphism]

And how does that not qualify as a proof?

 

[Gib Z]

Because that result is somewhat circular, it follows from the fact that sqrt 2 is not in Q.

 

[Hurkyl]

Because that result is somewhat circular, it follows from the fact that sqrt 2 is not in Q.

Can you not imagine any way to prove x²-2 is irreducible without invoking the irrationality of the square root of 2?

 

[Gib Z]

Can you not imagine any way to prove x²-2 is irreducible without invoking the irrationality of the square root of 2?

no :(

 

[Hurkyl]

no :(

Well, my first thought is the rational root theorem. Of course, this is really just a thin disguise of the proof using prime factorizations.

 

[morphism]

Maybe Eisenstein's criterion? Or the rational root theorem?

 

[Hurkyl]

Ooh, modular stuff. It's easy to show that x²-2 is irreducible modulo 3! And then irreducibility over Q follows from that.

 

[morphism]

You can also prove directly that Q[x]/<x^2 - 2> is a field, so <x^2 - 2> is a maximal ideal in Q[x], and thus x^2 - 2 is irreducible over Q.

 

[Hurkyl]

Is that easy to do without directly invoking the irrationality of the square root of 2?

 

[morphism]

I think so - here's a sketch:

- Q[x]/<x^2 - 2> =~ { a+bt : a,b in Q, t^2 = 2 }
- closure under products, addition and subtraction is easy to show, and 0,1 are in there
- 1/(a+bt) = (a-bt)/(a^2-2b^2)

Does that look OK to you?

Hmm. Maybe not. I think we need to be justify that a^2-2b^2 isn't going to be zero over Q.

 

[yasiru89]

I'm no great hand at number theory but I remember having read the following more algebraically minded proof;
Let p/q be a positive fraction in its lowest terms such that (p/q)^2 = 2

then p^2 = 2q^2
From this,
(2q - p)^2 = 2(p-q)^2

again 2 = (2q - p)^2 / (p - q)^2
the same property of the fraction p/q
since q<p<2q gives p-q<q
therefore we have another fraction with a smaller denominator and this contradicts our assumption, neatly establishing the irrationality of the positive solution of the equation
x^2 - 2 = 0