Various Proofs for Irrationality of sqrt2

1. Nov 1, 2007

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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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.

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Anyone else with alternative proofs please post here =]

Last edited: Nov 2, 2007
2. Nov 1, 2007

HallsofIvy

Staff Emeritus
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.

3. Nov 1, 2007

4. Nov 1, 2007

morphism

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

5. Nov 2, 2007

Gib Z

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

Thanks for the link neutrino =]

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

6. Nov 2, 2007

morphism

And how does that not qualify as a proof?

7. Nov 2, 2007

Gib Z

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

8. Nov 2, 2007

Hurkyl

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

9. Nov 2, 2007

Gib Z

no :(

10. Nov 2, 2007

Hurkyl

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

11. Nov 2, 2007

morphism

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

12. Nov 2, 2007

Hurkyl

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

13. Nov 2, 2007

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.

14. Nov 2, 2007

Hurkyl

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

15. Nov 2, 2007

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.

Last edited: Nov 2, 2007
16. Nov 13, 2007

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