Is there another proof about irrationality of $\sqrt{2}$ without using a contradiction?

Question

A well known proof to show the irrationality of $\sqrt{2}$ is by contradiction. Is there another method to show the irrationality of $\sqrt{2}$ better than by contradiction?

Answer 1

It's the matter of definition. Rational numbers are actually very rare particular case of real numbers.

Say, you talk of real numbers as simple continued fractions (which are by the way, an amazing topic to talk to high school students interested in math, it's very easy to understand and lots of fun to experiment with - since the OP apparently is asking in relation to teaching high school students?).

You just offer a definition of a SCF, as the following fraction:

$$x=\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+ \cdots}}}$$

Where all $a_j$ are positive integers.

And / or the second order recurrence relation for the denominators and numerators of the convergents (see Wikipedia and other sources).

Let the students play around with these definitions. Then explain that any such fraction converges in a sense that as you truncate it at a deeper and deeper levels, the larger and larger number of digits stays constant. And so on, CF are amazing topic.

Now, finally, tell them the most important thing:

By definition, rational numbers are precisely the real numbers with finite SCF expansion.

Again, this should be clear enough to a bright HS student.

And now show them how to get a SCF for $\sqrt{2}$:

$$x^2=2$$

$$x^2-1=1$$

$$(x-1)(x+1)=1$$

$$\sqrt{2}=x=1+\frac{1}{1+x}=1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+ \cdots}}}$$

It's quite easy to see that this continued fraction will never terminate, so by our own definition, this number can't be rational.

I'm aware, that this is not only proof based on contradiction, but not a proper proof at all. But again, if the goal is teaching high school students, this might work better.

Answer 2

By Eisenstein criterion the polynomial $x^2-2$ is irreducible over the integer ring, i.e. $2 \nmid 1$, and $2|-2, 2^2\nmid -2$, so by Gauss lemma it is also irreducible over rationals. Thus the solution is irrational (we know it is real) .

Answer 3

Proof 1: $\sqrt{2}$ is a non-integer root of a monic polynomial with integer coefficients, therefore irrational by the integer root theorem.

Proof 2: $\sqrt2$ is the limit of a sequence of rationals $\frac{p_n}{q_n}$ satisfying $|\sqrt2-\frac{p_n}{q_n}|=o(\frac{1}{q_n})$ which allows you to conclude as explained here.