3

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?

Shaun
  • 44,997
zeraoulia rafik
  • 7,055
  • 5
    Define "better". – law-of-fives Feb 04 '18 at 22:40
  • 1
    Hint : Use $x^2 - 2 = 0$ equation . – S.H.W Feb 04 '18 at 22:40
  • Why don't you like the standard proof ? Of course you could use the minimal polynomial $x^2-2$ or the concept of the $2$-adic valuation of a number, but the standard proof is easy and crystal-clear. – Peter Feb 04 '18 at 22:43
  • The standard proof is very nice , but for student it's hard for understanding it in the high school level in my country – zeraoulia rafik Feb 04 '18 at 22:45
  • http://jwilson.coe.uga.edu/EMT668/EMAT6680.2002/Rouhani/Essay1/pythagorastheorem.html – Dionel Jaime Feb 04 '18 at 22:46
  • 1
    @zeraouliarafik I think everyone should be able to understand this proof. The other possibilities are more complicated , to my opinion. – Peter Feb 04 '18 at 22:47
  • https://www.cut-the-knot.org/proofs/sq_root.shtml – Crostul Feb 04 '18 at 22:47
  • 5
    While there could be "nicer" proofs, in some sense they "have" to be by contradiction one way or another. After all, irrational means 'not rational', i.e. assume it is rational and arrive at a contradiction. Every other proof will still somehow do this, possibly in disguise. – Stefan Perko Feb 04 '18 at 22:48
  • 1
    Related: https://math.stackexchange.com/questions/1589603 – Watson Feb 04 '18 at 22:58

3 Answers3

5

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) .

sepehr333
  • 51
  • Mmm... Not very convincing. After all, the usual proof by contradiction of the irrationality of $\sqrt 2$ boils down to the fact that the ring $\mathbf Z$ is factorial. And Eisenstein's criterion to the fact that $A[X]$ is factorial if $A$ is. Same remark about the proof 1 of A. Mortier using the "integer root theorem". – nguyen quang do Feb 06 '18 at 08:17
3

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.

Yuriy S
  • 31,474
2

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.

Arnaud Mortier
  • 27,276