Question

Prove that $\sqrt{2}$ is an irrational number.

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Answer 1

ANSWER√

Consider the positive root of the equation x2−5=0 — we usually denote this by 5–√ . We transform this equation like so:

x2−4=1

(x−2)(x+2)=1

x−2=1x+2.

Now clearly 2<x<3 , so 0<x−2<1 . Let us write 5–√ as

5–√=x=2+(x−2)

=2+1x+2

=2+14+(x−2)

=2+14+1x+2

=2+14+14+(x−2)

=2+14+14+1x+2

=2+14+14+14+(x−2)

=2+14+14+14+1x+2

=2+14+14+14+14+⋱.

This is the simple continued fraction representation of 5–√ , which is clearly infinitely long — the number 4 will continue manifesting itself forever. If this continued fraction representation terminated, then 5–√ would have been rational. Since it doesn’t, 5–√ is irrational.

Strangely enough, then, 5–√ is irrational because 15–√−2=5–√+2.

Of course, by a slight (or, sometimes, not so slight; see the comments below) modification to this argument, we may use this proof to prove that n−−√ is irrational for any non-square number n .