# Sqrt

# Sqrt(2) is Irrational

The standard proof that
is irrational, which appears both in Rudin and on

Wikipedia, is (or can be written as) a reductio ad absurdum (RAA): if
is rational

then it cannot be expressed as a reduced form ( with all common factors divided
out);

since any actual rational number can be expressed in reduced form, it must be

that is not rational. More succinctly: if
is rational then it both has and does not

have a reduced form representation.

Below, I give a version of the standard proof. My version is somewhat longer

than the usual versions because I take care to spell out the RAA logic . I also

give a second proof, a contraposition, that takes the central idea, thatcannot

be expressed in reduced form, and presents it in a way that does not require first

assuming that is rational. Both proofs are
correct. Which proof you prefer is

largely a matter of taste. I prefer the second, less standard version; the
argument

seems more straightforward to me, despite the fact that it relies on a
construction

that may at first seem abstract.

Theorem 1. is not rational.

Proof. I give two proofs , the first is a form of reductio ad absurdum, the
second a

contraposition.

1. First Proof. Suppose that

Thencan be expressed as a ratio of natural
numbers. Consider any such

representation: a, b∈ N such that
= a/b. Therefore a^{2} = 2b^{2}. In particular,

a^{2}
is even. It is a theorem of number theory that this implies that a must

also be even, hence there is a ^{}
∈
N such that Hence
, which

implies . The
same argument, now applied to b, gives a
∈ N such

that b = 2, hence ^{
2} or

:

Since^{}
= a/2 and
= b/2, this implies that a=b is not in
reduced form. Since

a/b was an arbitrary ratio representation of,
this implies that there is no

ratio representation of in reduced form.

On the other hand, it is a theorem of number theory that any ratio of natural

numbers can be put into reduced form.

Therefore, the hypothesis that implies that
both can and cannot

be represented as a reduced form ratio of natural numbers. The conclusion

thatfollows by RAA.

2. Consider any set with the following
properties .

(a) There is an a, b∈ S such that

(b) If p, q ∈ S and p^{2} = 2q then there is a
∈ S such that p = 2.

This definition is not vacuous. For example, take S = R+.

Consider any a, b ∈ S such that
= a/b. Then a, b ≠ 0. Since 2b^{2} = a^{2},
the

construction of S implies that there is an a_{1} ∈
S such that a = 2a_{1}. Hence

, or b^{2} = 2a^{2}. Again by
construction of S, there is a b_{1}
∈ S such

that b = 2b_{1}, hence or

:

Continuing in this way yields sequences {a_{t} }, { b_{t}
} such that for all t, a_{t} / b_{t} =

and for t large enough, a_{t}, b_{t} < 1. Since a_{t},
b_{t } ≠ 0, this implies a_{t,} b_{t}
∈
N.

Thus S ≠ N.

Since S was arbitrary, this shows that N does not satisfy the two criteria

above: it fails either the first or the second (or both). It is a basic result
in

number theory that N satisfies the second condition (this was already invoked

in the previous proof ). Therefore it follows that N violates the first, which is

equivalent to saying that

Remark 1. Essentially the same argument shows that for any natural number n,

either n is the square of another natural number or

is irrational.

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