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The square root of 2 is not rational
There are many ways to see this fact, already accessible to us after the first lecture! (In case you are
wondering why this might be interesting, it is rumoured that Hippassus of Metapontum was killed over his
discovery of this fact, so this was a very astonishing and abrasive discovery at one time.)
Theorem 1. A real number x such that x2 = 2 is not a rational number.
We start with a proof using the well-ordering principle.
Proof 1. Suppose, towards a contradiction, such x were rational. Then the set
S := {n ∈ N : xn ∈ N}
is a subset of the natural numbers, and the well-ordering principle tells us there is a least element, say k.
But note that
k(x − 1) · x = kx2 − kx = 2k − kx ∈ N
as k ∈ S. Hence k(x − 1) ∈ S as 1 < x < 2. However,
k(x − 1) < k,
contradicting the choice of k. So the assumption that x is rational must be wrong, and x ∈
/ Q.
Next, we may use the principle of mathematical induction. (We will actually use strong induction.)
Proof 2. We will show that
2 6=
a2
b2
for any a, b ∈ N by using induction on a.
For the base case, a = 1, note that 1/b2 ≤ 1 for all b ∈ N, and so certainly does not equal 2.
For the strong inductive hypothesis, we suppose that for any a ∈ N with 1 ≤ a ≤ N we have
2 6=
a2
b2
for any b ∈ N. We wish to use this to show that
2 6=
(n + 1)2
b2
for any b ∈ N.
Suppose, towards a contradiction, that we have some b ∈ N for which 2 = (n+1)2 /b2 . Then (n+1)2 = 2b2 ,
and so (n + 1)2 – and hence n + 1 – is even. (To see this, note that any even number is of the form
2k for some k ∈ Z, and every odd number is of the form 2k + 1. Hence the square of an odd number
(2k + 1)2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1 is odd. Hence, if the square of a number is even, the number
itself must be even.)
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Write n + 1 = 2a0 . Then from above we see that b2 = 2(a0 )2 , or
2=
which contradicts the inductive hypothesis as b2 =
b2
(a0 )2
(n+1)2
,
2
so
n+1
b = √ < n + 1.
2
Hence, by the principle of strong induction, no rational number squares to 2.
Note that strong induction really was required above. Writing
2=
(n + 1)2
b2
we see that 2 must divide n + 1, and b, so we may cancel this factor to write the fraction in lower terms. We
only get b < n + 1, and not b = n, and so ‘regular’ induction would not conclude the proof at this point.
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