The irrationality of the square root of 2 (pdf)

Reading

Section 1.4

Practice Problems

1.4

1, 2, 5, 6, 8, 9, 13

Notes

• A discovery that shocked ancient greeks was that there are numbers, like \(\sqrt{2}\), that cannot be written as the quotient of two integers.
• This was a heavy blow to the Pythagorean school of thought, one of whose main goals was to derive everything starting from the natural numbers.
• Our proof will be a tad different from the book's. Compare with the book's proof and understand the subtle differences.

Proof that \(\sqrt{2}\) is irrational

• We start by employing contradiction. Suppose that \(\sqrt{2}\) was irrational.
• Then we can write it as \(\sqrt{2} = \frac{p}{q}\) where \(p,q\) are positive integers (how can we assume that?)
• Moving \(q\) over, we obtain \(q\sqrt{2} = p\).
• Squaring both sides gives us: \(2q^2 = p^2\). This means that \(p^2\) is even, therefore \(p\) must also be even.
• Therefore \(p = 2p'\) where \(p'\) is another positive integer smaller than \(p\).
• Then the equation becomes: \(2q^2 = (2p')^2 = 4(p')^2\). Which we can write as \(q^2 = 2(p')^2\).
• The same reasoning tells us that \(q^2\), and therefore \(q\), must be even. So \(q = 2q'\) where \(q'\) is a positive integer smaller than \(q\).
• We also see that \(\frac{p'}{q'} = \frac{2p'}{2q'} = \frac{p}{q} = \sqrt{2}\).
• So we were able to write \(\sqrt{2}\) as a quotient of two smaller positive integers.
• Continuing in this manner, we can keep writing \(\sqrt{2}\) as a quotient of smaller and smaller integers, each a factor of 2 less than the previous one.
• This process cannot continue forever. This is our contradiction.

Admittedly the last step is a tad "sketchy". It will be proven more precisely in the future. To do it properly, we will need one of the following two facts, both of which we will discuss a bit later:

• Every number can be written as a product of prime factors. Once all the factors of two have been removed by repeating the above process from \(p\) to \(p'\), we would be left with an odd number, so the process can't be repeated any more.
• Every non-empty set of natural numbers has a smallest element. Using this, if we consider the set of all numbers \(p\) such that \(\sqrt{2}\) can be written as a quotient \(p/q\), then this set is non-empty if \(\sqrt{2}\) is rational. If we consider the smallest such \(p\), then the above process from \(p\) to \(p'\) would give us a smaller \(p\), which is impossible.