2.2 Irrational Numbers
- An irrational number is a number which cannot be expressed in the form \(\frac {a}{b}\) for some integers \(a\) and \(b\).
- Decimal representation of an irrational number non terminating and non repeating.
Are there any? Nothing so far guarantees that a number failing to be rational exists at all. The next two results settle it.
Theorem 2.10. If \(P\) is an integer and \(P^{2}\) is divisible by \(2\), then \(P\) is divisible by \(2\).
Proof. Every integer is either even or odd, so \(P=2m\) or \(P=2m+1\) for some integer \(m\), and correspondingly \[P^{2}=4m^{2} \qquad \text {or}\qquad P^{2}=4m^{2}+4m+1 .\] The second is odd — it is \(2\left (2m^{2}+2m\right )+1\) — so it is not divisible by \(2\). Since \(P^{2}\) is assumed divisible by \(2\), the first case must hold, and then \(P=2m\) is divisible by \(2\). □
Note 2.11. This looks like a triviality and is not. The statement ”\(P^{2}\) even \(\implies \) \(P\) even” is what makes the next proof work, and its analogue is false for some other divisors: \(P^{2}\) divisible by \(4\) does not force \(P\) divisible by \(4\), as \(P=2\) shows. The property holds for prime divisors, which is why the argument below adapts to \(\sqrt {3}\) and \(\sqrt {5}\) but not to \(\sqrt {4}\).
Proof. By contradiction. Suppose \(\sqrt {2}\) is rational. Then there are integers \(m\) and \(n\), with \(n\neq 0\) and \(\dfrac {m}{n}\) in its lowest terms, such that \[\sqrt {2}=\frac {m}{n}.\]
Squaring both sides and clearing the denominator, \begin {equation} \label {eq:msquared} 2=\frac {m^{2}}{n^{2}}\qquad \implies \qquad m^{2}=2n^{2}. \end {equation} The right-hand side is a multiple of \(2\), so \(m^{2}\) is divisible by \(2\), and therefore \(m\) is divisible by \(2\) by Theorem 2.10. Write \[m=2k \qquad \text {for some integer } k.\]
Substituting this into (1), \[\left (2k\right )^{2}=2n^{2}\qquad \implies \qquad 4k^{2}=2n^{2} \qquad \implies \qquad n^{2}=2k^{2}.\] So \(n^{2}\) is divisible by \(2\), and therefore \(n\) is divisible by \(2\), again by Theorem 2.10.
But now \(2\) divides both \(m\) and \(n\), so \(\dfrac {m}{n}\) was not in its lowest terms after all — contradicting the way it was chosen. No such \(m\) and \(n\) exist, and \(\sqrt {2}\) is therefore irrational. □
Note 2.13. Notice where the contradiction comes from. Nothing is wrong with \(m\) and \(n\) both being even — \(\frac {4}{6}\) is a perfectly good fraction. The contradiction is with “in lowest terms”, a condition we were free to impose at the start because every fraction can be reduced. Choosing the right form of the assumption is what makes the proof work, and it is the step most often left out.
The same argument shows \(\sqrt {3}\), \(\sqrt {5}\) and \(\sqrt {p}\) for any prime \(p\) are irrational. Try it on \(\sqrt {4}\) to see where it must fail: the parallel step “\(P^{2}\) divisible by \(4\) implies \(P\) divisible by \(4\)” is simply untrue, and the proof stops dead — as it must, since \(\sqrt {4}=2\).
Proof.
Suppose that \(2+\sqrt {3}\) is a rational number. Then by definition there are integers \(a\) and \(b\) with \(b\neq 0\) such that
\(2+\sqrt {3}=\frac {a}{b}\).
\[\implies \quad \sqrt {3}=\frac {a}{b}-2\quad \implies \quad \sqrt {3}=\frac {a-2b}{b}\]
Since \(a\) and \(2b\) are integers then \(a-2b\) is also an integer, say \(m\). Thus \(\sqrt {3}=\frac {m}{b}\) where \(m\) and \(b\) are integers.
But LHS is irrational since \(\sqrt {3}\) is irrational while the RHS is rational. This is a contradiction since
there is no irrational number which is also rational. Therefore our assumption that \(2+\sqrt {3}\) is rational
is false.
Hence \(2+\sqrt {3}\) is an irrational number. □
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