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In the following proof that $\sqrt2$ is irrational, I cannot make sense of why $\frac{q a_n + p b_n}{q} \geqslant \frac{1}{q}$.
The sums corresponding to $a_n$ and $b_n$ are alternating sums, so why must $q a_n + p b_n \geqslant 1$ ?

Proof:

Suppose $\sqrt2 = \frac{p}{q} \in \mathbb{Q}$. On the one hand, $0 < (\sqrt{2} - 1)^{n} < 1$. Thus,

$$ \lim_{n \to \infty} (\sqrt{2} - 1)^{n} = 0 $$

On the other hand, the binomial expansion of $(\sqrt{2} - 1)^{n}$ is given by

$$ (\sqrt{2} - 1)^{n} = \sum_{k = 0}^{n} {n \choose k}(\sqrt{2})^{k}(-1)^{n-k} $$ $$ = \sum_{\substack{k = 0 \\ k \text{ even}}}^{n} {n \choose k}(2)^{\frac k2}(-1)^{n-k} + \sqrt2 \sum_{\substack{k = 0 \\ k \text{ odd}}}^{n} {n \choose k}(2)^{\frac{k-1}{2}}(-1)^{n-k} $$ $$ = a_n + b_n \sqrt2 $$

Finally, we see that

$$ (\sqrt{2} - 1)^{n} = a_n + b_n \sqrt2 = a_n + b_n \frac{p}{q} = \frac{q a_n + p b_n}{q} \geqslant \frac{1}{q} $$

Contradicting the fact that $(\sqrt{2} - 1)^{n}$ tends to $0$ as $n \to \infty$

QED

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  • $\begingroup$ Fun proof! Different than the usual one , even though I'd say not better $\endgroup$ – Ant Jul 4 '16 at 5:59
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As $(\sqrt{2}-1)^n > 0$ we know that $\frac{q a_n + p b_n}{q} >0$ and therefore $q a_n + p b_n>0$.

As $q a_n + p b_n$ is an integer we have $q a_n + p b_n \geq 1$

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  • $\begingroup$ Is a positive integer (which is clear, but perhaps it's better to state it explicitly). $\endgroup$ – egreg Jul 3 '16 at 15:57
  • $\begingroup$ So simple but so much headache. Thank you! $\endgroup$ – HoopaU Jul 3 '16 at 22:50
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The proof given is unnecessarily complicated, but never mind it is OK. The idea it is using is frequently seen in irrationality proofs.

In order to prove that a number $\alpha$ is irrational, we construct some expression dependent on $\alpha$ and some other parameters (in most case an integer $n$) say $f(\alpha, n)$ such that the expression $f$ is non-zero but can me made to be close to $0$ by choosing $n$ large (i.e. limit of $f$ as $n \to \infty$ is $0$). Moreover this expression $f$ is chosen to be such that is it a rational number with bounded denominator if $\alpha$ is rational (in most cases the denominator of $f$ is same as that of $\alpha$). This ensures that $|f|$ is a non-zero rational number with denominator no greater than say $q$. Then the numerator of $|f|$ is positive and thus $|f| \geq 1/q$ so that limit of $f$ cannot be $0$ and thus we get contradiction.

The expression $f$ here is $f(n) = (\sqrt{2} - 1)^{n}$ and it tends to $0$ as $n \to \infty$. Now if $\sqrt{2} = p/q$ then $$f(n) = \frac{(p - q)^{n}}{q^{n}}$$ and it appears that denominator of $f(n)$ is unbounded. Some ingenuity is needed to show that $f(n)$ indeed can be expressed as a rational number with denominator $q$ instead of $q^{n}$. Here we use the idea that a polynomial in $\sqrt{2} = \alpha$ with integer coefficients can be expressed as $a + b\alpha$ where $a, b$ are integers. Thus $f(n) = (\alpha - 1)^{n}$ being a polynomial in $\alpha$ can be expressed as $$a + b\alpha = \frac{aq + bp}{q}$$ if $\alpha = p/q$.


The idea descibed above is very simple and the real challenge is to find / invent an appropriate function $f$ with the desired property based on the number $\alpha$ under consideration. You can find a lot of such ingenious proofs by Ivan M. Niven in his book Irrational Numbers. Also see application of this technique for irrationality of $\zeta(2)$ and $\zeta(3)$.

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