Like the title says, I'm having trouble proving that there are no integers p and q such that

$|\sqrt5 - p/q| < 1/(7q^2)$. I was given the hint that $|(q\sqrt5 - p)(q\sqrt5 + p)| \geq 1$, but I don't quite know how that helps...


  • $\begingroup$ The title says you're trying to prove there's no $p,q$, but the body of your question says you're trying to prove it's true for all $p,q$, the direct opposite of the title. $\endgroup$ – Gerry Myerson May 9 '14 at 10:28
  • $\begingroup$ The title says show that it cannot be done but in the text you wan to prove that it can be done. The hint suggests it cannot be done (divide through by $q^2$ and think about how big p/q can be) $\endgroup$ – Paul May 9 '14 at 10:29
  • $\begingroup$ @Paul So then I'm left with $|5 - (p/q)^2| \geq 1/|q^2|$. Sorry but I don't see how this is much more informative. Are we trying to say that $1/|q^2|$ goes to 0 "faster" than $\sqrt5 - p/q$ goes to zero? $\endgroup$ – user121937 May 9 '14 at 10:51
  • $\begingroup$ You could use Siegel's theorem which states that if $\alpha\in \mathbb R$ is algebraic of degree $n$ and $\epsilon > 0$ is rational, then there are at most finitely many rational numbers $p/q$ such that $|\alpha - \frac{p}{q}| < \frac{\epsilon}{q^n}$ Here clearly you have $\sqrt{5}$ is algebraic of order $2$. $\endgroup$ – user88595 May 9 '14 at 11:08
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    $\begingroup$ @user88595, Siegel's Theorem is immensely overkill, and anyway if that 7 in the denominator were a 2 the statement would be false. $\endgroup$ – Gerry Myerson May 9 '14 at 13:09

Since $\sqrt{5}$ is not a rational number we know that $5 n^2-m^2$ is a nonzero integer for every integers $n$, $m$ with $q\ne0$ thus $$\forall\,(n,m)\in\Bbb{Z}^2\setminus\{(0,0)\},\quad 1\leq|5n^2-m^2|\tag{1}$$ Now, suppose that there is $(p,q)$ with $q\ne0$ such that $$\left\vert\sqrt{5}-\frac{p}{q}\right\vert\leq\frac{1}{7q^2}$$ It follows that $$\left\vert q\sqrt{5}-p\right\vert\leq\frac{1}{7|q|}\quad \hbox{and} \quad \left\vert q\sqrt{5}+p\right\vert\leq2|q|\sqrt{5}+\frac{1}{7|q|}$$ Thus $$\left\vert 5q^2 -p^2\right\vert\leq\frac{1}{7|q|}\left(2|q|\sqrt{5}+\frac{1}{7|q|}\right) = \frac{2\sqrt{5}}{7}+\frac{1}{49q^2} \leq \frac{2\sqrt{5}}{7}+\frac{1}{49}<\frac{6}{7}+\frac{1}{7}=1 $$ This contradicts $(1)$, and proves that no such $(p,q)$ exists.$\qquad\square$


I think you are my classmate, since we had submitted our homework a few hours ago, I would like to share my solution. Because we are studying continued fractions now, so my solution is based on it.

First, we know if we can find some $(p, q)$ pair, and $q \not= 0$, by Theorem 12.18 or Corollary 12.18.1 in the textbook(Elementary Number Theory by Kenneth H. Rosen 6ed), then the most possible solutions are the convergents $C_k = \frac{p_k}{q_k}, k = 0, 1, 2, ...$ of $\sqrt{5}$. Because they are the closest rational approximation if $q$ is given. But we can show no one in those convergents satisfy this inequality.

We can easily know $\sqrt{5} = [2;\bar{4}]$ and $2 + \sqrt{5} = [\bar{4}]$.

When $k = 0$ and $k = 1$ we can just do some calculations:

$k = 0$, $|\sqrt{5} - \frac{2}{1}| = \sqrt{5} - 2 > \frac{1}{7}$, since $5 > \frac{225}{49}$.

$k = 1$, $|\sqrt{5} - \frac{9}{4}| = \frac{9}{4} - \sqrt{5} > \frac{1}{112}$, since $\frac{251}{112} > \sqrt{5}$.

For $k > 1$, We know $\sqrt{5} = \frac{\alpha_{k+1}p_k + p_{k-1}}{\alpha_{k+1}q_k + q_{k-1}}$, so $$\begin{aligned} |\sqrt{5} - \frac{p_k}{q_k}| &= |\frac{\alpha_{k+1}p_k + p_{k-1}}{\alpha_{k+1}q_k + q_{k-1}} - \frac{p_k}{q_k}| \\ &= |\frac{\alpha_{k+1}p_kq_k + p_{k-1}q_k - \alpha_{k+1}p_kq_k - p_kq_{k-1}}{\alpha_{k+1}q_k^2 + q_{k-1}q_k}| \\ &= |\frac{-(p_kq_{k-1} - p_{k-1}q_k)}{\alpha_{k+1}q_k^2 + q_{k-1}q_k}| \\ &= \frac{1}{\alpha_{k+1}q_k^2 + q_{k-1}q_k} \\ &> \frac{1}{\alpha_{k+1}q_k^2 + q_k^2}\end{aligned}$$ We know $\alpha_{k+1} = [\bar{4}] = 2 + \sqrt{5}$ so $$|\sqrt{5} - \frac{p_k}{q_k}| > \frac{1}{(3 + \sqrt{5})q_k^2} > \frac{1}{7q_k^2}$$ Which completes the proof. $\blacksquare$

  • $\begingroup$ Interesting, I was trying to go this route but failed, probably because I quit too early. Our TA solved it using the solution below, so I do not think it was necessary to use theorems from the book. $\endgroup$ – user121937 May 11 '14 at 22:09

An alternative, we have $|q \sqrt{5} - p|< \tfrac{1}{7q}$ so if $ (q \sqrt{5}-p)(q \sqrt{5}+p) \geq 1 $ we can divide both sides

\begin{eqnarray} \\ |q \sqrt{5} + p| & \geq & 7q \\ | \sqrt{5} + \tfrac{p}{q}| &\geq & 7 \end{eqnarray}

This is impossible since $| \sqrt{5} + \tfrac{p}{q}| \leq \sqrt{5} + \tfrac{p}{q} \leq \sqrt{5}+1 < 4$.


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