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I'm trying to prove that the ratio of triangular numbers is dense on the interval from 0 to 1. More precisely, given $l\in[0,1]$ and $\varepsilon>0$, we can find $m,n \in \mathbb{N}$ so that $|\frac{m(m+1)}{n(n+1)}-l|<\varepsilon$.

I came up with an algorithm that seems to work, but I don't know how a proof would look.

Suppose $\varepsilon>0$ and $l\in[0,1]$ are given. Then use the following algorithm to determine $m$ and $n$:

Set $m_0=1$ and $n_0=2$.

Set $k=0$.

while $|\frac{m_k(m_k+1)}{n_k(n_k+1)}-l| \geq \varepsilon$

{

if $\frac{m_k(m_k+1)}{n_k(n_k+1)} and $n_k-m_k > 1$

Set $m_{k+1}=m_k+1$ and $n_{k+1}=n_k$

if $\frac{m_k(m_k+1)}{n_k(n_k+1)}>l$

Set $m_{k+1}=m_k$ and $n_{k+1} = n_k+1$

Otherwise:

Set $m_{k+1}=m_k+1$ and $n_{k+1}=n_k+1$

$k = k+1$

}

This seems to work. Could the algorithm be rewritten as a proof using the nested interval theorem? If so, I'm having a hard time seeing how. The idea would be to use the Archimedean principal to find a triangular ratio $p_0$ with $p_0 and a triangular ratio $q_0$ with $q_0>l$. Then created a sequence of nested intervals $I_n$ such that $I_n=[p_n,q_n]$ where $p_n$ is an increasing sequence of triangular ratios, and $q_n$ is a decreasing sequence of triangular ratios. Further, for any $I_k$, $I_{k+1}\subseteq I_k$.

  • 0
    I should clarify a bit about where $\frac{m(m+1)}{n(n+1)}$ came from. Triangular numbers are of the form $1+2+3+...+n$, which evaluates to $\frac{n(n+1)}{2}$. If we divide one triangular number by another, we get a number of the form $\frac{m(m+1)}{n(n+1)}$2012-03-13
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    You will find that $| l-\frac{m_k(m_k +1) }{n_k(n_k +1)} | \le \frac{2}{n_k}$, which will meet your needs2012-03-13
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    @Henry Hammering this out would make a great answer, since it specifically addresses the OP's algorithm.2012-03-13
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    @WimC: Feel free to do so - I am happy with joriki's answer2012-03-13

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