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    <p>Consider <code>S(n) = n^2</code> and <code>T(n) = n</code>. Then both <code>S</code> and <code>T</code> are <code>O(n^2)</code> but <code>S(n) / T(n) = n</code> which is not <code>O(1)</code>.</p> <p>Here's another example. Consider <code>S(n) = sin(n)</code> and <code>T(n) = cos(n)</code>. Then <code>S</code> and <code>T</code> are <code>O(1)</code> but <code>S(n) / T(n) = tan(n)</code> is not <code>O(1)</code>. This second example is important because it shows that even if you have a tight bound, the conclusion can still fail.</p> <p>Why is this happening? Because the obvious "proof" completely fails. The obvious "proof" is the following. There are constants <code>C_S</code> and <code>C_T</code> and <code>N_S</code> and <code>N_T</code> where <code>n &gt;= N_S</code> implies <code>|S(n)| &lt;= C_S * f(n)</code> and <code>n &gt;= N_T</code> implies <code>|T(n)| &lt;= C_T * f(n)</code>. Let <code>N = max(N_S, N_T)</code>. Then for <code>n &gt;= N</code> we have</p> <pre><code>|S(n) / T(n)| &lt;= (C_S * f(n)) / (C_T * f(n)) = C_S / C_T. </code></pre> <p><strong>This is completely and utterly wrong.</strong> It is not the case that <code>|T(n)| &lt;= C_T * f(n)</code> implies that <code>1 / |T(n)| &lt;= 1 / (C_T * f(n))</code>. In fact, what is true is that <code>1 / |T(n)| &gt;= 1 / (C_T * f(n))</code>. The inequality reverses, and that suggests there is a serious problem with the "theorem." The intuitive idea is that if <code>T</code> is "small" (i.e., bounded) then <code>1 / T</code> is "big." But we're trying to show that <code>1 / T</code> is "small" and we just can't do that. As our counterexamples show, the "proof" is fatally flawed.</p> <p>However, there is a theorem here that is true. Namely, if <code>S(n)</code> is <code>O(f(n))</code> and <code>T(n)</code> is <code>Ω(f(n))</code>, then <code>S(n) / T(n)</code> is <code>O(1)</code>. The above "proof" works for this theorem (thanks are due to <a href="https://stackoverflow.com/questions/7323349/division-operation-on-asymptotic-notation/7323520#7323520">Simone</a> for the idea to generalize to this statement).</p>
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    1. CO+1 for picking the exact same example as me at the same time ;-)
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    2. CO@sepp2k: Wow, five second difference on posting time! :-)
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    3. COExcellent answer Jason ... Thank u so much.
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