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Consider 2 integers such that $a/b \approx\pi$. Let $c=|\pi-(a/b)|$ As $a$ and $b$ grow $c \to 0$. Now consider $d=abc$. Do $a$ and $b$ exist such that $d$ is less than it would be for any other pair?

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so you want to know if there exist such $a$ and $b$ so that $d$ is smallest for every other pair? – dato datuashvili Oct 8 '12 at 6:12
smallest value would be so that $a/b=\pi$,except when either $a$ or $b$ is negative,in such case smallest would be negative infinite – dato datuashvili Oct 8 '12 at 6:13
dato, if $a,b$ are integers, then $a/b=\pi$ is impossible. – Gerry Myerson Oct 8 '12 at 6:16
aaaa yes i missed – dato datuashvili Oct 8 '12 at 6:18

For a good approximation we may expect $ab\approx \pi b^2$, so you are essentially looking for approximations that make $|\pi-\frac a b|\cdot b^2$ small. The best candidates for this are the continued fraction approximations for $\pi$, i.e $\frac31$, $\frac{22}7$, $\frac{333}{106}$, $\frac{355}{113}$, $\frac{103993}{33102}$ leading to $d$ values of $-0.425$, $0.195$, $-2.937$, $0.0107$, $-1.989$. Especially good are those aproximations before a big number shows up in the continued fraction. If instead of $\pi$ were asking for $\sqrt 2$, say, an explicit answer could be given. But for transcendentals like $\pi$ there are always surprises awating, that is new record-breaking approximations (=big numbers in the continued fraction) may lure "further down".

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Since $a$ is close to $b\pi$, $d$ is close to $\pi b^2|\pi-(a/b)|$. It is not known whether there are infinitely many pairs $a,b$ such that, say, $|\pi-(a/b)|\lt b^{-3}$. If there are, then your $d$ goes to zero as $a,b$ increase.

share|cite|improve this answer I think that helps the OP understand Gerry's answer better. – mick Oct 12 '12 at 13:58

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