Find the smallest positive integer $m$ such that there exists positive integer $n$ that satisfies that $\lvert {n\over m} - {2\over 5} \rvert\le {1\over100}$

I tried to simplify and turn it into the following $$\lvert {5n-2m\over5m} \rvert\le {1\over100}$$ $$100\cdot\lvert {5n-2m\over5m} \rvert\le1$$ Since $5m>0$ we get that $$100\cdot\lvert5n-2m\rvert\le5m$$ We can take cases now $$\begin{cases} 100(5n-2m)\le5m, & \text { if } & 5n-2m\ge0 \\[2ex] 100(2m-5n)\le5m, & \text { if } & 5n-2m\le0 \end{cases}$$ Unfortunately I am unable to continue from here. Any help would be appreciated.

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    $\begingroup$ What about $m=5$? $\endgroup$ – J. W. Tanner Sep 6 '19 at 11:57
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    $\begingroup$ Just try each case up to $m=5$ by hand. Try with $1\le n\le m$. $\endgroup$ – Peter Foreman Sep 6 '19 at 12:05


if $n\ge2$ then $|\frac n3-\frac25|=\frac n3-\frac25\ge\frac23-\frac25=\frac4{15}\gt\frac1 {100}$

if $n\le1$ then $|\frac n3-\frac25|=\frac25-\frac n3\ge\frac25-\frac13=\frac1{15}\gt\frac1 {100}$


if $n\ge2$ then $|\frac n4-\frac 25|=\frac n4-\frac25\ge\frac24 - \frac25=\frac1 {10}>\frac1{100}$

if $n\le1$ then $|\frac n4-\frac 25|=\frac 25-\frac n4\ge\frac25-\frac14=\frac3{20}>\frac1{100}$

Cases $m=1$ and $m=2$ are subsumed under case $m=4$, since $\frac n1=\frac {4n}4$ and $\frac n2=\frac {2n}4$.

Therefore $m=5$, where $|\frac25-\frac25|=0<\frac1 {100}$.


$|\frac{n}{m}\,-\,\frac{2}{5}|\,\leq\frac{1}{100} $ $$ |\frac{100n}{m}\,-\,40|\, \leq 1. $$ 39$\leq\frac{100n}{m}\,\leq41$ Therefore $\frac{n}{m}=\frac{2}{5}$ Hence 5n=2m hence min m will be 5 when n=2.

  • $\begingroup$ How do you know $\frac nm=\frac25$ from $39\le\frac{100n}m\le41$? There are other fractions $\frac{100n}m$ between $39$ and $41$, e.g., $\frac{900}{22}$. How do you know (as OP asked) that $5$ is the smallest positive integer $m$ satisfying $39\le\frac{100n}m\le41$? $\endgroup$ – J. W. Tanner Sep 6 '19 at 13:11

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