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Given:

$$\dfrac{2016}{2017}<\dfrac mn<\dfrac{2017}{2018}$$

Find the smallest value possible of the sum of the denominator and the numerator, i.e. $m+n$.

I don't know how to spot the very peculiar fraction with the minimum values of $m$ and $n$ in the domain $\left[\dfrac{2016}{2017},\dfrac{2017}{2018}\right]$.

Edit: thank you everyone for the answers! how can one prove that using the median operator the result that is strictly between the two fractions is the one with the minimized value possible for m+n.

I've done some work, I realized that I need to count the number of digits in a decimal number c, given x < c < y , let's say N, c= c*10^(N-1)/10^(N-1) m+n= c*10^(N-1)/(gcd(c*10^(N-1),10^(N-1))) + 10^(N-1)/(gcd(c*10^(N-1),10^(N-1))) m+n= (c+1)( 10^(N-1) / (gcd(c*10^(N-1),10^(N-1)) )

c is a variable that changes on the domain (x,y) and N is dependent on c, so N is also a variable, and then (m+n) is the last variable that is dependent on N and the gcd, which means on both N and c, to write N in terms of c for natural numbers, it's pretty easy and straight-forward: N= ceiling(log(c)) or N= floor(log(c))+1, now, since x and y are at least consecutive numbers, the variable c won't be a natural number and its length isn't easily given.

1) how can you be determined of the number of digits in any decimal number including the fractional part? 2) is it possible to have a function with two variables f(x,y), from which we can obtain the minimum value of m+n?

thanks a ton! I look forward for your answers.

Alex
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  • Are $m.n$ integers? Presumably they are natural numbers. – kingW3 May 13 '17 at 15:07
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    Clearly, $(m,n)=1$ How about calculating $$?<\dfrac{m-n}n<?$$ – lab bhattacharjee May 13 '17 at 15:08
  • @Iabbhattacharjee I can see how $\gcd(m,n)=1$ but how does $\frac{m-n}{n}$ help? – Shaun May 13 '17 at 15:14
  • How can you be certain that the fraction that is strictly between has the most minimized value of both the denominator and the numerator? – Alex May 16 '17 at 12:58
  • For the general problem of finding the simplest fraction in an interval, look at https://math.stackexchange.com/questions/115621. (Note that although the question asks about a different metric, it doesn't really affect the answer.) – tehtmi May 18 '17 at 06:34

4 Answers4

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$$\frac{1}{2016}>\frac{m}{n}-1>\frac{1}{2017}$$

$$\frac{1}{2016}>\frac{m-n}{n}>\frac{1}{2017}$$

$$2016<\frac{n}{m-n}<2017$$

So $n\ge 2\cdot2016+1=4033$ and $m-n\ge 2$. Hence $m\ge 4035$

The smallest value is $8068$.

CY Aries
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  • Thank you, that's really helpful, I'm trying to come up with a proof that the number at the exact middle such that x<(c/d)<y, hence c+d >= (x+y)+2 – Alex May 15 '17 at 13:36
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I believe that you will have to use the mediant operator, which does the following to two fractions in simplest form: $$\frac{a}{b}\oplus \frac{c}{d}=\frac{a+c}{b+d}$$ if you apply this to your fractions, you will get $$\frac{2016}{2017}\oplus \frac{2017}{2018}=\frac{4033}{4035}$$ And the answer should be $4033+4035=8068$. The mediant gives the fraction that is strictly between two other fractions, and I believe that it minimizes the numerator and denominator while doing so. To read more about the mediant, see this page.

Franklin Pezzuti Dyer
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  • that's an elegant solution, how to be certain that the the fraction that is strictly between two fractions has the minimum possible value for both the numerator and the denominator? – Alex May 15 '17 at 13:38
  • You can be sure it is minimized by trying it with a denominator of $4034$, and noticing that $\frac{4033}{4034}$ is too big and $\frac{4032}{4034}$ is too small. – Franklin Pezzuti Dyer May 16 '17 at 19:19
  • The mediant doesn't always give the "simplest" fraction for any $a$, $b$, $c$, $d$. According to Wikipedia, this is only necessarily true for $bc - ad = 1$ (which is true here). I also find that trying $4034$ is not a convincing argument for minimality of $4035$. For example, $\frac{2999}{3000}$ is between the two given candidates of $\frac{4032}{4034}$ and $\frac{4033}{4034}$, so why can we skip considering it? – tehtmi May 18 '17 at 06:23
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Hint: $$\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$$

Jose Arnaldo Bebita Dris
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If $\frac{c}{d}< \frac{m}{n} < \frac{a}{b}$, then the integral point $(m,n)$ has direction between those of $(c,d)$ and $(a,b)$.

If moreover, $a d - b c = 1$, then there are no integral points inside the parallelogram determined by $(a,b)$, $(c,d)$. It follows that $m\ge a + c$, $n\ge b + d$.

Here is also an analytic proof: there exist $u$, $v$ positive such that

$$(m,n) = u(a,v) + v(c,d)$$

Solving the system with Cramer's rule ( the determinant of the system is $1$) we get $u$, $v$ integers $\ge 1$.

orangeskid
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