Finding a Fraction Between Two Fractions

[mak_st]

I think the last step should add something like this:
Let (b-5a)/(6a-b) = p/q for some integers p, q with 11/5 < p/q < 43/9, which simplifies to
a/b = (p+q)/(6p+5q)
Now we want to minimize 6p + 5q.
If q = 1, then p = 3 or 4
If q >= 2, then p > 11q/5 >= 22/5, hence p >= 5 which clearly makes 6p+5q larger than the previous case
Hence, we can conclude that q = 1 and p = 3 is the minimum, that is, b = 23

[amitsrivastava]

Can you please explain why you consider only integer values for (b - 5a)/ (6a -b) ? I am not getting the rationale here. Am I missing out on something critical ?

[田村博志-zy]

By the Euclidean method, we have
1 = 74*16 - 13*91
= 57*13 - 10*74
= 40*10 - 7*57
= 23*7 - 4*40
= 6*4 - 1*23.
We have a chain of rational numbers
52/303 < 4/23 < 7/40 < 10/57 < 13/74 < 16/91.
1/6 is inappropriate.

Then we have to prove that there is no fractional number a/b such that
b < 23,
52/303 < a/b < 4/23.
We may prove that easily because 4/23 is very close to 52/303.

[andreben]

This is really nice. I believe this is deeply related to continued fraction expansion of rational numbers ^-^

[oldjoec]

A far faster (but not-so-elegant) solution, would be to find the smallest value of "a", by guess-and-check, starting at a=2. Since the fractions are a bit larger than 1/6, just use b=6a-1. Takes a very short time to reach 4 as the smallest "a", with b=23, which is also, then, the smallest "b". i.e. Values 2/11, and 3/17 don't work, but 4/23 appears with less than a minute's work, and no algebra.

[caesar_cipher]

letsthinkcritically : please explain why u say that values of 3 and 4 for the ratio will minimise the value of b.
This is the most critical logical step, so incumbent upon you to explain it properly before making another video

[davidtomasella]

You can directly add 1/4 everywhere after the first reciprocal,
And you get:
95/16 < (4b+a)/(4a) < 316/52

Then: you choose (4b+a)/(4a) = 6 and find a/b = 4/23

[sswy]

52/303 > 52/312, which is 4/24. 16/91 < 16/88, which is 4/22. and we can easily check 4/23 satisfy the condition. and if a < 4 ie. 1 or 2 or 3, can easily check no answer.

basically the idea is we see common factors of (52, 16) is 4, so try to make both the numerators equal to 4, and then find b between the denominators.

[_Possible]

Why did we need to choose integer values for the fraction at the end?

[stevenwilson]

I love your method for finding the fraction but I don't see how you validated that this is the minimum b.

[tianqilong]

actually it could be much simpler, after reciprocal, considering that the gap is kinda large, just let a = 1, 2, 3, 4, respectively, whatever a that fits in first, that corresponding b will be the ans. In this case, a = 4, so it is not much work tbh

[PyarMatKaro]

When choosing the integer at the last step, the smallest integer in the interval should always be chosen. This minimises both the numerator and the denominator of the answer

[jeanmarcbonici]

We ask
\[Alpha]=52/303 and \[Beta]=16/91
The algorithm :
b=1 ;
While[Ceiling[\[Alpha], 1/b] >= \[Beta] && \[Alpha] >= Floor[\[Beta], 1/b], ++b] ;
written in Wolfram-Language
provides
b=23

[noahtaul]

A proof I’d put on a contest: Farey fractions say that 1/6, 4/23, and 3/17 have no fractions of any smaller denominator between them. But 52/303 is the mediant of 16/96 and 36/207, which are equal to 1/6 and 4/23, and 16/91 is the mediant of 4/23 and 12/68, which are equal to 4/23 and 3/17. So 1/6<52/303<4/23<16/91<3/17, with no smaller denominator fractions between the outermost, so 4/23 is the fraction we’re looking for.

Of course, this presupposes you knew the answer already, but it’s a nice, clean, simple proof that doesn’t give anything away except the answer and enough to prove that the answer is correct.

[wesleydeng]

I did this way: notice that gcd(52, 16) =4. Then 52/303 = 4/23.30 and 16/91=4/22.75. So a/b = 4/23 is a solution. Easy to check no solutions for a=1, 2, 3.

[khamza]

You can tell 1/6 is closest
to up it a little you go 2/(6*2-1)
then 3/(6*3-1)
and finally 4/(6*4-1) which is 4/23

[bait]

Interrsting div 4 and change a/4 to a, get the answer.quicker

[wiggles]

An easy way is to just let "b" be the sum of the numerators, and "a" be the sum of the denominators. So, let b=91+303=394, and a=16+52=68. Then, b/a=394/68 is between 91/16 and 303/52.

[זאבגלברד]

7/40 , 9/52 , 10/57 are 3 more possibilities. 4/23 is the smallest using a C program. So something is not 100% right , I think.

[advaykumar]

This is how I did it-
I took common numerator which is 208/1212 and 208/1183
Then I calculated min of (deno/(hcf(208 and denominator ))) using python