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Number Theory Fraction between 96/35 & 97/36 with minimum denominator

sweatingbear

Member
May 3, 2013
91
Q: Find a fraction between \(\displaystyle \frac {96}{35}\) and \(\displaystyle \frac {97}{36}\) that has the smallest possible denominator.

My thoughts: Let's assume this fraction is of the form \(\displaystyle \frac pq\) where \(\displaystyle \gcd(p,q) = 1\). Since \(\displaystyle \frac{96}{35} > \frac{97}{36}\), we can write

\(\displaystyle \frac{97}{36} < \frac {p}{q} < \frac{96}{35} \, .\)

By multiplying by an appropriate factor in each given fraction respectively and thus obtaining common denominators, we equivalently have

\(\displaystyle \frac{3395}{1260} < \frac {p}{q} < \frac{3456}{1260} \, .\)

In total, there exist 60 different fractions in the interval above which \(\displaystyle \frac {p}{q}\) can equal. Therefore, intuitively, if we can find a \(\displaystyle p\) between \(\displaystyle 3395\) and \(\displaystyle 3456\) which cancels as many prime factors in \(\displaystyle 1260\), we will be able to determine the fraction with the smallest denominator.

This is where I am stuck; surely one could resort to a brute-force method, but I am hoping to solve it in more algebraic terms. Anyone got a clue?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
It would be easier if you wrote \(\displaystyle \displaystyle \begin{align*} \frac{96}{35} = 2\,\frac{26}{35} \end{align*}\) and \(\displaystyle \displaystyle \begin{align*} \frac{97}{36} = 2\,\frac{25}{36} \end{align*}\), because then it reduced the problem down to finding a fraction between \(\displaystyle \displaystyle \begin{align*} \frac{25}{36} \end{align*}\) and \(\displaystyle \displaystyle \begin{align*} \frac{26}{35} \end{align*}\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Q: Find a fraction between \(\displaystyle \frac {96}{35}\) and \(\displaystyle \frac {97}{36}\) that has the smallest possible denominator.

My thoughts: Let's assume this fraction is of the form \(\displaystyle \frac pq\) where \(\displaystyle \gcd(p,q) = 1\). Since \(\displaystyle \frac{96}{35} > \frac{97}{36}\), we can write

\(\displaystyle \frac{97}{36} < \frac {p}{q} < \frac{96}{35} \, .\)

By multiplying by an appropriate factor in each given fraction respectively and thus obtaining common denominators, we equivalently have

\(\displaystyle \frac{3395}{1260} < \frac {p}{q} < \frac{3456}{1260} \, .\)

In total, there exist 60 different fractions in the interval above which \(\displaystyle \frac {p}{q}\) can equal. Therefore, intuitively, if we can find a \(\displaystyle p\) between \(\displaystyle 3395\) and \(\displaystyle 3456\) which cancels as many prime factors in \(\displaystyle 1260\), we will be able to determine the fraction with the smallest denominator.

This is where I am stuck; surely one could resort to a brute-force method, but I am hoping to solve it in more algebraic terms. Anyone got a clue?
You are making the assumption that the lowest possible denominator divides $1260$, but there is no guarantee that it does.

The best I can suggest is to try each possible denominator, starting from the lowest.
That is, first try denominator 2, then 3, then 4, etcetera.
I'm not aware of any algebraic theorem that would speed up the process.

It turns out you're done much quicker than you'd think.
 

Petrus

Well-known member
Feb 21, 2013
739
You are making the assumption that the lowest possible denominator divides $1260$, but there is no guarantee that it does.

The best I can suggest is to try each possible denominator, starting from the lowest.
That is, first try denominator 2, then 3, then 4, etcetera.
I'm not aware of any algebraic theorem that would speed up the process.

It turns out you're done much quicker than you'd think.
Hi,
Does it not work to calculate SGD of them both and then you rewrite the SGD with prime number and the lowest prime number that both got is the answer?
edit: I notice I did missunderstand the question:(
\(\displaystyle |\pi\rangle\)
 
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kaliprasad

Well-known member
Mar 31, 2013
1,322
we need to find p/q to be between 26/35 and 25/36

so let 25/36 < p/q < 26/35

ow 25 q < 36 p
35 p < 26 q

we need to find integers p and q such that q is lowest

some one should be able to proceed. I am not able to do so.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
we need to find p/q to be between 26/35 and 25/36

so let 25/36 < p/q < 26/35

ow 25 q < 36 p
35 p < 26 q

we need to find integers p and q such that q is lowest

some one should be able to proceed. I am not able to do so.
First, notice that $\frac{25}{36}\approx 0.694$ and $\frac{26}{35}\approx0.743$. Now follow I like Serena's excellent hint in comment #3 above:

Can we find a fraction in the interval $[0.694, 0.743]$ with $q=3$ as its denominator? – No, because $\frac23 \approx 0.667$ is too small.

Can we find a fraction in the interval $[0.694, 0.743]$ with $q=4$ as its denominator? – No, because $\frac34 = 0.75$ is too big.

$\vdots$

Continue like that with $q= 5,6,\ldots$ until you get lucky.
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
continuing my method approach I get 7/10
between 0.694 and 0.743.
3/4 is .75 too large
3/5 too small
4/5 too large
4/6 too small so on
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
By some programming I get

\(\displaystyle \frac{5}{7} \approx 0.714 \)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
By some programming I get

\(\displaystyle \frac{5}{7} \approx 0.714 \)
Yes! (Happy), but I'm fighting the impulse to say something sarcastic about young people these days needing a computer program in order to evaluate 5/7. (Lipssealed)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Yes! (Happy), but I'm fighting the impulse to say something sarcastic about young people these days needing a computer program in order to evaluate 5/7. (Lipssealed)
I totally agree with you , this is a bad habit. We rely on technology to solve lots of easy problems which might effect us . In the future , computations will become your nightmare if you don't have a calculator .
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
I totally agree with you , this is a bad habit. We rely on technology to solve lots of easy problems which might effect us . In the future , computations will become your nightmare if you don't have a calculator .
I keep explaining to people I tutor how to add fractions and how to multiply them.
If they ever learned how to, they've forgotten, relying on their calculators.
Which is just fine... until they have to learn more advanced math that has unknowns in numerators and denominators.
The advanced math does not appear to be a problem, but somehow their regular teachers never take the time how to calculate with fractions. (Lipssealed)
 

sweatingbear

Member
May 3, 2013
91
Well if one must resort to a brute-force method, then here is a solution:

\(\displaystyle \frac{97}{36} < \frac {p}{q} < \frac{96}{35} \ \Longleftrightarrow \ \frac{97}{36} \cdot q < p < \frac{96}{35} \cdot q \, . \)

Since \(\displaystyle p \in \mathbb{N} \) we need to find a value for \(\displaystyle q\) such that the interval above in which \(\displaystyle p\) lies allows \(\displaystyle p\) to take on integral values.

By brute-force, the first value of \(\displaystyle q\) for which this is possible is \(\displaystyle q = 7\). This yields

\(\displaystyle \frac{97}{36} \cdot 7 < p < \frac{96}{35} \cdot 7 \, ,\)

which approximately is equivalent to

\(\displaystyle 18.9 < p < 19.2 \, .\)

Now, since \(\displaystyle p \in \mathbb{N}\) we can conclude that \(\displaystyle p\) must equal \(\displaystyle 19\). Therefore the fraction with minimum denominator is

\(\displaystyle \frac {p}{q} = \frac {19}{7} \, .\)

Sure it works but I really did not like the brute-force approach. Instead I would love to see an algebraic approach but anyways, fair enough.
 
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