MHB Fraction between 96/35 & 97/36 with minimum denominator

[SweatingBear]

Q: Find a fraction between $$\frac {96}{35}$$ and $$\frac {97}{36}$$ that has the smallest possible denominator.

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

$$\frac{97}{36} < \frac {p}{q} < \frac{96}{35} \, .$$

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

$$\frac{3395}{1260} < \frac {p}{q} < \frac{3456}{1260} \, .$$

In total, there exist 60 different fractions in the interval above which $$\frac {p}{q}$$ can equal. Therefore, intuitively, if we can find a $$p$$ between $$3395$$ and $$3456$$ which cancels as many prime factors in $$1260$$, 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]

It would be easier if you wrote [math]\displaystyle \begin{align*} \frac{96}{35} = 2\,\frac{26}{35} \end{align*}[/math] and [math]\displaystyle \begin{align*} \frac{97}{36} = 2\,\frac{25}{36} \end{align*}[/math], because then it reduced the problem down to finding a fraction between [math]\displaystyle \begin{align*} \frac{25}{36} \end{align*}[/math] and [math]\displaystyle \begin{align*} \frac{26}{35} \end{align*}[/math]

 

[I like Serena]

Q: Find a fraction between $$\frac {96}{35}$$ and $$\frac {97}{36}$$ that has the smallest possible denominator.

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

$$\frac{97}{36} < \frac {p}{q} < \frac{96}{35} \, .$$

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

$$\frac{3395}{1260} < \frac {p}{q} < \frac{3456}{1260} \, .$$

In total, there exist 60 different fractions in the interval above which $$\frac {p}{q}$$ can equal. Therefore, intuitively, if we can find a $$p$$ between $$3395$$ and $$3456$$ which cancels as many prime factors in $$1260$$, 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]

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:(
$$|\pi\rangle$$

 

[kaliprasad]

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]

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]

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]

continuing my method approach I get 7/10

That's one possibility, but it's not the best. Any advances on 7/10?? (Wondering)

 

[alyafey22]

By some programming I get

$$\frac{5}{7} \approx 0.714 $$

 

[Opalg]

By some programming I get

$$\frac{5}{7} \approx 0.714 $$

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)

 

[alyafey22]

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 .

 

[I like Serena]

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]

Well if one must resort to a brute-force method, then here is a solution:

$$\frac{97}{36} < \frac {p}{q} < \frac{96}{35} \ \Longleftrightarrow \ \frac{97}{36} \cdot q < p < \frac{96}{35} \cdot q \, . $$

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

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

$$\frac{97}{36} \cdot 7 < p < \frac{96}{35} \cdot 7 \, ,$$

which approximately is equivalent to

$$18.9 < p < 19.2 \, .$$

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

$$\frac {p}{q} = \frac {19}{7} \, .$$

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.