How to calculate the ratio of two primes in a given set of primes?

Hi all,

I'm working on a personal project here. My goal is this: Fix some natural number n (greater than or equal to 3) and look at the set primes = primes(n). This is the set of all primes less than or equal to n. From here we take any two primes, p_i and p_j and look at the ratio (p_i)/(p_j) = r_ij. If r_ij is less than 1, we "keep" this value and put it into a set. Otherwise (that is, if r_ij >= 1) we ignore it.

Lastly, as this is my ultimate goal, I want to take any r_ij value that is less than 1 and convert it into base 3. From here I will check certain things about the base 3 decimal expansion of any r_ij value and if it satisfies certain things, sort said value into a set, and if it doesn't, another. Ideally, I would be able to see back in base 10 what these values are in each set respectively.

The entire motivation of this little experiment comes from the Cantor set. In my Measure Theory class we were talking about it and I noticed a peculiar pattern. That is, I couldn't find a lot of prime fractions that were in the Cantor set. I know something is in the Cantor set (call it C) if in its base 3 decimal expansion, it includes multiple 1s. This is what I want (or rather, trying) to check. Take any two primes less than or equal to n, compute their fraction. If greater than or equal to 1, ignore said ratio, but if less than one, convert it to base 3, look at its expansion, and if it has zero or one 1's in its decimal expansion, put it in C and have it be presented back as a readable base 10 fraction. Otherwise, put said ratio into a set (call it Nc--"not Cantor") as a readable base 10 fraction. I hope this makes sense. I apologize for the long winded post.

Currently, here is all I have:

n = input('Enter value for n: '); if floor(n) == n primes = primes(n) else error('Enter a positive integer value for n.') end

I am stuck on making the for loop that takes any two elements in my set denoted as "primes" and computing their ratio. I would love some input for this. Thanks!

[KALYAN ACHARJYA]

We prefer Maths Explanation better than too long text.

[Walter Roberson]

If the ratio of the two primes is greater than 1, why not just exchange the roles of the two primes?

[Guillaume]

To me the hardest would be to compute the base 3 expansion of the numbers (not sure what base 3 decimal expansion is, how does it differ from normal base 3?) particularly since I have a feeling that the base 3 expansion of fractions of prime are going to be infinite strings.

[Matthew Graham]

@Walter Roberson. Yeah, I did not think of that at first, but that is a good idea. I've done some more thinking and I'm not sure that will be the best way to approach it anymore for reasons @Guillaume described. Because a prime fraction in base 3 is infinite I would have to truncate the decimal expansion and so when I try to convert it back to a readable base 10 number I would have issues. I think the best way to go about doing it is to fix some prime, p1, and then let the denominator p2 vary over the set and then compute each number in base 3. Now the only problem with this is that if the set of primes has a large cardinality, then just fact checking which prime fractions are/aren't in C is going to be very tedious.

Answer 1

[John D'Errico]

As long as the set is sorted, you never need to worry about the ratio being greater than than 1.

How do you compute a base 3 expansion? That too is easy enough. I'm not sure why you want to do that. But it is not too difficult.

P1 = 5;
P2 = 7;
5/7
ans =
         0.714285714285714

The trick is to convert it to an integer. Then just use dec2base.

Tern = dec2base(floor(P1/P2*3^33),3,33)
Tern =
    '201021201021201021201021201021201'

(Tern stands for a Ternary expansion.) So if we take a few terms...

2*1/3 + 1/27+ 2*1/243 + 1/729
ans =
       0.713305898491084

2*3^-1 + 1*3^-3 + 2*3^-5 + 1*3^-6
ans =
       0.713305898491084

We can get the complete approximation as:

dot(Tern - '0',3.^(-1:-1:-33))
ans =
       0.714285714285714

As you can see, this is the ternary expansion of the fraction 5/7. This is essentially as good as we can do in terms of double precision arithmetic. I used 33 terms in that expansion because 3^33 is the largest integer power of 3 that is less than 2^53-1.

In fact, dec2base even gets the leading zeros in that expansion correct.

Can you do better than that? Well, yes. But it would require the use of higher precision. I'll use my vpi toolbox here to give you 100 ternary digits in the expansion of 13/17.

char(vpi2base(vpi(vpi(3)^100*13/17),3)+'0')
ans =
  '2021221102010011202122110201001120212211020100112021221102010011202122110201001120212211020100112021'

Be careful, in that vpi2base will miss any leading zeros as I did it here. That is easily fixed. But I'm not really sure why you need more ternary digits than a double can provide.

[Matthew Graham]

Thanks! The reason I want to look at base 3 expansion of prime fractions is because the cantor set C can be represented in terms of base three decimals where there are no ones in the decimal. That is, C can be written as the intersection of all numbers between [0,1] whose decimal expansion is consists only of 0s and 2s.