Modulus of a fraction

How do I calculate the modulus of a fraction? For example if I want to calculate mod(33/4,5) -- I know the result is 2, because 33/4 = 132/16 and when I calculate modulus 5 that gives me 2/1=2. Can anyone please help me?

Cheers!

[Walter Roberson]

I don't think I've ever seen modulus of a fraction be said to have that meaning.

How would you calculate mod(33/2,6) ?

[Andrew Newell]

The MATLAB builtin modulus function gives

mod(33/4,5)

ans =

3.2500

[Walter Roberson]

I can see a certain logic in Thomas's calculation, but that logic would fail if the denominator and the modulus are not relatively prime.

[Tanveer ul haq]

@Walter Roberson: mod(a/x,y) is possible only when x and y is relatively prime.

Answer 1

[Tanveer ul haq]

Download both the .m files and the run "fraction_modulo.m" to get your desired modulus.

Answer 2

[John D'Errico]

I assume you are talking about the ring of integers modulo 5. In that case, 33/4 can be thought af as 33 times the mulitplicative inverse of 4, in that ring.

As long as the modulus n is prime, then all numbers from 1 to n-1 have a multiplicative inverse. (If the modulus is composite, then SOME numbers will still have a multiplicative inverse, though many will not.) 4 for example is its own inverse modulo 5, because we see that

mod(4*4,5)
ans =
     1

And therefore we can compute what you want using code I've posted on the file echange in my VPI toolbox.

inv4 = minv(4,5)
inv4 =
     4

And now we see the result as 2, as you expected.

mod(33*inv4,5)
ans =
     2

Do you need minv? Well, no. You can use gcd. As long as 4 and 5 are relatively prime, 4 has a multiplicative inverse in the generated ring of integers modulo 5. We can both test that fact, and get the inverse directly from gcd.

[G,C] = gcd(4,5)
G =
     1
C =
    -1

Q will always be 1 as long as it is true they are relatively prime. C gives you the inverse, as

inv4 = mod(C,5)
inv4 =
     4

So there is no need for any special tools. Not even my own minv utility.