# Recurring decimals in prime fractions

1. Apr 2, 2006

### Oxymoron

When I write out the decimal expansion of $1/p$ where $p$ is a prime, it is always a recurring decimal with a period $per(p)$. I was thinking why inverting a prime number should always give a recurring decimal but could not think of a reason other than it has to be something to do with our base 10 system of counting.

Some prime reciprocals, such as 1/7, 1/17, and 1/19 have periods which is one less that the prime. That is, $per(p) = p-1$ for some primes. And in fact, for all primes $p$, $per(1/p)\,|\,(p-1)$. These are all interesting but there is something which I cant seem to explain.

Take the prime 11. Then the period of 1/11 is 2. Take some other prime p. Then

$$per\left(\frac{1}{11}\times\frac{1}{7}\right) = 2per\left(\frac{1}{p}\right)$$

so long as $p < 31$.

After looking at 1/11, I started looking at other prime reciprocals, hoping to find other relationships between the periods of 1/p, 1/q and 1/pq. Unfortunately, the only thing I could find was that

$$per\left(\frac{1}{p}\times\frac{1}{q}\right) = \infty$$

whenever $p=q$ and the period of 1/p is odd.

I was hoping for a nice clean relationship between 1/p,1/q, and 1/pq but I could not find one at all. So I tried checking the web and again, noone seems to have found any relationships. So I came here. If anyone has seen, or done any work on comparing periods of different prime reciprocals, could you please share your knowledge.

Cheers.

Last edited: Apr 2, 2006
2. Apr 2, 2006

### shmoe

These, and the relation you want for 1/pq, can be explained by the relation between the period of 1/n and the order of 10 mod n (you're interested in the case gcd(10,n)=1 here, though the other case is worth looking at).

This isn't true, the period of 1/7 is 6, the period of 1/77 is also 6.

infinity? This can't be true either, it will always be finite as 1/pq is a rational number, and hence its decimal either terminates or repeats.

3. Apr 3, 2006

### Oxymoron

What is n here? Does n = pq?

Last edited: Apr 3, 2006
4. Apr 3, 2006

### shmoe

Sure, if you like. It doesn't have to be the product of 2 primes though, there's a nice relation between the period of a number n with gcd(10,n)=1 and the order of 10 mod n.

5. Apr 3, 2006

### Oxymoron

That got me thinking...

The period of 1/p is the order of 10(mod p) right? so we have a congruence relation:

$$10^r \equiv 1(\mod p)$$

The period of 1/q is the order of 10(mod q). So:

$$10^s \equiv 1(\mod q)$$

And the period of 1/pq is the order of 10(mod pq). So:

$$10^t \equiv 1(\mod pq)$$

...

6. Apr 3, 2006

### Oxymoron

No matter what prime p we take, p is always relatively prime to 10 (since p has no divisors anyway). Take a prime. Then suppose that p divides $10^r-1$, that is, $nm = 10^r - 1$ for some integer $1\geq m < 10^r$. Then

$$\frac{1}{n} = \frac{m}{10^r-1} = m(10^{-r} + 10^{-2r} + \dots)$$

If our integer $m = a+0 + 10a_1 + 10^2a_2 + \dots 10^{r-2}a_{r-2} + 10^{r-1}a_{r-1}$ for decimal digits $a_0,\dots,a_{r-1}$, then the decimal expansion of 1/p will be

$$\frac{1}{p} = 0.a_{r-1}a_{r-2}\dots a_1a_0a_{d-1}a_{d-2}\dots a_1a_0\dots$$

which will repeat with a period that will always divide r.

Then the period of 1/p is the smallest r such that p divides 10^r-1. In other words, the period of 1/p is the smallest d such that

$$10^r \equiv 1(\mod p)$$

But this is the definition of $\mbox{ord}_p(10)$.

So the period of 1/p is the order of 10(mod p).

Last edited: Apr 3, 2006
7. Apr 3, 2006

### Oxymoron

Which should mean that

$$\mbox{per}\left(\frac{1}{pq}\right) = \mbox{ord}_{p+q}(10)$$

8. Apr 3, 2006

### shmoe

Should be the order of 10 mod pq, not p+q. Now do you know how to find the oder of 10 mod pq from the orders mod p and mod q?

9. Apr 3, 2006

### Oxymoron

Of course, pq, nor p+q, I dont know why I wrote that.

I think I may know, although I kind of jumped at that conclusion without properly running through it.

$$\mbox{per}(1/p)\times\mbox{per}(1/q)$$

is the order of 10(mod pq). If you let n = pq (although I may now understand what you mean by this, n can be any integer so long as it is relatively prime to 10?) then

$$\mbox{per}(1/p)\times\mbox{per}(1/q) = \mbox{per}(1/n)$$

which is equal to the order of 10(mod n) = 10(mod pq). Which is the same thing as

$$\mbox{per}(1/pq) = \mbox{ord}_{pq}(10)$$

Last edited: Apr 3, 2006
10. Apr 3, 2006

### shmoe

Yes. If it's not relatively prime, then it's a terminating decimal.

Remember th 1/7 and 1/11 example? The period of 1/11 was 2, the period of 1/7 was 6, the period of 1/77 was 6, so it's not the product of the periods.

Try a few more examples to see if you can spot a relationship. You might want to think in terms of the order of 10 mod p and q here. Specifically if r is the order of 10 mod p and 10^t=1 mod p then r divides t. Also, if 10^t=1 mod pq then 10^t=1 mod p.

11. Apr 3, 2006

### Oxymoron

For example

1/7. Find $\mbox{ord}_7(10)$

$$10^2 \equiv 2(7)$$
$$10^3 \equiv 6(7)$$
$$10^4 \equiv 4(7)$$
$$10^5 \equiv 5(7)$$
$$10^6 \equiv 1(7)$$

and we have exhausted mod 7. So infact,

$$\mbox{ord}_7(10) = 6$$

p=17. Find $\mbox{ord}_17(10)$.

$$10^2 \equiv 15(17)$$
$$10^4 = 15^2 \equiv (-2)^2 \equiv 4(17)$$
$$10^8 = 4^2 \equiv 16(17)$$

Thus we can see that $\mbox{ord}_{17}(10)$ is not 1,2,4, or 8. But we must remember that $\mbox{ord}_{17}(10)$ divides $\phi(17) = 16$. The divisors of 16 are 1,2,4,8, and 16. Therefore $\mbox{ord}_{17}(10) = 16$.

Now find $\mbox{ord}_{7\cdot 17}(10)$

$$\mbox{ord}_{7\cdot 17}(10) \Rightarrow 10^{rs} \equiv 1(7\cdot 17)$$

But is there a quicker way of doing this other than working them all out?

Last edited: Apr 3, 2006
12. Apr 3, 2006

### shmoe

Suppose the order of 10 mod 7*17 is t. Then 10^t=1 mod 7*17. So we have both:

10^t=1 mod 7
10^t=1 mod 17

What does this tell you about t compared to 6 and 16 (the orders of 10 mod 7 and 17).

13. Apr 3, 2006

### Oxymoron

But surely t cant take on two values at the same time?

14. Apr 3, 2006

### Oxymoron

So 7 divides t-1 and 17 divides t-1. So 6 divides t and 16 divides t. So t=48 works but t=48 = lcm(6,16).

Last edited: Apr 3, 2006
15. Apr 3, 2006

### shmoe

Nope. You maybe mistyped?

Which shows lcm(6,16) divides t. You should also be able to show t divides lcm(6,16), then you'd have equality. Of course showing 10^48=1 mod 7*17 directly would work as well, but you can prove this lcm relation in general.

16. Apr 3, 2006

### Oxymoron

Yeah, I meant to write

7 divides 10^t - 1 and 17 divides 10^t -1

=>

6 divides 10^t and 16 divides 10^t

which just comes from the definition of

$$10^t \equiv 1(\mod 7)$$

and

$$10^t \equiv 1(\mod 17)$$

17. Apr 3, 2006

### Oxymoron

Which makes it that

$$\mbox{lcm}(6,16) \mbox{ divides } 10^t$$

18. Apr 3, 2006

### shmoe

7 divides 10^t - 1 implies 6 divides t, not 10^t (as 6 is the order of 10 mod 7). Similiar for 16, which is why we have lcm(6,16) divides t.

19. Apr 3, 2006

### Oxymoron

Ahah! of course! of course! (I was lucky getting the right answer from incorrect calculations).

So finally we have that if

$$\mbox{ord}_{pq}(10) = t$$

then

$$t = \mbox{lcm}(p-1,q-1)$$

20. Apr 3, 2006

### Oxymoron

and since

$$\mbox{ord}_{pq}(10) = \mbox{per}(1/pq)$$

we have

$$\mbox{per}(1/pq) = \mbox{lcm}(p-1,q-1)$$