# Are lowest/irreducible fraction forms unique?

1. Apr 26, 2010

### holezch

I thought they were unique, but given a fraction a/b, couldn't you always write it as the (decimal representation of a/b without the decimal place)/ 10^(n) ?
thanks

2. Apr 26, 2010

### Hurkyl

Staff Emeritus
Every rational number does indeed have a unique representation of the form a/b, where a is an integer and b is a positive integer.

If what you suggest both makes sense and results in something that is in lowest terms, then you have merely recomputed the form a/b.

3. Apr 26, 2010

### holezch

thanks, but for example, you can write 121/123 = 0.983739837398373983739837398373983739837398373983739837398373983739837398373983739837398373983739837398373983739837398373983739837398373983739837398374
.....
or you can write (n being the number of digits)
983739837398373983739837398373983739837398373983739837398373983739837398373983739837398373983739837398373983739837398373983739837398373983739837398374 / 10^(n)

these are both in lowest forms right? but they have different integers a, b for a/b
and I'm curious, I've never seen a proof for that statement

4. Apr 26, 2010

### Hurkyl

Staff Emeritus
Erm... exactly how many digits do you think are in the decimal expansion of 121/123?

5. Apr 26, 2010

### holezch

Well, a lot.. But the problem is that I can't find an example where the number of digits is finite. But if I could, you could apply the same idea?

6. Apr 26, 2010

### Hurkyl

Staff Emeritus
Of course. 1/2 and 127/1000 have only finitely many digits.

(ignoring the infinitely many trailing zeroes, and assuming you choose that decimal representation rather than the one ending in infinitely many trailing nines)

7. Apr 26, 2010

### holezch

D'oh of course, 1/2.. But now in this example, 5/10 is indeed just 1/2.. How can I prove that these irreducible forms are unique? Will it be an easy proof? (I'm guessing that it will be something like supposing you have a' and a then showing that they are equal, a usual strategy)
thanks

8. Apr 26, 2010

### Hurkyl

Staff Emeritus
It follows more or less directly from prime factorization of integers.

9. Apr 27, 2010

### HallsofIvy

Well, first only a small portion of the rational numbers have decimal representations that terminate after a finite number of places (those with only powers of 2 and 5 in their denominators) and those are the only ones for which that is possible. And, of course, that would, in general, NOT be irreducible.

For example, 1/4 = 0.25 so your expression would be 25/100 which is not irreductible because both 25 and 100 are divisible by 25. It reduces to, no surprise, 1/4!