# I would be ever so greatful if (finite decimal expansions)

1. Nov 6, 2005

### Natasha1

... Someone could help me with this. I need to prove that

Finitedecimal = p/(2^a)(5^b) where either a or b could be 0 and p is a prime.

Basically this formula comes from...

Finitedecimal = (a1/10)+(a2/10^2)+....+(an/10^n)

2. Nov 6, 2005

### HallsofIvy

If you are referring to the "a" in p/2a5b, then a and b can be any whole numbers.
If you are referring to a1, a2, etc. they are whatever the digits are in the decimal form of the number. For example, suppose x= 0.21573 (I've just made up those digits- the crucial point is that there are only a finite number of them). Here a1= 2, a2= 1, a3= 5, a4= 7, and a5= 3. Noting that there are only 5 digits in the number, if we multiply both sides by 105= 100000, 100000x= 21573 so x= 21573/100000. Of course, 100000= 105= (25)(55). Since 21573 is not even, it does have a factor of 2 and since it does not end in 0 or 4, it does not have a factor of 5 and so the fraction cannot be reduce further. "a" and "b" are both 5.
For the general proof, let x= (a1/10)+(a2/10^2)+....+(an/10^n)
and multiply both sides by the highest power of 10: 10n. You should be able to see that the right hand side is now an integer.

There is one error in your statement: while every finite decimal can be expressed as a fraction p/(2a5b) p does not have to be prime- it only has to be a number not divisible by 2 or 5. For example, the number above, 21573= 27(17)(47).

3. Nov 6, 2005

### Natasha1

what a boss! Thanks gourou :!!)