Why can any fraction whose denominator has only 2s and 5s in its prime factorization can be written as a terminating decimal?

fraction whose denominator has only 2s and 5s in its prime factorization
can be produced by multiplying together numbers with a terminating decimal part of .5, .2 or .0

I would refer you to Hardy & Wright's "Theory of Numbers" Oxford University Press chapter 9 page 107 and sequence. The proof of the following theorem is laid out there and is a bit too long winded to reproduce here

Theorem 135

The decimal for a rational number p/q between 0 and 1 [ie: a fraction] is terminating or recurring, and any terminating or recurring decimal is equal to a rational number. If (p,q)=1 [ie: relatively prime] and q=2^a5^b [2 to the power alpha and 5 to the power beta] and the max(a,b)=µ, then the decimal terminates after µ digits. If (p,q)=1 and  q=2^a5^b*Q where Q>1 and (Q,10)=1, and v is the order of 10 mod Q, then the decimal contains µ non-recurring digits and v recurring digits.

The comments in square brackets [] are mine. The proof requires another rather complicated theorem (88) given on page 71.

The proof is NOT a 7th grade maths question.

If we use the idea that the product of 2 non-repeating numbers is non-repeating then

(1) Since 1/2 =0.5  

   then 1/2^n   is non-repeating

   Similarly

   Since 1/5 =0.2  

   then 1/5^m   is non-repeating

(2) Since   1/2^n   and  1/5^m are non repeating

   then   1/(2^n*5^m) is no repeating

(3) Since 1/(2^n*5^m) are non-repeating then, where A is an integer,

   A/(2^n*5^m) is non-repeating


Hence all fractions whose denominators are of the form 2^n*5^m are non-repeating.

For this to be anywhere close to a Grade 7 question it would have to be of the form:-

(Q)   Show that 1/(2^6*5^6) is a non-repeating fraction.

(A)   1/(2^6*5^6) = 1/10^6 = 0.0000001


or possibly

(Q)   Show that 1/(2^8*5^6) is a non-repeating fraction.

(A)   1/(2^8*5^6) = 1/(2^2*10^6) = (1/4)*(1/10^6)

                                 = 0.25*0.0000001 = 0.000000025