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

Posted on 2012-12-30
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Last Modified: 2012-12-31
Why can any fraction whose denominator has only 2s and 5s in its prime factorization can be written as a terminating decimal?

This is 7th grade math question.  It's not a homework problem.


Thank you!
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Question by:naseeam
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14 Comments
 
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Expert Comment

by:John Hurst
ID: 38731762
Any number you can divide by 2 or divide by 5 terminates (does not repeat).

.... Thinkpads_User
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LVL 84

Expert Comment

by:ozo
ID: 38731764
A denominator of 10^n has only 2s and 5s in its prime factorization.
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phoffric earned 300 total points
ID: 38731794
Notice that 1/10 is half of 1/5 (since 1/5 = 2/10). If you have any number 1/(5*5*...*5), then you can convert that to (2*2*...2)/(10*10*...10); and numbers that have only 10's in the denominator just move the decimal point of the integer numerator.

Notice that 1/(2*2*...*2) = (1/2)*(1/2)*...*(1/2); and each factor is just 0.5, which when all multiplied together is just a non-repeating decimal. Or, alternatively, as above, notice that:
   1/(2*2*...*2) = (5*5*...*5)/(10*10*...10), and again we are just moving the decimal point of the numerator.
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LVL 95

Expert Comment

by:John Hurst
ID: 38731800
If this is a 7th grade problem, I assume that naseeam wants a very simple answer as first provided. ... Thinkpads_User
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Assisted Solution

by:ozo
ozo earned 50 total points
ID: 38731825
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
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Expert Comment

by:aburr
ID: 38731995
"Why can any fraction whose denominator has only 2s and 5s in its prime factorization can be written as a terminating decimal?"
-
To be perfectly correct you must define "fraction" a bit narrower.
9/3 terminates
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Expert Comment

by:ozo
ID: 38732000
3 does not have only 2s and 5s in its prime factorization, so whether 9/3 terminates is irrelevant.
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Expert Comment

by:aburr
ID: 38732006
"3 does not have only 2s and 5s in its prime factorization, so whether 9/3 terminates is irrelevant. "
true
But it is relvant when considering the converse (I know, that was not asked)
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LVL 27

Assisted Solution

by:BigRat
BigRat earned 50 total points
ID: 38732980
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.
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Assisted Solution

by:GwynforWeb
GwynforWeb earned 100 total points
ID: 38733110
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.
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Expert Comment

by:John Hurst
ID: 38733120
All covering the same ground over and over again here.

.... Thinkpads_User
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Assisted Solution

by:GwynforWeb
GwynforWeb earned 100 total points
ID: 38733151
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
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Expert Comment

by:aburr
ID: 38733219
... Thinkpads_User said

"All covering the same ground over and over again here."   Not quite

you said
"Any number you can divide by 2 or divide by 5 terminates (does not repeat). . "

but that is not really an answer to the question asked (which was Why?

furthermore consider a number (you said ANY number so try 1/3)

(1/3)/2 = 0.16666666....

does not terminate.   so at best you have to expand your answer a bit.
The other answers tried (and eventually did) answer the why?

.
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LVL 95

Expert Comment

by:John Hurst
ID: 38733248
Touche.  I meant and should have said any non-repeating number divided by 2 or 5 will terminate and not repeat. Carelessness on my part.

... Thinkpads_User
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