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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?
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!
This is 7th grade math question. It's not a homework problem.
Thank you!
Asked:
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If this is a 7th grade problem, I assume that naseeam wants a very simple answer as first provided. ... Thinkpads_User
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
can be produced by multiplying together numbers with a terminating decimal part of .5, .2 or .0
"Why can any fraction whose denominator has only 2s and 5s in its prime factorization can be written as a terminating decimal?"
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To be perfectly correct you must define "fraction" a bit narrower.
9/3 terminates
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To be perfectly correct you must define "fraction" a bit narrower.
9/3 terminates
"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)
true
But it is relvant when considering the converse (I know, that was not asked)
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.
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.
(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
(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
... 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?
.
"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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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),