1but before checking for 2 and 5 it shud b in lowest form
obviously!!!
How do you decide if or not a fraction is terminating?
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11 Answers
36if the denominator of a fraction can be written in the form of 2m.5n where m,n belongs to positive integers... then the fraction is terminating or else non-terminating.....
49@Rahul: 31 is also a fraction! [3]
and u said:
"......in the form of 2m.5n where m,n belongs to positive integers......"
Isn't that an integer?
1@SUBHOMOY
bhaiya i don't think 3/1 is a fraction
yes its a rational number
the very time in junior classes i heard fraction it was not 3/1!!
we can write it that way but its rational no..
i don't think we need to worry about that 3/1
(BUT I GUESS NISHANT SIR EXPECTED A BETTER ANS)
30Let the fraction in its lowest form be of the form xy, x and y being integers.
Now let y = 2m.5n , m and n are also non-negative integers.
multiplying Numerator and denominator by 2n.5m
then = xy = x.2n.5m 10m+n which clearly terminates.
In any other case, where the denominator has a factor pr, where p is a prime other than 2 and 5 and r is a positive integer, then the denominator cannot be represented in the form 10k, and the fraction will be recurrent,non-terminating decimal.
11The proof of this lies in the fact that every terminating decimal
has the form n/10^e, for some e >= 0 (e is the number of places to
the right that the decimal point must be moved to give you an integer,
and n is that integer), and every fraction of that form has a
terminating decimal found by writing down n and moving the decimal
point e places to the left. Now when you cancel common factors from
n/10^e = n/(2*5)^e = n/(2^e*5^e), it may reduce the exponents
in the denominator
62A simple part left to prove..... .
How do you prove your claim that every terminating number has to have only 10^k as the denominator?
21Let the fraction be ab ,where gcd(a,b) = 1. this will terminate <=> there exists some integer k>0 such that 10kab is an integer. which means b|10k. bcz gcd(a,b) =1. We get Rahuls answer