36
rahul
·2011-06-12 05:16:19
if 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.....
1
jee12
·2011-06-12 05:19:01
but before checking for 2 and 5 it shud b in lowest form
obviously!!!
36
rahul
·2011-06-12 05:27:42
thanks fr adding the most imp part...
49
Subhomoy Bakshi
·2011-06-12 05:45:12
@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
jee12
·2011-06-12 07:23:27
@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)
49
Subhomoy Bakshi
·2011-06-12 12:11:43
oops! i had intended to write 32
62
Lokesh Verma
·2011-06-12 12:22:51
Good work rahul... Neone for a proof?
30
Ashish Kothari
·2011-06-12 23:40:21
Let 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.
11
Khilen Khara
·2011-06-13 01:52:15
The 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
62
Lokesh Verma
·2011-06-13 10:53:07
A simple part left to prove..... .
How do you prove your claim that every terminating number has to have only 10^k as the denominator?
21
Shubhodip
·2011-06-13 11:53:17
Let 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