How do you decide if the number terminates?

but before checking for 2 and 5 it shud b in lowest form

obviously!!!

thanks fr adding the most imp part...

@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)

Let the fraction in its lowest form be of the form xy, x and y being integers.

Now let y = 2^m.5ⁿ , m and n are also non-negative integers.

multiplying Numerator and denominator by 2ⁿ.5^m

then = xy = x.2ⁿ.5^m 10^m+n which clearly terminates.

In any other case, where the denominator has a factor p^r, 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 10^k, and the fraction will be recurrent,non-terminating decimal.

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

Let the fraction be ab ,where gcd(a,b) = 1. this will terminate <=> there exists some integer k>0 such that 10^kab is an integer. which means b|10^k. bcz gcd(a,b) =1. We get Rahuls answer