If 9/10 < a/b < 10/11, where a and b are positive integers, then find the smallest possible value of b.
put the fractions in a common denominator (110)
99/110 < a/b < 100/110
does not help much, huh? There are no integers between 99 and 100 sooooo: Let's double everything (the next common denominator)
198/220 < a/b < 200/220 Now we might guess b to be least at 220 (and a to be 199)
Doing some bashing we get the answer a/b as 19/21 which leaves the answer of b as 21.
This is how Gor.. came up with this answer.
9/10 < a/b < 10/11
18/20 < a/b < 20/22
There is a fair bet that 19/21 will be between the 2. (Maybe it is certainty, but I wouldn't want to say that without proving it)
Check by using decimals
9/10 = 0.9
10/11 = 0.909090....
19/21 = 0.90476
And it is between the 2 as predicted.
The smallest value of b is 21