   # Which number is greater - (99^999 + 1)/(99^1000 + 1) OR (99^1000 +1)/(99^1001 + 1)

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Which number is greater - (99^999 + 1)/(99^1000 + 1) OR (99^1000 +1)/(99^1001 + 1) posted Jun 9, 2016

First number is greater, here is the explanation:

for - (99^999 + 1)/(99^1000 + 1)
consider it as (99^999 + 1)/99^1000 , 99^1000 is big enough that adding 1 does not make a lot of difference
(99^999 + 1)/99^1000= (99^99/99^1000)+(1/99^1000)= (1/99)+1/99^1000 ......(1)

for-(99^1000 +1)/(99^1001 + 1)
consider it as (99^1000 +1)/99^1001
(99^1000 +1)/99^1001=(1/99)+(1/99^1001) ..........(2)

on comparing 1 and 2 we get (99^999 + 1)/(99^1000 + 1) as bigger one. answer Jun 10, 2016

99 ^ 999 is too big of a number, close to infinite. Adding one 1 to both numerator or denominator does not make any difference, therefore, we can write
99 ^999 / 99 ^ 1000 = 1 / 99 and
99 ^1000 / 99 ^ 1001 = 1 / 99
Then , both numbers are the same answer Jun 14, 2016 by anonymous
–1 vote

The second number is greater than the first one. answer Jun 9, 2016
–1 vote

The second number is greater than the first one coz the first number is negative while the second one is positive.

And positive number is always greater than negative number. answer Jun 10, 2016
that is not a minus sign

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