There is no unique pair of numbers that satisfies these requirements.
Suppose a and b is such a pair, and sqrt(105) = x
then you want a < x < b
But a < (a+x)/2 < x < (b+x)/2 < b
So that (a+x)/2 and (b+x)/2 are a closer pair.
and you can then find a closer pair still - ad infinitum.
The question can be answered (sort of) if it asked about "integers" rather than "numbers".
100 < 105 < 121
Taking square roots, this equation implies that
10 < sqrt(105) < 11
so the answer could be 10 and 11.
But (and this is the reason for the "sort of") the above equation also implies that
-11 < sqrt(105) < -10
giving -11 and -10 as a pair of consecutive integers.
So, an unambiguous answer is possible only if the question specifies positive integers.
35.9999999999 and 36.0000000001
5 and 6
114 and 116
4 and 5 because the square root of 18 is about 4.242640687
-6 and -5 or 5 and 6
35.9999999999 and 36.0000000001
5 and 6
114 and 116
They are; 8 and 9 but the square root of 68 is about 8.246211251
4 and 5 because the square root of 18 is about 4.242640687
The square root of 103 is roughly 10.14889157, so it is between root 100 (10) and root 121 (11).
every other number in existance wow you are stupid
-6 and -5 or 5 and 6
9 and 10 9 x9 =81 10 x 10 = 100
17 and 19
53 and 55.
A square number will have one factor pair that consists of the same number (the square root). In the list of factors, that number will be written once.