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
10 & 11 Because 10^(2) = 100 11^(2) = 121 And 103 fall between 100 & 121
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.