Let be a divisor of . Then is also a divisor of ; if both are less than we have which is clearly impossible.
Here is the problem. Let n be a positive integer, not = 1, and not prime, prove n has a divisor d such that 1<d<sqrt(n)
The way i began attacking this was by saying n is either even or odd. If n is even and not prime then n >= 4 and 2|any even integer and 1<2<sqrt(4). They problem im having is on the odd case there is no number like 2 that will divide all odd numbers so i dont know how to show a divisor would fall in this interval
As Shanks said, this is wrong, the interval should be , otherwise perfect squares ( , , , ...) fall short of the interval. I'll use the interval I mentioned.
Let us conjecture that is not prime, greater than one, and has not got any factor (divisor, whatever) such as .
Then, since is not prime, it is composite and clearly has two divisors. Denote its two divisors and , that is, ( ).
Since and are not in , we must conclude that and . Thus we must conclude that , that is, . However we know that , so there is a contradiction.
composite has at least one divisor in . Thus, if is any divisor of , .
EDIT : actually this proof is complete, because and needn't be prime, and any composite number can be represented as the product of two composite numbers. Correct me if I am wrong.