*Originally posted by mtthw*

**But that's not a contradiction, as k*sqrt(n)-k*floor(sqrt(n)) is not a multiple of sqrt(n).
**

You haven't used any property of sqrt(n) in your proof.

Why does k*sqrt(n)-k*floor(sqrt(n)) need to be a multiple of sqrt(n)?

Of course it is a contradiction:

k*sqrt(n)-k*floor(sqrt(n)) is a positive integer, since k*sqrt(n) and k*floor(sqrt(n)) are integers and k*floor(sqrt(n)) < k*sqrt(n).

It is smaller than k, since sqrt(n)-floor(sqrt(n)) < 1.

When multiplied by sqrt(n), the result is an integer:

[k*sqrt(n)-k*floor(sqrt(n))]sqrt(n)=kn-(k*sqrt(n))(floor(sqrt(n)))=difference between two integers=integer.

But k was supposed to be the SMALLEST positive integer that gives an integer when multiplied by sqrt(n).

Contradiction...

Now I ask if this proof can be generalized to roots other than the square root.