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Recently I was given the following problem: find all positive real numbers x such that
4*flor(x)=3x{x}
In this expression flor(x) denotes the floor function: flor(x)=m∈Z : m≤x<m+1, and {x} denotes the fractional part of x: {x}=x-flor(x).
Some values of x can be eliminated right away. When x is integral, {x}=0, making the right side of the equation 0; if x<1, flor(x)=0, making the left side of the equation 0. Thus we only need consider n<x<n+1, n any positive integer.
For any such n, flor(x)=n and {x}=x-n. Our equation then becomes
4n=3x(x-n)
This is trivial to put in terms of n:
4n=3x^2-3nx
3x^2-3nx-4n=0
x=(3n+sqrt(9n^2+48n))/6
Most students would stop at this point. We've found an expression for x in terms of n, and that's all we were looking for, right? Not quite; these kinds of problems reveal their subtleties in their boundary conditions. It is established that n<x<n+1, so (3n+sqrt(9n^2+48n))/6 must be between n and n+1. Is this boundary condition always satisfied? Is it ever satisfied?
The lower bound is easy to confirm:
(3n+sqrt(9n^2+48n))/6>(3n+sqrt(9n^2))/6=(3n+3n)/6=n
What about the upper bound? We want to know if our expression is ever greater than or equal to n+1:
(3n+sqrt(9n^2+48n))/6≥n+1
sqrt(9n^2+48n)≥3n+6
9n^2+48n≥9n^2+36n+36
n≥3
For any n≥3, our answer exceeds n+1! It turns out the expression doesn't exceed the boundary by very much (the value of x-n asymptotically approaches 4/3), but a deviation greater than or equal to 1 will not work. Thus n=1,2 and our corresponding values of x are
x=(3+sqrt(57))/6, (3+2sqrt(33))/3
When I was a student, I frequently gave into the temptation to not check the validity of a solution. I would simply derive an expression and submit it as my answer. When solving real-world problems, however, ensuring the sanity of an answer is often critical. It's a skill that deserves more emphasis in education than it usually gets.
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[Last updated: 2021-12-27]