The number that exceeds its square by the greatest amount is:

Let the number be $x$. The amount by which it exceeds its square is $y=x-x^2$.To find the maximum, we find the derivative and set it to zero:$\frac{dy}{dx}=1-2x$$1-2x=0⟹x=\frac{1}{2}$ Check second derivative: $\frac{d^{2}y}{d{x}^2}=-2<0$ (so it is a maximum).The number is $\frac{1}{2}$.