The equation $\mid x \mid+\sqrt{\lfloor x \rfloor+\sqrt{1+\{x\}}} = 1$ has been analyzed with the conclusion that $x = 0$ is a valid solution. The discussion establishes that for values of $x$ in the range of -1 to 0, the absolute value simplifies to $-x$, the floor function $\lfloor x \rfloor$ equals -1, and the fractional part $\{x\}$ is expressed as $x + 1$. This leads to two additional solutions, one straightforward and one more complex, confirming the behavior of the floor function in this context.
PREREQUISITESMathematicians, students studying real analysis, and anyone interested in solving complex equations involving floor and fractional functions.