Let $K$ be the ring of all real functions and let $f \in K$. Suppose that f is not a zero-divisor. Prove that $f \in U(K)$.

Question

Let K be the ring of all real functions and let $f \in K$. Suppose that $f$ is not a zero-divisor. Prove that $f \in U(K)$.

Answer 1

It is enough to show that $E=\{x:f(x)=0\}$ is empty. Suppose $E$ is not empty $f(x)=0$, define $g(x)=1, g(y)=0, y\neq x, fg=0$. Contradiction $h(x)={1\over{f(x)}}$ is the inverse of $f$.