15. Given that $a$ and $b$ are real numbers, and $ab=1, a \neq 1$, let $M=\frac{a}{a+1}+\frac{b}{b+1}, N=\frac{1}{a+1}+\frac{1}{b+1}$. Then the value of $M-N$ is $\qquad$.
15.0.

Given $a b=1, a \neq 1$, so,
$$
\begin{array}{l}
M=\frac{a}{a+1}+\frac{b}{b+1}=\frac{a}{a+a b}+\frac{b}{b+a b} \\
=\frac{a}{a(1+b)}+\frac{b}{b(1+a)}=\frac{1}{a+1}+\frac{1}{b+1}=N .
\end{array}
$$

Thus, $M-N=0$.