## Thomas' Calculus 13th Edition

For a product, we have at least two factors (a product of one term makes no sense), so we prove that the formula is true for any $n\geq 2$ $1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ Check that the formula holds for $n=2$. We are given by the problem text that $f(x_{1}\cdot x_{2})=f(x_{1})+f(x_{2})$ $\qquad$ Yes, it is true when $n=2.$ $2. \ (\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{p})$ Prove that IF the formula holds for any positive integer $n=k$, THEN it also holds for the next integer, $n=k+1$. Let us assume that for $n=k$, $f(x_{1}\cdot x_{2}\cdots\cdot x_{k})=f(x_{1})+f(x_{2})+\ldots+f(x_{k})$ (we call this the step hypothesis) For $n=k+1$, the LHS equals $f(x_{1}\cdot x_{2}\cdots\cdot x_{k}\cdot x_{k+1})$, which we rewrite as $LHS= f[(x_{1}\cdot x_{2}\cdots\cdot x_{k})\cdot(x_{k+1})]$ $f$ of a product of TWO factors, for which we use the given formula $=f(x_{1}\cdot x_{2}\cdots\cdot x_{k})+f(x_{k+1})$ the first term is, by the hypothesis, equal to $f(x_{1})+f(x_{2})+\ldots+f(x_{k})$ $=f(x_{1})+f(x_{2})+\ldots+f(x_{k})+f(x_{k+1})=RHS$ for $n=k+1$ which makes the formula true for $n=k+1.$ By mathematical induction, the formula is true for any integer $n\geq 2.$