= Solution
{c}
Start with $[a_0,b_0]=[a,b]$. At stage $n$, form the <midpoint> $c_n=(a_n+b_n)/2$. If $f(c_n)=0$, stop; for a <sequence> defined for all later indices, repeat $c_n$ forever. Otherwise retain the half-<interval> with opposite endpoint signs: set $[a_{n+1},b_{n+1}]=[a_n,c_n]$ if $f(a_n)f(c_n)<0$, and $[a_{n+1},b_{n+1}]=[c_n,b_n]$ otherwise. This is the <bisection method>.
If it never stops, the <closed intervals> are nested, retain $f(a_n)f(b_n)<0$, and have lengths
$$
b_n-a_n=\frac{b-a}{2^n}\longrightarrow0.
$$
By the <nested interval theorem>, there is a unique real $\xi$ in their intersection. In particular, $a_n\to\xi$, $b_n\to\xi$, and the <sequence> of <midpoints> satisfies
$$
|c_n-\xi|\leq\frac{b-a}{2^{n+1}}\longrightarrow0.
$$
Since $f$ is <continuous>, $f(a_n)f(b_n)\to f(\xi)^2$. All these products are negative, so their <limit of a sequence> is nonpositive. Therefore $f(\xi)^2\leq0$, forcing $f(\xi)=0$. We have proved that the <sequence> is a <convergent sequence> whose <limit of a sequence> is a <zero of a function>:
$$
\boxed{c_n\longrightarrow\xi,\qquad f(\xi)=0,\qquad |c_n-\xi|\leq\frac{b-a}{2^{n+1}}.}
$$
The <error bound> applies at every stage before exact termination; after termination the error is zero. The <intermediate value theorem> also guarantees a <zero of a function> in each retained <interval>, but the nested-<interval> argument identifies the particular one to which the <sequence> of <midpoints> converges.
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