Use the Nested Interval Theorem to find a point x where f(x)=0 Suppose f is continuous on [a,b], and f(a)<0<
f(b). Then either f((a+b)/2)=0 or f has different signs at the end points of the interval [a,(a+b)/2] or f has different signs at the end points of [(a+b)/2,b]..Either f is 0 at the mid point or f changes sign on one of the two intervals. Let I2 be that interval. Continue in this way, to define In for each n. Use the Nested Theorem to find a point where f(x)=0
I generally understand what the question is saying. And my understanding is that it is pretty analogous to the dichotomy in algorithm. But since it is a maths problem.. I feel I have trouble to prove this problem using mathematic language... 
 A: b) Without loss of generality, suppose we have a continuous function $%
f:[a,b]\rightarrow 
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\mathbb{R}
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$ such that $f(a)<0<f(b)$ Let $I_{0}=[a,b]$. Let $m_{2}$ denote the midpoint of $I_{1}.$
Now either $f(m_{1})<0,$ $f(m_{1})>0,$ or $f(m_{1})=0$ . If $f(m_{1})=0$ we
are done. If not, we let $x_{1}=m_{1}$ $y_{1}=b$ if $f(m_{1})<0$ or $x_{1}=a,
$ $y_{1}=m_{1}$ if $f(m_{1})>0.$ We then define $I_{1}=[x_{1},y_{1}].$ Now
suppose that $I_{k}=[x_{k},y_{k}]$ has been defined. Let $m_{k+1}$ be the
midpoint of $I_{k}.$ Then again we have either  $f(m_{k+1})<0,$ $%
f(m_{k+1})>0,$ or $f(m_{k+1})=0.$ If $f(m_{k+1})=0$ we are done. If not we
define $x_{k+1}=m_{k+1}$ $y_{k+1}=y_{k}$ if $f(m_{k+1})<0$ or $x_{k+1}=x_{k},
$ $y_{k+1}=m_{k+1}$ if $f(m_{k+1})>0.$ Then let $I_{k+1}=[x_{k+1},y_{k+1}].$
This recusively defines the family of nested intervals $\{I_{n}\}$. By the
Nested interval theorem,this family has a non empty intersection. Let $c$ belong to the intersection. Now by construction, $f(x_{n})\rightarrow 0$ and $f(y_{n})\rightarrow 0$ as well as $(x_{n})\rightarrow c$ and $%
(y_{n})\rightarrow c.$ Since $f$ is continuous, it follows that $f(c)=0.$ 
