Let's let f (a) < 0 <; F(b). Let c = (a+b)/2. If f (c) = 0, the conclusion holds.

If f(c)>0, then [a 1, b 1] = [a, c]; If f (c)

Remember c1= (a1+b1)/2 again. If F (C 1) = 0, the conclusion holds;

If f (c1) >; 0, then [A2, B2] = [A 1, c1]; If f (c 1)

Go on, or at a certain step, f(ck)=f[(ak+bk)/2]=0, and the conclusion holds.

Or this process can go on indefinitely, so as to get an interval nested sequence {[an, bn]}, which satisfies:

(1), [a 1, b 1] contains [a2, b2] contains [a3, b3] contains. ...

(2), bn-an = (b-a)/2 n tends to 0, when n tends to infinity;

(3)，f(an)& lt； 0 & ltf(bn)，n= 1，2，3，...。

According to the closed interval set theorem, there exists C in all intervals, that is, an.

And both an and bn tend to C. The continuity of f(x) in c is

f(c)= lim f(an)& lt； =0,

f(c)=lim f(bn)>=0,

So f (c) = 0. Obviously due to f (a)

Complete the certificate.