Denote c = sup S and suppose f ( c ) < 0 or f ( c ) > 0 . In both cases you'll get a contradiction. So, necessarily f ( c ) = 0 .
Letf be a continuous function on the closed interval [0; 1], and suppose that f(0) < 0
and that f(1) > 0. By considering the set S= {x E [0,1] :f(x)_<0) prove that there exists
a x' such that f(x')=0
S is non empty as x=0 is a member and bounded above by 1, so by the completness axiom Sup(S) exists. Sup(S)_> x E S and Sup(s) _< 1. Not sure where to go from here. Thanks in advance.
ok I'm in a better maths mood today. I got frustated by this yesterday.
assume F(c)<0 which implies the theorem.
c<_1 so F(c)<0 implies c is a member of the set
So for all m E [0,1] with m > c implies that F(m) > 0 but get stuck here. Its sort of intuitive that it can't be because of the density of reals but I don't know how to prove it convincingly.