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Problem: Let $f$ be a continuous function whose domain includes the closed interval [$a$,$b$]. If $f(a)$ $\lt$ $0$ and $f(b)$ $\gt$ $0$, then there is a number $x$ between $a$ and $b$ such that $f(x)$ = $0$

Proof:

Let $f$ be a continuous function whose domain includes the closed interval [$a$,$b$]

Let $f(a)$ $\lt$ $0$ and $f(b)$ $\gt$ $0$

Then if $S$ is any open interval containing the number $f(x)$, then there is an open interval $T$ containing the number $x$ such that if $t$ $\in$ $T$, and $t$ is in the domain of $f$, then $f(t)$ $\in$ $S$ (our class' defintion of continuous)

Then it has a left-most point $a$ and a right-most point $b$

I'm not sure where to go from here. I need to show that there is a number $x$ between $a$ and $b$ such that $f(x)$ = $0$

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    $\begingroup$ Let $c = \sup\{x \in [a,b] : f(x) < 0\}$ and see what $f(c)$ should be. $\endgroup$ – mathworker21 Dec 1 '17 at 5:25
  • $\begingroup$ This is my first theoretical math class and my teacher has not defined "sup" ever. Could you elaborate please? $\endgroup$ – Propaloo Dec 1 '17 at 5:29
  • $\begingroup$ The problem you have mentioned is famous by the name Intermediate Value Theorem and you can find ample proofs and discussions on this topic via Google and by searching on this site. You may also have a look at my blog post. Ideally this should not be given as a problem, but rather discussed in class with all the details. $\endgroup$ – Paramanand Singh Dec 1 '17 at 7:33
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According to your "class" definition of continuous, you could come up with the following solution :

Since the set $C = \{x\in [a,b] : f(x) < 0\}$ is bounded, it must have a least upper bound(roughly, a number which is greater than every element of $C$, but is the smallest number with that property). If you like I can elaborate here.

Let $c$ be that least upper bound(also called the supremum and denoted $c = \sup C$). We claim $f(c) = 0$.

What if $f(c) < 0$? Then let $S$ be the interval $\left(\frac{3f(c)}{2},\frac{f(c)}{2}\right)$. Note that $S$ contains strictly negative numbers. Now, we know that there is an interval $T$ around $c$ which is such that if $t \in T$ then $f(t) \in S$. Now, an interval around $c$ contains at least one point larger than $c$, say $c_0$, but then $f(c_0) < 0$ so $c_0 \in C$. This contradicts the fact that $c$ was the supremum of $C$.

Similarly, I leave you to see that $f(c) > 0$ will also lead to a contradiction. Hence, you can conclude that $f(c) = 0$. Clearly, $c \in (a,b)$ since $c < b$.

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  • $\begingroup$ Just to add, as Paramanand Singh mentioned earlier, this is a very classical result, caled Bolzano's theorem or the precursor to the intermediate value theorem, and should be discussed in class. Also, the supremum of a upper bounded set, is the smallest number with the property that it is larger than every number of that set. For example, the supremum of $(1,2)$ would be $2$, since it's larger than every number in $(1,2)$, but is also the smallest number with this property. Similarly, $\sup\{1,2,3,4\} = 4$. But the supremum of, say $\mathbb N$ does not exist. $\endgroup$ – астон вілла олоф мэллбэрг Dec 2 '17 at 3:08

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