Proof of Mean Value Theorem on an Open Interval

By An Open Interval is an Interval All of Whose Points Are Interior the set $(p,q)$ is an interval all of whose points are interior points of it. Claim numbers refer to Elementary Order Arithmetic in an Ordered Field.

Step 1 (the slope). From $a<b$, claim 1 gives $0<b-a$, so $b-a\ne0$ and $(b-a)^{-1}$ exists. Put

and let $\ell:(p,q)\to\mathbb{R}$ be given by $\ell(s)=\lambda\,(s-a)$.

Step 2 ($\ell$ is differentiable with constant derivative $\lambda$). Let $c\in(p,q)$ and let $k\in\mathbb{R}$ satisfy $k\ne0$ and $c+k\in(p,q)$. In the field $\mathbb{R}$,

so the difference quotient equals $\lambda\,k\,k^{-1}=\lambda$. Hence for every $\varepsilon$ with $0<\varepsilon$ the number $\delta=1$, which satisfies $0<\delta$ by claim 6, witnesses the defining condition of differentiability for the candidate value $\lambda$, since the quantity to be estimated is $|\lambda-\lambda|=|0|=0$, and $0<\varepsilon$. So $\ell$ is differentiable at every point $c$ of $(p,q)$ with $\ell'(c)=\lambda$.

Step 3 (Rolle applied to $\varphi=g-\ell$). By Derivative of a Sum and of a Difference, the function $\varphi=g-\ell$ on $(p,q)$ is differentiable at every point $c$ of $(p,q)$, with

Moreover $\varphi(a)=g(a)-\lambda\,(a-a)=g(a)$, and

so $\varphi(a)=\varphi(b)$.

By Rolle's Theorem on an Open Interval applied to $\varphi$ on $(p,q)$ with the points $a<b$, there exists $c\in(a,b)$ with $\varphi'(c)=0$, that is $g'(c)=\lambda$.

Step 4 (conclusion). Multiplying $g'(c)=\lambda$ by $b-a$ gives

which is the assertion.