In my bachelor courses, we have defined the space $C^1(\mathbb{R})$ as $$C^1(\mathbb{R}) = \{f: \mathbb{R} \rightarrow \mathbb{R}~|~f \text{ is differentiable}~\land~f' \text{ is continuous}\}.$$ However, I really do not understand why it is required that $f'$ must be continuous. Indeed, if $f$ is differentiable at $a \in \mathbb{R}$, then the limit $$\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$$ does exist. Isn't it equivalent to the fact that our function $f'$ is continuous in $a$ ? I asked my teacher for an example of a function that would be differentiable at a point but whose derivative would not be continuous at this point. He gave me the function $$f(x) = |x| \quad \text{at zero.}$$But I disagree with him, it's not that the derivative is not continuous in zero, it doesn't even exist because these two limits don't agree $$- 1 = \lim_{x \rightarrow 0^-} \frac{f(x) - f(0)}{x - 0} \neq \lim_{x \rightarrow 0^+} \frac{f(x) - f(0)}{x - 0} = 1.$$ Therefore, $f$ is not differentiable at $0$. Could someone explain to me the definition of $C^1(\mathbb{R})$? Is there an example of a differentiable function whose derivative is not continuous? Thank you very much!
Suppose $f$ is continuous on $[a,b]$ and differentiable on (a,b). Does it follow that $f'$ is continuous on $(a,b)$?
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All of my degrees (Bachelor's, Master's, PhD) are in Statistics, and I am just wondering if it would be ethical to just list my Bachelor's and Master's degree on my resume, and indicate my experience as a RA while I was a PhD student as my "work experience"? (Since this Statistical Programmer position only requires Master's degree plus some years of experience)  I am asking this question because I do not want to do anything unethical/cause me troubles later on. Proving $(\forall a\in\mathbb{Z^+})(m\in\mathbb{Z^+}\to a^m\equiv a^{m-\phi(m)}\pmod{m})$ Problem: $(\forall a\in\mathbb{Z^+})(m\in\mathbb{Z^+}\to a^m\equiv a^{m-\phi(m)}\pmod{m})$  My work: Start by letting $m=p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}$. If $(a,p_i)=1$ for some integer $i$, then we have that  $$ a^{\phi(p_i^{a_i})}\equiv 1\pmod{p_i^{a_i}} $$ by Euler's theorem. Since $\phi(p_i^{a_i}) \mid \phi(m)$, we have that $p_i^{a_i}\mid (a^{\phi(m)}-1)$ if $(a,p_i)=1$.   Beyond this, I am stuck at the moment. 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