I'm working (solo) through Thomas Judson's Abstract Algebra and I'm looking at Example 1.22. I'd like to know whether my proof is complete (it differs from his):
Problem
Suppose $f$ and $g$ are differentiable functions on $\mathbf{R}$. Define $\sim$ as $f(x) \sim g(x)$ if $f'(x) = g'(x)$. Prove that $\sim$ satisfies the properties of an equivalence relation.
My Proof
Reflexivity
Need to prove $f(x) \sim f(x)$:
$f'(x) = f'(x)$, so $f(x) \sim f(x)$
Symmetry
Need to prove $f(x) \sim g(x) \Rightarrow g(x) \sim f(x)$):
If $f(x) \sim g(x)$, then $f'(x) = g'(x)$
Then $g'(x) = f'(x)$ and therefore $g(x) \sim f(x)$
Transitivity
Need to prove $f(x) \sim g(x) \space and \space g(x) \sim h(x) \Rightarrow f(x) \sim h(x)$
If $f(x) \sim g(x)$, then $f'(x) = g'(x)$
If $g(x) \sim h(x)$, then $g'(x) = h'(x)$
$f'(x) = g'(x) = h'(x)$, so $f'(x) = h'(x)$, so $f(x) \sim h(x)$
Are there any problems with this proof? In the book (viewable at the link above) Judson says "it is clear that $\sim$ is both reflexive and symmetric and he gives a different, more difficult to understand (in my opinion) proof of transitivity. It makes me think my proof might be missing something...