Prove $(cd^{-1})^{-1} = c^{-1}d$ I'm working my way through Michael Spivak's Calculus.
There's something I don't quite get about proving in general:
I have to prove that:
1) $(cd^{-1})^{-1} = c^{-1}d$ 
2) $(cd^{-1})^{-1} \cdot cd^{-1} = c^{-1}d \cdot cd^{-1}$ 
3) $1 = c^{-1} c d^{-1} d $ 
4) $1 = 1$ 
2: I multiply by $cd^{-1}$ on both sides. 
3: The left side will equal 1, due to the extension of multiplicative inverses ($a^{-1}a = 1$). The commutative law of multiplication ($ab = ba$) allows me to rearrange the factors on the right side, such that each factor will be multiplied by its inverse. The right then, will equal 1.
4: This is where I get into trouble - I have reduced the expression and found, that $1=1$. How do I interpret and formulate this reduction, so it makes a conclusion to my proof (provided that the proof itself is correct) ?
 A: You should write it the other way, like this:

Clearly, $1=1$. Therefore $c^{-1}cd^{-1}d=1$. Also $aa^{-1}=1$, for any $a$, so $(cd^{-1})^{-1}cd^{-1}=1=c^{-1}cd^{-1}d$. By communativity, $(cd^{-1})^{-1}cd^{-1}=1=c^{-1}dcd^{-1}$. Dividing by $cd^{-1}$, we get $(cd^{-1})^{-1}=c^{-1}d$.

Start by what you know, and then write what that implies. Of course, the proof by Bhaskar is better than this one, but this is a valid proof, provided that we know what we mean by dividing, but that is clear for numbers in $\mathbb R_{\neq 0}$. 
A: To show $(cd^{-1})^{-1} = c^{-1}d$, all you gotta do is check $$cd^{-1} \times c^{-1}d=1$$ which is clear as $c$ and $d$ commute and thus $cd^{-1}.d.c^{-1}=c.1.c^{-1}=1$

Also check $c^{-1}d \times cd^{-1} =1$ which is clear as $c$ and $d$ are numbers
A: Your proof is almost correct, but if want to be totally rigorous, you have to pay attention to a few issues. Don't worry, I am really overly picky to explain every single point but nobody will ever ask you so much details.
Issue 1. You should define carefully the setting of your question (see the first comment). Actually your answer "$c$ and $d$ are numbers" is not precise enough and should be "$c$ and $d$ are nonzero numbers" or even better "$c$ and $d$ are nonzero real numbers".
Issue 2. In your proof of $(2)$, you need to say that $c$ and $d$ commute. By the way, this would be false for matrices, for instance.
Issue 3. Your proof consists to prove that $(1) \iff (2) \iff (3) \iff (4)$.
But actually, you rather prove the sequence $(1) \implies (2) \implies (3) \implies (4)$. I agree that the other direction is easy, but for $(2) \implies (1)$ you have to simplify by $cd^{-1}$ and hence you should make clear that $cd^{-1}$ is nonzero.
Issue 4 (your original question). When you arrive to (4), you know that (4) is true and if you proved that (1) is equivalent to (4), (1) is true also.
Last point. A better answer would be $(cd^{-1})^{-1} = dc^{-1}$. Not only because it also works in a noncommutative setting (in any group, to be precise) but also because it is shorter to prove: $(cd^{-1})(dc^{-1}) = c(d^{-1}d)c^{-1} = cc^{-1} = 1$.
