Calculus Spivak. Chapter 1. Question 1. (i) or are there many ways of skinning a cat I'm taking on Spivak's Calculus a little later on in life via self-study as i'm looking to improve my CS abilities and have always been interested in Maths but unfortunately didn't have the chance when I was younger. I recently finished a course in College Algebra and have a pretty good grounding with Trig already.
I understand that Spivak is a challenging book, but i'm having trouble getting my head around what an acceptable answer might look like.
For example:
Prove: if $ ax = a $ for some number $a \ne 0$, then $x = 1$
My "proof" is:
$$
ax + (-a) = a + (-a) = 0
$$
$$
a(x+(-1)) = 0
$$
$$
x+(-1) = 0
$$
$$
x+((-1) + 1) = 0 + 1
$$
$$
x=1
$$
Whereas Spivak's is:
$$
1 = a^{-1}a=a^{-1}(ax)=(a^{-1}a)x=1•x=x 
$$
Much neater, and far more elegant. But is mine wrong? Why is it wrong?
My hunch is that I need to be much more judicious in using only P1-P12s in my proofs and that each stage of the proof must use one of these dozen properties, or a property (theorem?) derived from them, and that's how I need to approach it going forward, but I want to be sure as I continue.
How did Spivak approached the problem? Did he recognise it as analogous to P6 (existence of a multiplicative entity) and work backwards from there? Would be great to get some commentary on the steps taken.
I'm having the difficulty making the jump to proofs it would seem. I have a copy of Velleman's How to prove it, would it be worth going through this first?
 A: I do not have the book, but yours appears to be correct if you have that there are no zero divisors other than $0$. Otherwise it does not follow from $a\neq 0$ and $ab=0$ that $b=0$.
A: Problems
First the problems stem more from spivak's version than yours. The reason why there might be problems is that there are different types of algebraic structures in mathematics. Which in itself is a whole field (abstract algebra). The comments you are getting seem to stem from the fact that you don't reference the fact that you are using a field. Which by definition has no elements $a\neq 0$ such that $ab=0$ for some $b\neq 0$. Having a field allows you to do all kinds of neat little tricks with multiplication without worrying about terms just dropping off.
Too see why not having a field might be a problem look at $Z_6=\{0,1,2,3,4,5,6\}$ where you are adding and multiplying via modulus (i.e. $a+b=(a+b)\mod6$ and $ab=(ab)\mod6$). In this example $2*3=6=0\mod 6$ so in spivak's example $a=2$ there would be no inverse ($2x\neq 1$ for any $x\in\{0,1,2,3,4,5\}$).
Why the differences?
Your proof works. It just seems that given the extra structure (field) Spivak's feels the need to shorten the proof a bit by using the extra structure (i.e. everything that isn't 0 has a multiplicative inverse). Differences in proof's aren't bad, sometimes they hint at new an interesting discoveries sometimes they might show something can be proven with less assumptions. So your doing pretty good.
