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In my Math book (Elementary Analysis by Kenneth Ross), it says to prove Theorem 3.1 (ii), that $a * 0 = 0$ for all a. However, the solution in the book is:

A3: $a + 0 = a$ for all a

DL: $a(b + c) = ab + ac$ for all a,b,c.

(i) $a + c = b + c$ implies $a = b$

We use A3 and DL to obtain $a * 0 = a * (0 + 0) = a * 0 + a * 0$, so $0 + a * 0 = a * 0 + a * 0$. By (i) we conclude $0 = a * 0$.

I understand how to prove (i). However, what confuses me is the starting point for this next proof. How do you know and where do you start in a mathematical proof? I've seen other proofs proving this theorem on here at StackExchange, but I want to understand how my book is proving it. One way I've seen that makes sense to me is starting with the distributive law:

$a(b + c) = ab + ac$

Let b and c = 0

$$ \begin{align*} a(0 + 0) &= a * 0 + a * 0 \\ a * 0 &= a * 0 + a * 0 \\ (a * 0) + (-a * 0) &= (a * 0 + a * 0) + (-a * 0) \\ 0 &= a * 0 + (a * 0 + (-a * 0)) \\ 0 &= a * 0 + 0 \\ 0 &= a * 0 \\ \end{align*} $$

This makes sense. However, in the above expression provided by the book, they start from $a * 0$ and get to $a * 0 + a * 0$. This makes sense. But, I don't understand where they're getting $0 + a * 0$ from. Also, is there a general guideline or tip to help one know where to start with a proof? Do we start with the equation/expression itself that we want to prove and manipulate it with the known axioms? Or do we start with an axiom and try to work our way back to the proposition/theorem?

EDIT: Is there any recommended books to help me learn proofs as a total beginner? It feels like every math book in college is written for people who already understand how proofs work and such.

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  • $\begingroup$ well $0$ plus anything is itself right? $0$ is the additive identity, so that step seems logical $\endgroup$ – J. Marx-Kuo Sep 13 '16 at 22:24
  • $\begingroup$ I guess what I'm confused about is that since we start with the expression $a * 0$ and manipulate THAT particular expression down to $a * 0 + a * 0$, where does the $0 + a * 0$ come from on the LEFT side? After the sentence "So $0 + a * 0 = a * 0 + a * 0$, isn't the expression we just created the RIGHT side? Where did the LEFT side randomly come from? $\endgroup$ – user83903 Sep 13 '16 at 22:28
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    $\begingroup$ Ok, so they definitely have $a * 0 = a* 0 + a * 0$, and you agree with this. But isn't $a*0 = a*0 + 0 = 0 + a*0$ also true? Then you equate these two sets of equations to get what they have. This is how the $0 + a*0$ appears on the left side. What part of this are you confused about? $\endgroup$ – J. Marx-Kuo Sep 13 '16 at 22:33
  • $\begingroup$ I'm confused at the part where they write $a * 0 = a(0 + 0) = a * 0 + a * 0$ Is this RS = RS = RS? Or is this LS = RS = RS? Because how did they get $0 + a * 0 = a * 0 + a * 0$ $\endgroup$ – user83903 Sep 13 '16 at 23:32
  • $\begingroup$ so $0$ is special in that it is the identity element, which means that for any $a$ (which could be equal to $0$ but does not have to be), we have that $$a = a + 0 = 0 + a$$ This is axiomatic in the definition of $0$ and that addition is commutative (which I assume is another axiom implicitly assumed in your textbook). Therefore $$a * 0 = a * (0) = a * (0 + 0)$$ follows from the above identity property of $0$, in that we can add it to anything, even the $0$ in the parentheses. The fact that $$a* (0 + 0) = a*0 + a*0$$ is the distributive law. $\endgroup$ – J. Marx-Kuo Sep 13 '16 at 23:39

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