Prove if a = ab then b =1 I am self-studying Theodore Faticoni's 'The Mathematics of Infinity: A Guide to Great Ideas' (John Wiley and Sons, 2nd edition, 2012) as an introduction to set theory. I have close to no experience writing proofs and am having trouble with an early exercise (p. 14):
"Give a highly detailed proof that if a=ab then b=1"
I'm tempted to divide both sides by 'a' and call it good, but I'm assuming this isn't "highly detailed." How do I make this rigorous?
So far no axioms/postulates have been presented (these are from the first chapter, simply called 'Logic'). Not really sure what I can assume. Existence of multiplicative inverse? Existence of multiplicative identity?
Another similar question: Give a highly detailed proof that If x^2 - 1 = 0, then x is a member of {-1,1}.
EDIT
Ok I want to see if I can do a solid (logical rather than arithmetic) proof for the second problem I mentioned in the original post.
Given the statements
P: x^2-1 = 0, and
Q: x is a member of {-1,1};
Prove: If P then Q.
We will begin with a logically equivalent statement: (not P) or Q, which can be read: either x^2 - 1 ≠ 0, or x is a member of {-1,1}.
We assume x^2 - 1 = 0 is true, hence, (not P) is false.
Thus, Q must be true by definition of the connective 'or.'
If P and Q are both true, then P implies Q.
P and Q are both true.
P implies Q.
It seems to me these two problems are not formulated well. But I think this is a sufficient solution. It surely is not 'highly detailed' in the arithmetic sense, but I am hoping that as a solution to a logical problem it is sound?
 A: You cannot prove something from nothing. Every proof is based on axioms, definitions and assumptions, and without those there is nothing one can prove. If no context is given (like here), then no proof can be given either.
Especially an exercise that asks you to prove that "if $a=ab$, then $b=1$" requires to state specifically what the symbols mean. Notably, it cannot be proved for the most elementary interpretation, where we are talking about natural (or real) numbers, since it is actually false: $0=0\cdot 0$ is a counterexample. If we let $a=0$ and $b=0$, then $a=ab$ is true, but $b=1$ is obviously false.
A: A student's rule of thumb: A sure hint to an exercise is the chapter and the section it is located.
The exercise you have mentioned is at the end of the section on tautologies of the first chapter of the book introducing elementary logic. Evidently, basic arithmetical content is assumed - not Peano Arithmetic, etc.)
For the task: Each proposition has to be dealt separately. Otherwise, it would not be a proposition of the calculus and the proof would be directly arithmetical without an appeal to logical form.
Also check for a tautology equivalent to the form of the statement in question to make the task easier.
As an example (you choose the notation you find appropriate):
An equivalent tautology is '(not P) or Q'. Thus, we have
$``\text{not } a = ab"\text{    or    }``b=1"$
Take the proposition '(not P)' by itself:
If $a \neq ab$ then either $a > ab$, which is not arithmetically possible (as in the example 1.4.3, we take $a, b\in\mathbb{N}$ as nonzero), or $a < ab$. Then, $1 < b$.
We have been given that $a = ab$. Then, Q is true: $b=1$
There is a hidden lesson in this exercise: In mathematical logic, some statements appear so obvious, even trivial; however, its proof turns out to be very hard or, the statement itself turns out to be false. Hence, the utmost rigour is required.
