Induction for statements with more than one variable. I'm going through the first chapters of Tao's Analysis text and I'm not entirely sure about one thing, namely why we're allowed to 'fix' variables when inductively proving statements pertaining to more than one variable. This is not explained in the text. For instance, the very first exercise in the book asks to prove associativity of addition of natural numbers: $(a+b)+c=a+(b+c), \ a,b,c \in \mathbb{N}$. It says to fix two of them and induct on the third. But how do I know such a proof is really correct? The axiom of induction only talks about properties of the form $P(n)$. And here we have kind of a family of properties indexed by the remaining variables: $P_{a,b}(c)$. Is this the right way to think of it?
I found this linked in a similar question: http://www.mathblog.dk/proof-method-multidimensional-induction/ where it says it's necessary to do induction on each variable. Is this tacitly omitted in Tao's text because in most of these examples the induction looks identical on both variables?
 A: Yes, it is tacitly omitted in Tao's text because the induction on one variable replicates what induction on the others will yield. So inducting on one variable, say $c$, and letting the other two remain "fixed" at $a, b$ suffices to cover what induction on each and every one of them will yield. 
For the base case, we have to prove that for all $a, b$, it holds that $P(a, b, 0)$, which may entail an inductive proof.
Then, what you call proving $P_{a, b}(c)$ by induction on $c$, with $a, b$ held "fixed" does in fact prove for all $a, b, c$, $P(a, b, c)$ holds.
ADDED: Please also read the answer to this post about which variable to induct on, when given two, and whether or not induction is needed on each. (The posted question also originates from the questioner's study of Tao's Analysis.
A: Fixing $a,b$ and inducting on $c$ means proving what you called $P_{a,b}(c)$ by induction on $c$, for all $a$ and $b$ simultaneously.  So you're actually doing infinitely many induction proofs at once, but, since they are all exactly the same except for having different values of $a$ and $b$, you can write them all as a single proof, by writing variables $a,b$ instead of specific values.
A: There's nothing wrong with this approach as long as you state the proposition correctly. In your case, $P(c)$ is the proposition that $\forall a, b \in \mathbb{N}, (a+b)+c=a+(b+c)$. This is a perfectly valid proposition; there are no "free variables" because of the $\forall$ quantifier.  (You may need to use induction on $a$ or $b$ for the base case, but you might not need to.)
A: To "fix" a number is to make an argument involving that number without ever specifying which it is. The result is that your final argument will work regardless of any particular number you apply it to.
Logically, when we fix a number, we create a universal quantification of the rest of the theorem. To give a simple example, "Fix $a$. Then $a + 0 = a$ by computation". This happens to be a (trivial) proof of the proposition $\forall a. a + 0 = a$. 
A subtler thing that "fix" does for you is to bring a variable 'into existence'. When you say "fix an integer $a$", now you know that there is a variable $a$ which is an integer.
(Synonyms for "fix" include "let" or "assume". But they all do the same thing).
So we want to prove associativity of addition: $\forall a b c \in \mathbb{N}. (a + b) + c = a + (b + c)$. This starts out with three universal quantifiers. Two of them will come from "fixing" $a$ and $b$. The last will come from induction.
The proposition $P$ which we induct over will actually depend on $a$ and $b$. We write $P(c) = (a + b) + c = a + (b + c)$. Note that $a$ and $b$ are constants in this proposition, and this is a level move because they were fixed outside the definition of $P$. (This is, in fact, the definition of a constant).
Our base case and inductive cases are then:
$P(0) = (a + b) + 0 = a + (b + 0)$
Given $(a + b) + c = a + (b + c)$, we have $(a + b) + S(c) = a + (b + S(c))$
And these can both be proven using the peano axioms.
