How is induction on $p-q > 0$ used correctly here? I'm self learning Rotman's Algebraic topology and I've come across this theorem and proof.  I have two questions:

$(1.)$  How did the author arrive at $H_n(X^p, X^{q+1}) = 0$ if $q+1 \ge n$?
$(2.)$  How is induction used properly here?  Normally induction is done on a base case, $p=q$ in this example, and then assume it's true for $p-q = k$ and show that it's true for $p-q = k+1$.  But I don't see anywhere an expression of the form $p-q = k+1$ being shown to be true.  Can someone explain how induction is used here to show where $k+1$ is true?



You have
$$(p',q') = (p,q+1)$$
where 
$$p-q >0,\;\;\text{and either}\;\;n > p\;\;\text{or}\;\;q \ge n$$
To apply the inductive hypothesis, you need to have
\begin{align*}
&{\small{\bullet}}\;\;p'-q' \ge 0\\[4pt]
&{\small{\bullet}}\;\;p'-q' < p-q\\[4pt]
&{\small{\bullet}}\;\;\text{either}\;\;n > p'\;\;\text{or}\;\;q' \ge n\\[4pt]
\end{align*}
Check the conditions one at a time . . .
\begin{align*}
&p-q > 0 \implies p-(q+1) \ge 0 \implies p'-q' \ge 0\\[4pt]
&p'-q'=p-(q+1) < p-q\\[4pt]
\end{align*}
To show $n > p'\;\;\text{or}\;\;q' \ge n$, consider two cases . . .

$\qquad$Case $(1)\,\!\!:\;n > p$.$\;\,$Then $n > p \implies n > p'$.

$\qquad$Case $(2)\,\!\!:\;q \ge n$.$\;\,$Then $q \ge n \implies q+1 > n \implies q' > n \implies q' \ge n$.

Thus, in each of the two cases, the requirements on $(p',q')$ are met.

Therefore the inductive hypothesis can be applied.
