TITLE: What do we actually prove using induction theorem?
QUESTION [8 upvotes]: Here is the picture of the page of the book, I am reading:


$$P_k: \qquad 1+3+5+\dots+(2k-1)=k^2$$
  Now we want to show that this assumption implies that $P_{k+1}$ is also a true statement:
  $$P_{k+1}: \qquad 1+3+5+\dots+(2k-1)+(2k+1)=(k+1)^2.$$
  Since we have assumed that $P_k$ is true, we can perform operations on this equation.
  Note that the left side of $P_{k+1}$ is the left side of $P_k$ plus $(2k+1)$.
  So we start by adding $(2k+1)$ to both sides of $P_k$:
  \begin{align*}
1+3+\dots+(2k-1) &= k^2  &P_k\\
1+3+\dots+(2k-1)+(2k+1) &= k^2+(2k+1) &\text{Add $(2k+1)$ to both sides.}
\end{align*}
  Factoring the right side of this equation, we have
  $$1+3+\dots+(2k-1)+(2k+1) =(k+1)^2 \qquad P_{k+1}$$
  But this last equation is $P_{k+1}$. 
  Thus, we have started with $P_k$, the statement we assumed true, and performed valid operations to produce $P_{k+1}$, the statement we want to be true. 
  In other words, we have shown that if $P_k$ is true, then $P_{k+1}$ is also true. 
  Since both conditions in Theorem 1 are satisfied, $P_n$ is true for all positive integers $n$.

In the top line there is written that we have to show that 'the assumption $P_k$ is true implies $P_{k+1}$ is true'.
And what I think is that: as long as we know that the state of a proposition being true for any positive integer $k$ after Base number implies that proposition is true for integer $k+1$, we have to show that $P_k$ is true. However I don't have an idea yet how to show the truth of $P_k$.
So, my first question is that who is right? My book or me? And if I am right then how can I show the truth of $P_k$?
2) This may be taken as the second question but it is also bit annoying.
In the second paragraph it is written, "Since we have assumed that $P_k$ is true, we can perform operations on it". 
Why is it necessary for an equation to be true for performing operations on it? Well, this is not much annoying as compared to the first question because whole the induction theory depends upon that.

REPLY [3 votes]: There is another way of thinking of this.
Throw out the standard induction theorem, and replace it with this one:

Any nonempty set of natural numbers has a smallest element.

This seems obviously true, right?  This is called the well-ordering principle, and it is equivalent to induction.  Here's how you do induction with the well-ordering principle (WOP for short):

Prove that $P_0$ holds (or $P_1$, if you don't consider zero to be a "natural number").
Take this set:
$$
S = \{x | x \in \mathbb{N} \wedge \neg P_x \}
$$
That's the set of natural numbers for which our proposition does not hold.
Assume $S$ is non-empty.  Then, by WOP, it has a least element, which we can call $k+1$.  In particular, we know that $\neg P_{k+1}$ (we know that the proposition does not hold for $k+1$, since it's an element of $S$).
By step (1), we know $k+1 > 0$ (or $k+1 > 1$, if you started with $P_1$).  So $k$ is a natural number.  If we had not proven step (1), $k$ could be equal to negative one (or zero), which would invalidate the rest of the proof.
Prove that $\neg P_{k}$; that is, prove that the proposition is not true for $k$, based on our knowledge that it is not true for $k+1$.
Since $k < k + 1$ and $k \in S$, we have a contradiction ($k + 1$ isn't the smallest element of the set).  That means our assumption in step (3) must be false.  The set of counterexamples is empty, meaning the proposition must hold for every number.

Note that these statements are equivalent:
$$
P_k \implies P_{k+1} \\
\neg P_{k+1} \implies \neg P_k
$$
If you can prove one, you can prove the other quite easily.  Standard induction asks you to prove the first, while WOP-based induction requires you to prove the second.
In the end, it's just a matter of notation.  But you might find one form of induction more intuitive than the other.  If you are going to use this method, you should note that we usually talk about $k$ and $k - 1$ instead of $k+1$ and $k$; I used the latter because it corresponds more obviously to standard induction.