Mathematical induction question: why can we "assume $P(k)$ holds"?
Solution 1:
The "inductive step" is a proof of an implication: $$\mathbf{if}\ P(k),\ \mathbf{then}\ P(k+1).$$ So we are trying to prove an implication.
When proving an implication, instead of proving the implication, we usually assume the antecedent (the clause after "if" and before "then"), and then use that to prove the consequent (the clause after "then"). There are several reasons why this is reasonable, and one reason why it is valid.
Reasonable:
An implication, $P\to Q$, is false only in the case where $P$ is true but $Q$ is false. In any other combination of "truth values", the implication is true. So in order to show that $P\to Q$ is valid (always true), it is enough to consider the case when $P$ is already true: if $P$ is false, then the implication will be true regardless of what $Q$ is.
More informally: in proving $P\to Q$, we can say: "if $P$ is false, then it doesn't matter what happens to $Q$, and we are done; if $P$ is true, then..." and proceed from there.
Why is it a valid method of proof?
There is something called the Deduction Theorem. What it says is that if, from the assumption that $P$ is true, you can produce a valid proof that shows that $Q$ is true, then there is a recipe that will take that proof and transform it into a valid proof that $P\to Q$ is true. And, conversely, if you can produce a valid proof that $P\to Q$ is true, then from the assumption that $P$ is true you can produce a proof that shows that $Q$ is true.
The real interesting part of the Deduction Theorem is the first part, though: that if you can produce a proof of $Q$ from the assumption that $P$ is true, then you can produce a proof of $P\to Q$ without assuming anything about $P$ (or about $Q$). It justifies the informal argument given above.
That's why, in mathematics, whenever we are trying to prove an implication, we always assume the antecedent is already true: the Deduction Theorem tells us that this is a valid method of proving the implication.
Solution 2:
You have to prove a logical implication in the process of using induction.
We can express this implication as $P \to Q$, where $P,Q$ are statements.
To prove that $ P \to Q$ is true, we consider whether $P$ is true or false. If $P $ is false, then the entire implication is true. Ex falso quodlibet.
So we assume that $P$ is true and deduce that $Q$ is also true. This is the standard way to prove a logical implication directly.
Solution 3:
You are almost correct. But observe:
You proved it was true for $n=1$; thus, using 2. you know it's true for $n=2$.
Well, it's true for $n=2$. So, using 2., it's true for $n=3$.
$\ \ \ \vdots$
Solution 4:
It is like a self proving prophecy. This is the standard process. Assume for a sec there was a flaw in the process and you were assuming P(n) was true but expected the property as (0(0+1))/3 instead of (0(0+1))/2. You would not be able to solve for the n + 1 case. Try it.