A Question regarding mathematical induction.

Question

The principle of mathematical induction (as I understand stand it) states the following -

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Consider a property $P(n)$ defined for the positive integers $n \in \mathbb{Z}^{+}$. Suppose the following two statements are true

1. $P(1)$ is true.

2. For all $k \geq 1$, if $P(k)$ is true then $P(k+1)$ is also true.

Since $P(1)$ is true, and $\forall k \in \mathbb{Z}^{+}$ $P(k+1)$ is true whenever $P(k)$ is true, $P(n)$ is true for all $n \in \mathbb{Z}^{+}$.

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I understand everything except the point "If $P(k)$ is true for all $k \geq 1, k \in \mathbb{Z}^{+}$". How can we make such an assumption? I could have also assumed that $P(k)$ is true for all $k \geq 2, k \in \mathbb{Z}$. On what grounds are these assumptions valid? It seems to me that you pick any integer for which $P(k)$ is true. I understand that assuming $P(k)$ is valid for $k \geq 1$ allows us to use the base case and conclude that $P(n)$ is true for all $n \in \mathbb{Z}$, but what I cannot understand is why this is a valid assumption.

This post isn't the same as this one Mathematical induction question: why can we "assume $P(k)$ holds"?. That post asks "how can we assume $P(k)$ holds and then show that $P(k+1)$ follows without ever proving the truth of P(k)$. My question is different. It asks why $P(k), k \geq 1$ is assumed to be true and not $P(k)$ for some other positive integer other than $1$.

Answer 1

You can assume anything you want! And that assumption, by itself, doesn't mean anything, so none of this should trouble you.

However, the fact that we can infer something (call it 'bla bla') from an assumption does mean something: it means that the conditional statement "if [assumption], then [bla bla] " is true.

With induction, the inductive step shows that for any $k$: if $P(k)$, then $P(k+1)$. To show that, you typically pick some arbitrary $k$, assume $P(k)$, and try to show $P(k+1)$. That may look as if it is assuming that $P(k)$ is true for any $k$, but it isn't.

Compare: Suppose I ask you to prove that any number divisible by $4$ is an even number. To do this, I assume you would start with: "OK, let's assume that $n$ is some arbitrary number divisible by $4$". But note: that is totally not the same as assuming that any arbitrary number is divisible by $4$. Likewise, when in the inductive step you assume that $k$ is some arbitrary umber for which $P(k)$ is true, you are not assuming that $P(k)$ is true for any number $k$.

Also note that the inductive step by itself also does not mean that $P(k)$ is true for any $k$. What the inductive step does mean is that if $P(1)$, then $P(2)$. And it also means that if $P(2)$, then $P(3)$. And that if $P(3)$, then $P(4)$. Etc. ... which does not mean $P(k)$ for any $k$.

But now throw in the base case $P(1)$, and suddenly you now can conclude $P(2)$, ... and therefore $P(3)$, ... and therefore $P(4)$, etc. ... meaning that between the inductive base and the inductive step, you obtain that $P(k)$ for any $k$.

Answer 2

I understand everything except the point "If $P(k)$ is true for all $k \geq 1, k \in \mathbb{Z}^{+}$". How can we make such an assumption? I could have also assumed that $P(k)$ is true for all $k \geq 2, k \in > \mathbb{Z}$. On what grounds are these assumptions valid?

Hint: If you want to prove that $A$ implies $B$, you can start by first assuming that $A$ is true and then prove that $B$ must also be true. Then you can conclude that $A$ implies $B$.

(Maybe proofs in propositional and predicate logic should be taught before proofs by induction.)