Formal logical form of mathematical induction

Question

I don't understand induction. I first describe what I know and then I ask a set of conceptual questions.

From Terence Tao's Analysis I book (3rd edition), we have the following description of mathematical induction:

Axiom 2.5 (Principle of mathematical induction). Let $P (n)$ be any property pertaining to a natural number $n$. Suppose that $P (0)$ is true, and suppose that whenever $P (n)$ is true, $P (n+1)$ is also true. Then $P (n)$ is true for every natural number $n$.

To put it formally:

$$(P(0) \; \wedge\; \forall k {\in} \mathbb{N}(P(k) \implies P(k + 1))) \implies \forall n {\in} \mathbb{N}(P(n)).$$

When I use induction, I do the following: I show that the base case $P(0)$ holds. I proceed to an inductive assumption: $\forall k {\in} \mathbb{N}(P(k))$ is true. I then realize I cannot assume that $\forall k {\in} \mathbb{N}(P(k))$ since this is precisely what needs to be proved, however this is not a problem since analysing the formal logical statement of mathematical induction shows that it does not matter whether our inductive assumption is false or true:

1. If $\forall k {\in} \mathbb{N}(P(k))$ is false, then the implication $\forall k {\in} \mathbb{N}(P(k) \implies P(k + 1))$ is true, and, therefore, once $P(0)$ is true, the whole hypothesis $P(0) \; \wedge\; \forall k {\in} \mathbb{N}(P(k) \implies P(k + 1)$ is true, and, therefore, we obtain that $\forall n {\in} \mathbb{N}(P(n))$ is true, as desired.

2. If $\forall k {\in} \mathbb{N}(P(k))$ is true, then we show that $\forall k {\in} \mathbb{N}(P(k + 1))$ is true. Once this has been shown, the implication is true, and once $P(0)$ is true, we obtain that $\forall n {\in} \mathbb{N}(P(n))$, as desired.

This closes induction.

Here's what I'm struggling with:

1. Consider case $1$ above. We have that $\forall k {\in} \mathbb{N}(P(k))$ is false and $\forall n {\in} \mathbb{N}(P(n))$ is true, which is a contradiction. Where am I wrong in my reasoning?
2. Some authors express induction formally as $$(P(0) \; \wedge\; \exists k {\in} \mathbb{N}(P(k) \implies P(k + 1))) \implies \forall n {\in} \mathbb{N}(P(n)).$$ a) Is this correct? b) If this is correct, why do we have two inequivalent statements for one concept? c) And if this is correct, which statement should I use: with a universal quantifier or with an existential quantifier?

Answer 1

This is a very good question, because these are all common misconceptions, and I probably had the same questions as a student.

Let $P (n)$ be any property pertaining to a natural number $n$.

To put it formally:

$$\Big(P(0) \; \wedge\; \forall k {\in} \mathbb{N}\,\big(P(k) \implies P(k + 1)\big)\Big) \implies \forall n {\in} \mathbb{N}\:P(n).$$

Yes.

I proceed to an inductive assumption: $\forall k {\in} \mathbb{N}\,P(k)$ is true.

No: assuming $\forall k {\in} \mathbb{N}\, P(k)$ is equivalent to assuming $\forall n {\in} \mathbb{N}\, A(n),$ begging the question (circular reasoning)! Rather, in the induction hypothesis, $k$ is just an arbitrary natural number; at this point, we haven't invoked Universal Generalisation yet.

If $\forall k {\in} \mathbb{N}\,P(k)$ is true, then we show that $\forall k {\in} \mathbb{N}\,P(k + 1)$ is true.

No, the induction step is showing not $$\Big(\forall k {\in} \mathbb{N}\;P(k)\Big) \implies \Big(\forall k {\in} \mathbb{N}\;P(k + 1)\Big),$$ but $$\forall k {\in} \mathbb{N}\,\big(P(k) \implies P(k + 1)\big),$$ here finally introducing the universal quantifier, thereby discharging the arbitrary constant (akin to eventually discharging every provisional assumption in a proof). For absolute clarity, instead of recycling the induction hypothesis's arbitrary constant $\boldsymbol k$ as the induction-step's conclusion's dummy variable, it may be more elucidating to write $$\forall \boldsymbol m {\in} \mathbb{N}\,\big(P(\boldsymbol m) \implies P(\boldsymbol m + 1)\big),$$ or even $$\forall \boldsymbol n {\in} \mathbb{N}\,\big(P(\boldsymbol n) \implies P(\boldsymbol n + 1)\big),$$ instead.

Some authors express induction formally as $$\Big(P(0) \; \wedge\; \exists k {\in} \mathbb{N}\,\big(P(k) \implies P(k + 1)\big)\Big) \implies \forall n {\in} \mathbb{N}\:P(n).$$

Incorrect (and not even mathematically true). Unfortunately, many authors write the induction hypothesis as “Suppose that $P(k)$ for some natural number $k\text”,$ which, although not definitively wrong, is potentially misleading or confusing.

Better: “Suppose that $P(k)$ for some arbitrary natural number $k.\text”$

Best: “Suppose that $P(k)$ for an arbitrary natural number $k.\text”$

Incidentally, the abovementioned misconception that the induction hypothesis is $\forall k {\in} \mathbb{N}\, P(k)$ may be due partly to the egregious phrasing “Suppose that $P(k)$ for any natural number $k\text”,$ whose meaning is at best ambiguous. It can be improved as “Take any (even better: take an arbitrary) natural number $k,$ and suppose that $P(k).\text”$

In any case, the statement $\exists k {\in} \mathbb{N}\,\Big(P(k) \implies P(k + 1)\Big)$ actually means that some natural number $k$ either satisfies $P(k+1)$ or fails to satisfy $P(k);$ clearly, this is weaker than what the induction step shows.

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Reply to comments

Why is not this circular reasoning: “suppose $P(k)$ for an arbitrary $k\text”?$ What is the reason behind this supposition?

First, click on the first link above. To explain it another way: if, for an arbitrary element $k$ of a set $S,$ we have subsequently derived the result $(P(k)\Rightarrow P(k+1)),$ then, since $\boldsymbol k$ has no special feature besides being $\boldsymbol {S'}\textbf s$ member (by the definition of $k$ being an arbitrary element of $S),$ we can conclude that $(P(n)\Rightarrow P(n+1))$ is actually true for every member of $S,$ in other words, that $\forall n{\in}S\,(P(n)\Rightarrow P(n+1))$ or, equivalently, $\forall k{\in}S\,(P(k)\Rightarrow P(k+1)).$

To be clear: as an arbitrary constant, $k$ has an arbitrarily fixed value for each iteration of the induction step, in which it doesn't alternately carry different values.

to finally clarify, am I right that it does not matter whether $P(k)$ is true or false for a particular but arbitrarily chosen $k∈\mathbb N,$ since even if $P(k)$ is false the implication follows vacuously?

Yes.

The formalisation at the top of this post shows that mathematical induction proofs are logically valid; thus, if the two premises of an induction proof have been proven true—that is, if its base case has been proven and its induction step has successfully reached its desired conclusion—then it is a sound argument, which entails that its conclusion $\forall n\,P(n)$ must be true.