Existential crisis about proof by induction

Question

I recently read this post on this website and it gave me a bit of an existential crisis about proof by induction. My understanding is that, in proof by induction, we show two things to be true: $P(0)$ for some base case $0$ (not necessarily the number $0$), and then show that $P(n) \implies P(n+1)$ for all $n$. The combination of these show that the statement is true for all $n$.

However, in many of the "false" induction proofs (the example that really gives me issues is the all horses are the same color proof), the issue is that an incorrect base case was used. In the case of this example, $P(n) \implies P(n+1)$ is only true for $n > 1$, so the induction fails when trying to show $P(1) \implies P(2)$.

This is a problem, in my eyes. The whole point of induction is to prove a single base case and then prove a statement which carries out the whole chain of implications. But now, it seems as though there could potentially be a problem at each implication $P(k) \implies P(k+1)$ (for some $k$). How do I know for sure that at some $k = 10^{100}$, the implication does not fail, like it did for $k=1$ in the example? In the example, $P(2) \implies P(3)$ is true due to a special property of $P$, namely that dividing $3$ horses into $2$ sets of $2$ guarantees an overlap, while $P(1) \implies P(2)$ does not have this property. But maybe it is possible to give an example where something "special" happens at a much higher $k$ that is not as obvious to detect.

To me, it seems like to be completely sure the induction proof works, we need to prove $P(n) \implies P(n+1)$ for all $n$ individually, which defeats the whole purpose of induction as a method to begin with. But this cannot be the case because induction has been used to prove mathematical statements. So where is the gap in my logic?

Answer 1

To evaluate whether a proof... is valid for all $n$, you need to check....

No, you have misunderstood the foundation of proof by induction.

To use induction, you have to demonstrate two things:

• There exists an integer $n_0$ such that $P(n_0)$.

• For all integers $n$ $~\color{red}{\text{such that} ~n \geq n_0}$,
  $P(n) \implies P(n+1).$

Assuming that the constraints in both of the bullet points above are validly met, this will establish that the assertion $p(N)$ is true for all $n \in \Bbb{Z},$ such that $n \geq n_0.$

The whole point of the horses example is (I am speculating) that the base case was $n_0 = 1$, while the implication [$P(n) \implies P(n+1)$] failed at the point $n = n_0 = 1$, but succeeded for each $n > n_0.$

Per the 2nd bullet point above, this represents an invalid use of Induction, which (intuitively) explains why this invalid approach is able to prove a false result.

Your posting suggests that because of the complication of having to observe both of the bullet pointed constraints, proof by Induction is inherently unworkable, so that the implication [$p(n) \implies p(n+1)]$ must be manually checked for each individual integer $n \geq n_0.$

I disagree and offer the following counter example that shows a valid proof by induction:

$\underline{\text{Theorem}}$
For all $~n \in \Bbb{Z^+},~$ the $~\displaystyle \left[\sum_{k=1}^n k\right] = \frac{n(n+1)}{2}.$

Note:
It is irrelevant that the theorem is elementary, or that there exists alternative proofs. The point is still that proof by induction works.

Take $n_0 = 1.$
Then $~\displaystyle \left[\sum_{k=1}^1 k\right] = 1 = \frac{1(1+1)}{2}.$
So, the Theorem is established in the base case.

Now, assume that the Theorem has been proven for a specific $N \in \Bbb{Z^+}$, such that $~N \geq n_0 = 1.$

Then, by inductive assumption, you can reason that :

$$\left[\sum_{k=1}^{N+1} k\right]$$

$$= \left[\sum_{k=1}^N k\right] + \left[\sum_{k=N+1}^{N+1} k\right]$$

$$= \frac{N(N+1)}{2} + (N+1)$$

$$= \frac{N(N+1)}{2} + \frac{2N+2}{2}$$

$$= \frac{N^2 + 3N + 2}{2} = \frac{(N+1)(N+2)}{2}.$$

So, the Inductive proof has been completed, because it has been shown that if the proof is true for the value $N$, then the theorem is also true for the value $N+1$. Further, this analysis holds for each value of $N \in \Bbb{Z^+}$ such that $N \geq n_0 = 1.$

That is, there isn't any hole in the above analysis which would prevent the implication from holding, for each value of $N \in \Bbb{Z^+},$ such that $N \geq n_0.$

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It is true that it is feasible that an invalid proof by induction, that looks valid, has a specific logic hole. Personally, I once posted an invalid proof by Induction answer on this forum that had such a logic hole. I had to scramble to remedy the proof.

This doesn't mean that proof by Induction is unworkable. It just means that you have to be careful.

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Addendum
I relegated philosophy to this section of the answer.

An objection to my position could be made that any (human) application of Induction could have an inherent flaw, because the human might well overlook a logic hole.

That is, there isn't any way for the proof composer to ensure that re-reading the proof will uncover a subtle logic hole. It is entirely plausible that the proof composer, who has made an invalid assumption while composing the proof, will (wrongly) retain this invalid assumption during the proof reading.

While the scenario in the above paragraph is certainly plausible, the same objection could be levied against any Mathematical proof. The pitfall is not limited to proofs based on Induction.

So, with respect to comparing proof by induction with other methods of proof (implemented by a human), proof by induction is no more unworkable than any other such method.

Answer 2

I see where you are coming from with the horses example, but this is not a problem with induction, rather it is a problem with a false proof in general. Like with a lot of false proofs, you are making an implicit assumption about the problem that is not true. This isn't a problem specifically with induction, rather it is a problem with making proofs without understanding exactly what is going on, any proof technique can make compelling arguments if you miss cases.

I often show my students the "proof" $$ x^2 = x+x+\ldots+x \text{(x times)} $$ $$ \frac{d}{dx}x^2 = \frac{d}{dx}(x+x+\ldots+x \text{(x times)}) $$ $$ 2x = 1+1+\ldots+1 \text{(x times)}) $$ $$ 2x = x $$ $$ 2=1 $$

Clearly this is false, does this mean that all direct proofs are inherently containing potential holes we'll never know about? No, rather it just means we need to be careful that what we're doing really is allowed under the properties of our problem

Answer 3

Your central question is, "How can we tell if a proof of $P(k)\Rightarrow P(k+1)$ is valid without doing it individually for every $k$?".

The answer (as wjmccann said) has nothing really to do with induction. It's a question of "how can we avoid mistakes in reasoning that involves variables (such as $k$)?".

And there is no general answer. Let's look at the horses example, and by way of contrast, that traditional proof by induction, the formula $1+2+\cdots+n=n(n+1)/2$.

In the horses example, we let $P(k)$ be "any set of $k$ horses all have the same color". We then consider a set of $k+1$ horses, put them in some order, and let $A$ be the first $k$ horses, $B$ be the last $k$ horses. By inductive hypothesis, all horses in $A$ have the same color, likewise all horses in $B$. So far there is no logical error.

The next step involves looking at a horse in $A\cap B$. Now, $A\cap B$ can be empty or nonempty. If $k=1$, then $A\cap B=\varnothing$, so there is no horse to look at. The fallacious proof simply overlooks this case. This oversight has nothing to do with induction.

To make this clearer, suppose you broke it out as a lemma: "If an ordered set has $k+1$ elements, and $A$ are the first $k$ elements and $B$ are the last $k$, then $A\cap B\neq\varnothing$." The lemma fails for $k=1$ and holds for $k>1$, but induction is not involved.

By contrast, if $P(k)$ says that $\sum_{i=1}^k i=k(k+1)/2$, then one does a little algebra to show that $P(k)\Rightarrow P(k+1)$. I won't reproduce the algebra; I'm sure you've seen somewhere. The point is, though, that there are no errors in the algebra. It is correct for all integers $k\geq 1$.

You could imagine a proof by induction of some other formula, let's say one where at some point you had an equation $$(k-2)A(k)=(k-2)B(k)$$ and in the next step the $(k-2)$ factors were canceled, giving $A(k)=B(k)$. That would be an algebraic error, somewhat like the "horses" error. The proof of induction step would be valid only for $k\neq 2$. But again, that would be a mistake in algebra (dividing by zero), not a problem with induction.

Does this have anything to do with induction? Just this: the proof of the implication $P(k)\Rightarrow P(k+1)$ must hold for every $k$ from some point (say $k_0$) onward, and also the base case $P(k_0)$ must hold. Then induction allows you to conclude that $P(k)$ holds for all $k\geq k_0$.

The interesting thing about the "horses" example is that on the one hand, the reasoning for $P(k)\Rightarrow P(k+1)$ holds for all $k\geq 2$, but the "base case" holds only for $k=1$. So it's a "near miss": if $P(2)$ were just true, induction would work for all $k\geq 2$, and if $P(1)\Rightarrow P(2)$ just held, then induction would be fine for all $k\geq 1$.

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I want to add one thing, with regard to your statement:

To me, it seems like to be completely sure the induction proof works, we need to prove $P(n)\Rightarrow P(n+1)$ for all $n$ individually...

and you wonder about the case where the general proof, with a variable $n$, fails for some large $n$, like $n=10^{100}$.

Proofs involving variables are the heart of mathematics. (And all sorts of variables, not just for numbers.) Of course mistakes can and do happen, but that's just life.

On the other hand, I think it's much more likely one would make an error dealing directly with the $n=10^{100}$ case than handling it symbolically.

Consider the implication $P(n)\Rightarrow P(n+1)$ for the $\sum_{i=1}^n i=n(n+1)/2$ example. Say you tried to prove this directly for $n=10^{100}$. Isn't it much more likely you'd make an arithmetic error, working with those very many digits, than an algebraic error?

If the $P(n)$ proposition involved some system of $n$ objects, then this applies with even more force. Doing the case $n=10^{100}$ "by hand" is out of the question, and even a value of $n$ in the thousands would be highly error-prone for humans.

Answer 4

First things first. I hope that you can agree that as long as you can prove $P(0)$ and $P(n) \to P(n+1)$, then you have proven $P(n)$ for any $n$. So yes, there are some tricky 'false induction' proofs, but none of those take away from induction as a valid proof technique: just make sure that the proof for $P(0)$ is valid, and just make sure that the proof for $P(n) \to P(n+1)$ is valid, and you're good.

Still, you ask: but how can we make sure that the proof for $P(n) \to P(n+1)$ is valid? Does it really work for any $n$? In response to that I want to point out that the inductive step is really not different from any other universal proof. That is, what we are proving is $\forall n (P(n) \to P(n+1))$, and that is not any different from other universal proofs where we prove $\forall n \ \phi(n)$ for some formula $\phi(n)$. So, your worry is really about universal proofs, not about induction specifically.

Indeed there are plenty of other proofs that contain hard to catch mistakes, and those tricky mistakes have nothing to with induction. And please note those hard to catch mistakes often involve certain 'edge' cases, e.g. where $n=0$ or $n=1$ ... it's really rare (though of course not impossible) that $n=17$ would amount to some kind of special case ... but the nature of the problem should make it quickly very clear where the edge cases are. Note that in the case of the horses problem, we are also clearly dealing with a special edge case where $n=1$.

Finally, if you are still not convinced that some proof really works, you can always try and convert it to a formal proof. And if you were to do this with the horses problem, you'd indeed run into a situation where you no longer can make the inference that you need to make to complete the proof.

In short, there is nothing 'fishy' about induction as a proof technique. You just need to be careful with proving each part. But that is the same for any math proof, induction or not.

Answer 5

This isn't really a question about induction, although an induction proof is where it has raised its head for the OP.

The key point is

But maybe it is possible to give an example where something "special" happens at a much higher k that is not as obvious to detect.

And the answer is that yes, this sometimes happens. And by dumb luck, John Baez posted an article today that catches one such "bug" right in the act! It happens in a computer program that enumerates some configurations (details are in the linked post), and when $n = 7$,

I made a mistaken assumption in my code to try to save computation time — instead of trying every way of orienting the N rectangles, it always orients the last rectangle the same way, under the assumption that the same arrangement but flipped will be generated anyway. ... The issue is that “last” is in the order the rectangles are visited, which may change when the square is flipped!

and so $1$ out of $1372$ configurations fell through a hole in the proof. (Some might argue that it was a program, not a proof, but I say that the distinction is immaterial in context of this question.)

So yes, proofs are created by people, induction proofs and all sorts of other proofs, and we are relying on each other to not make mistakes, and to catch each other's mistakes when we do make them. I know there are personality types who are attracted to math in whom this induces unsettling feelings. All I can say is that they should probably not think too hard about where the ingredients for the dinner they ate tonight came from - that would really unsettle them.

Answer 6

You don't need to prove $P(n)$ in order to prove $P(n)\Rightarrow P(n+1)$. The implication is totally independent of the veracity of $P(n)$.

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Induction asserts that given

• $P(0)$ and
• $\forall n\geq0\,\,\,(P(n)\Rightarrow P(n+1))$,

we can deduce $\forall n\geq0\,\,\,P(n)$.

Recall that if $x$ is true, then $(x\Rightarrow y)\iff y$. If $x$ is false, then $x\Rightarrow y$ is universally true. That is to say, if any particular $P(n)$ was false, it wouldn't affect the implication $P(n)\Rightarrow P(n+1)$ whatsoever! So each (hypothetically different) proof that $P(n)\Rightarrow P(n+1)$ may proceed like

If $P(n)$ then ... $P(n+1)$.

or, equivalently

Let $P(n)$. Then ... and so we have $P(n+1)$.

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I believe the confusion comes from phrasing the proof of $P(n)\Rightarrow P(n+1)$ as a direct proof which presumes the correctness of $P(n)$: the implication does not rely on $P(n)$ being true, even though the first step is to presume it.