1

I'm having trouble understanding the following situation: Given an identity P(n) that is wrong (but I don't know whether it's right or wrong), I am trying to check whether this identity can be proven using induction. I find that the base case of n=k is to be true for P(k). Now I attempt to show that P(n) -> P(n+1) and since P(n) was a false statement, it turns out that implication comes out to be true (since if P is false P->Q can still be true). Now I have shown that base case for n=k is true and P(n)->P(n+1) is true. So by induction, I conclude that P(n) is true (which is actually wrong). Why is this paradox occurring?

RobPratt
  • 45,619
Mardia
  • 325
  • 1
    From a false premise, anything follows: see this post. This is called the principle of explosion. – Toby Mak Mar 21 '22 at 23:26
  • If you are trying to prove $P(n)$ for all $n$ above some $n_0$, define $Q(n)\iff P(n+n_0)$, so now prove by induction they $Q$ holds for all $n$. – Asaf Karagila Mar 21 '22 at 23:36
  • Did you prove $P(n)\implies P(n+1)$ just by observing that identity $P(n)$ is wrong? A wrong identity means $P$ is wrong for some $n$ (possibly not all $n$), and your base case proof (that $P(k)$ is true) means you should also consider the implication with a true $P(n)$. – peterwhy Mar 21 '22 at 23:45
  • @peterwhy I was assuming that I don't know whether identity P(n) is wrong or right. I came up with this situation because I know a false statement can give us true implication. (If I'm understanding correctly about induction) If I show a specific case being valid for P(k) and if I can show that the P(n)->P(n+1) is valid then P(n) is true for all n>=k. But this is a contradiction. I started with a false statement P(n) that is true for some cases but is not true for all case and P(n)->P(n+1) still being valid. – Mardia Mar 21 '22 at 23:50
  • @Mardia "If I show a specific case being valid for $P(k)$ and if I can show that the $P(n)\implies P(n+1)$ is valid", then by induction $P(n)$ is true for all $n\ge k$. $P(n)$ may still be false for $n < k$. Then this satisfies that $P(n)$ is sometimes true and sometimes false. – peterwhy Mar 22 '22 at 00:01
  • @Peterwhy Is it not possible that statement P(n) was so wrong that there is still a false case j > n? Where can I find the logical proof of this? – Mardia Mar 22 '22 at 00:02
  • 1
    @Mardia If you can prove both: 1. $P(k)$, and 2. for all $n\ge k, P(n)\implies P(n+1)$, then step by step, you can prove $P(k+1)$, then $P(k+2)$, then $P(k+3)$, ..., eventually $P(j)$ where $j>k$. ($k,n,j$ are integers) – peterwhy Mar 22 '22 at 00:11

3 Answers3

2

I think you are confused about what the "base case" means.

Suppose (for example) that $P(n)$ is the statement $$ n = n+1 . $$

The assertion $$ P(n) \text{ implies } P(n+1) $$ is true for two good reasons. The first, as you you note, is that the antecedent is false, so the implication is true. But that does not imply the truth of $P(n+1)$, just the truth of the implication.

The second reason is simple algebra. Add $1$ to both sides to arrive at $$ n = n+1 \implies n+ 1 = (n+1) + 1. $$

So the inductive step is fine. But there is no true "base case" to get things started: $P(1)$ is false, so chain of correct implications $$ P(1) \implies P(2) \implies P(3) \implies \cdots $$ is useless.

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
  • Is it not possible to have a statement P(n) such that P(n) holds for some n=k but the implication is still true? – Mardia Mar 21 '22 at 23:42
  • If $P(k)$ is true for some particular $k$ and the inductive argument is correct then $P(n)$ will be true for all $n > k$. It can't "suddenly become false again". – Ethan Bolker Mar 21 '22 at 23:45
  • 1
    @Mardia Something here might help: https://matheducators.stackexchange.com/questions/10021/why-are-induction-proofs-so-challenging-for-students/10057#10057 – Ethan Bolker Mar 21 '22 at 23:48
  • But if P(n) was a contradictory statement or logically invalid statement, isn't possible that P(k) is true for some particular k and is still correct on the implication P(n)->p(n+1)? Since logical steps on top of logically invalid statement can lead to correct implication? Also, is there a proof for the statement you just said? "can't suddenly become false again." – Mardia Mar 21 '22 at 23:54
  • 1
    That it can't suddenly become false again is precisely what the principle of induction guarantees. That principle can't be proved - it's essentially an axiom about how the natural numbers behave. Suppose $P(n)$ is the statement "$n > 10$", Clearly if $n>10$ then $(n+1)>10$ so the induction step $P(n) \implies P(n+1)$ is always correct, but $P(n)$ is true only from $11$ on. – Ethan Bolker Mar 22 '22 at 00:03
2

I was initially really confused by your claim that '$P(n)$ is wrong' in conjunction with your claim that 'the base case $P(n)$ is true for $n=k$'. That is: how can you say that $P(n)$ is wrong ... and yet it is also (at least for the case $n=k$ also true?

But I am guessing that when you say that '$P(n)$ is wrong', you have something in mind like:

$P(n):$ $n = n^2$

which is indeed wrong in general, but still is true for the base case $n=k=0$

So: when you say that '$P(n)$ is wrong', you really mean to say that $\forall n \ P(n)$ is wrong!

But that is a totally different claim from saying that the $P(n)$ as part of the claim $P(n) \to P(n+1)$ is wrong.

That is, when we do an inductive proof, and make the inductive hypothesis/assumption that $P(n)$ is true, what we are really claiming that $P(n)$ is true for some arbitrarily chosen number $n$. And note: in a case like this, whether $P(n)$ is actually true depends on what this arbitrarily chosen $n$ is: if $n=0$, then $P(n)$ is true, but for $n \neq 0$, it is not true.

Therefore, when you say that we can set $P(n) \to P(n+1)$ to true 'because $P(n)$ is wrong', you are making a mistake: For $n=0$, we have that $P(n)$ is true. And, since in that case $P(n+1)$ is false, we in fact have that $P(n) \to P(n+1)$ is false, breaking the inductive proof ... and thus resolving the paradox that you perceive.

But yes, the basic mistake you are making is that you are confusing $P(n)$ with $\forall n \ P(n)$

And this is in fact a really common confusion I see when it comes to induction. That is, I frequently see people claim that induction is circular by claiming that the inductive assumption is exactly the same as the ultimate claim that we are trying to prove. But again, the former is the claim that $P(n)$ is true for some arbitrarily picked $n$, whereas the latter is the claim $\forall n \ P(n)$, which are two different things.

Bram28
  • 100,612
  • 6
  • 70
  • 118
  • Thank you for your answer. So what you have answered, I understand. But what I'm unsure is the fact that there won't be a statement P(n) such that for sequences of n's, P(n) will be oscillating between truth and false. These are the assumption I have. 1) There are infinitely many P(n)'s. 2) Applying logical steps to logically/axiomatically wrong P(n) can still lead to a true statement Q(n). Assuming these, let us assume we don't know if P(n) is true or not. We check that for n=k, P(k) is true. Now I apply logical steps and find that P(n) does in fact can lead to P(n+1). (continuing) – Mardia Mar 22 '22 at 16:45
  • But when we plug in k+j to P(n), P(k+j) is actually false. Is there no such statement P(n) where this is possible? If not, why? If it's possible for a completely false statement to lead to a true statement, why isn't possible that P(n)->P(n+1) is true for wrong P(n) for some n's giving true value to P(n)? – Mardia Mar 22 '22 at 16:51
  • @Mardia You're welcome! :) OK, so if I understand you correctly: What if you have a $P(n)$ that is, say, true for all even $n$'s, but false for all odd $n$'s? Well, then for any even $n$: $P(n) \to P(n+1)$ is false ... and that will break the inductive step. And yes, $P(n) \to P(n+1)$ would be true for any odd $n$, but what the inductive step needs to show is that $P(n) \to P(n+1)$ is true for every $n$, i.e. the step needs to show that $\forall n \ (P(n) \to P(n+1))$. So as soon as there is an exception to that, the inductive step cannot be proven. – Bram28 Mar 22 '22 at 17:17
  • @Mardia I am wondering if some of the misunderstanding is from the particular way induction has been presented to you. That is, it is typical to present the inductive step as: Show $P(n) \to P(n+1)$ ... but that's a little imprecise. What you really have to show is $\forall n (P(n) \to P(n+1))$. Also, if you have shown that, then it automatically follows that $P(n) \to P(n+j)$ for any $n$ and $j$, because $P(n)$ implies $P(n+1)$, which implies $P(n+2)$, which implies $P(n+3)$ which implies .... .... which implies $P(n+j)$. So what you are trying to do is impossible. – Bram28 Mar 22 '22 at 17:25
  • So even when P(n) is a complete nonsense, it's impossible that it leads to P(n)->P(n+1)? I tried to come up with an example but have failed. I think what's throwing off is the fact that if P(n) is a false statement, implication can still be true. I agree with this. Implication can indeed be true. But can we show there exists no such false P(n) such that no matter what you do, you can never show for all n, P(n)->P(n+1)? I'm wondering what's the structure underneath the math that makes it possible to do this. – Mardia Mar 22 '22 at 17:35
  • Let's say P(n) is a set of all n's where P(n) is true (n are natural numbers) and let's assume P(n) is true. What building blocks of math makes it possible that when we apply mathematically logical steps to the statement P(n), for all n, P(n)->P(n+1) is only satisfied when P(n) is true? I agree that if P(n) is a true statement, applying correct logical steps will cause no contradiction. But what makes it possible that applying these steps to flawed P(n) guarantees that for all n, P(n)->P(n+1) is not possible? – Mardia Mar 22 '22 at 17:35
  • @Mardia If $P(n)$ is total nonsense, by which I assume you mean to say that $P(n)$ is false for any $n$, then automatically $P(n) \to P(n+1)$ is true (exactly because oif what you say: if the ';if' part is false, then automatically the whole 'if ... then ..' is true). In fact, if $P(n)$ is false for every $n$, then $P(n) \to P(n+1)$ will be true for any $n$, and so the inductive step is true. But now of course you can't prove the base case, and so the inductive proof as a whole will still fail. – Bram28 Mar 22 '22 at 17:40
  • Right, I agree. Okay, Let me try to say it differently. Let's say P(n) is a complete nonsense but still holds for some k (so P(k) is true). But P(n) is so wrong that P(n+1) is completely not related to P(n+1). What is the axiom that guarantees us that this not being possible? Because if P(n) and P(n+1) is not related, it should be possible that P(k) is true but P(k+1) is not even when you can show P(n)->P(n+1) being true since P(n) is a complete B.S. I'm sorry if this doesn't make sense. I asked the same question about 2 years ago but I was still unsatisfied so I came back to it. – Mardia Mar 22 '22 at 17:46
  • @Mardia As such, you therefore can have $P(n) \to P(n+1)$ be true for all $n$, even though it is not true that $P(n)$ is true for all $n$. – Bram28 Mar 22 '22 at 17:46
  • There is no axiom that assures there is a relation between $P(n)$ and $P(n+1)$. Whether there is such a relationship depends on the nature of $P(n)$. For example, with $P(n)$ being '$n$ is an even number', there clearly is no such relationship. In fact, in that case, $P(n) \to P(n+1)$ would be false for any even $n$. Now when you say that $P(n)$ is "complete B.S." ... all I can interpret that as is: $P(n)$ is false for any $n$ .. but then $P(n) \to P(n+1)$ is automatically true again ($F \to F = T$) – Bram28 Mar 22 '22 at 17:51
  • Okay, I tried to find the core idea of my misunderstanding. I think this question will answer everything for me. Theoretically, is it always possible to show inductive step for a true statement P(n)? It's okay if it's extremely difficult to show this step, as long as it's possible. If so, what guarantees us this? (Edit: Okay, If I'm correct, I think it is. Can you check my if my reasoning is correct? Is it possible to message you or should I start a new problem and @ you to verify?) – Mardia Mar 22 '22 at 18:15
  • @Mardia By 'true statement $P(n)$' do you mean that $\forall n \ P(n)$ is true? Or that $P(n)$ is true for some particular $n$? Likewise, when you 'show inductive step', do you mean to show $P(n) \to P(n+1)$ for some particular $n$, or for all $n$, i.e. show that $\forall n (P(n) \to P(n+1))$? As I explained in my Answer, those are all different things! Also, for an inductive proof, you need to show $\forall n (P(n) \to P(n+1))$. And if all you have is that $P(n)$ for some particular $n$, then there is no guarantee that you can show that. – Bram28 Mar 22 '22 at 18:50
  • But if you know that $\forall n P(n)$, then showing that $\forall n (P(n) \to P(n+1))$ is trivial: if all number have property $P$, then that includes the number $n+1$ for any number $n$. So that's certainly the case ... though not very interesting. So what typically happens is: you don't already know that $\forall n \ P(n)$, but you now prove it using induction. And yes, theoretically, if $\forall n \ P(n)$ is provable at all, then you can also prove it using induction (and, many times, much easier than without using induction). I think the latter is what you mean .. and yes, it's true. – Bram28 Mar 22 '22 at 18:53
  • Oh! So maybe this is what you're getting at. Suppose we have something like $P(n)$: "the number $n$ can be written as the sum of $4$ square numbers" (which actually turns out to be true, but the proof is difficult). Now, if we were to try and prove this using (weak) induction, i.e. base case together with inductive step going from $P(n)$ to $P(n+1)$, we will quickly find that there is no direct relationship between $P(n)$ and $P(n+1)$. That is, assuming that $P(n)$ is true for some arbitrary $n$ really doesn't give us any further insight as to whether $P(n+1)$ is true. (...cont'd) – Bram28 Mar 22 '22 at 19:00
  • So, can we really even do an inductive proof for this theorem then? Well, the answer is still yes. Since we can prove this 'Four squares theorem', we can therefore also prove that it is true for any number $n+1$ ... even if in that proof we make no use of the assumption that $P(n)$ is true at all. But that's ok, since if $P(n+1)$ is true, then $P(n) \to P(n+1)$ is true as well. Hence, we will have shown $P(n) \to P(n+1)$ for any arbitrary $n$, and thus we complete our inductive step. It feels a little like cheating, sure, in that we never use the inductive assumption, but it is a proof! – Bram28 Mar 22 '22 at 19:04
  • Let us continue this discussion in chat. – Mardia Mar 22 '22 at 19:12
  • I have written my understanding now. Can you verify if I'm on the right track? – Mardia Mar 22 '22 at 19:56
1

After having proved that $P(k)$ holds, the next step is to prove that, for every $n\geqslant k$, $P(n)\implies P(n+1)$. Simply proving that for one specific $n$ you have $P(n)\implies P(n+1)$ (which is what you did) is not enough.

José Carlos Santos
  • 427,504
  • I meant P(n) -> P(n+1) for every n>=k is true. What if I get this implication to be true and there is a base case k holding true? Isn't it possible to get this result when statement P was flawed in the first place? – Mardia Mar 21 '22 at 23:32
  • @Mardia in the first step if you have $P(1)$ false and $P(2)$ true then false implies true. After that everything is true regardless of what $P(n)$ is. This means $P(n) \implies P(n+1)$ can be true but it's entirely optional. – CyclotomicField Mar 21 '22 at 23:37
  • @CyclotomicField What if there are infinitely many cases where P holds and fails all together? Is this not possible? – Mardia Mar 21 '22 at 23:39
  • @Mardia it's possible. The truth value of $P(n)$ doesn't matter at all if the base case is false. It can be true, it can be false, and it can switch between them. – CyclotomicField Mar 21 '22 at 23:41
  • @CyclotomicField Assuming P(n) was a wrong statement to begin with, let's say I've shown that P(1) holds true and P(n)->P(n+1) . P(n) fails at n = 10000. Such a case is possible? – Mardia Mar 21 '22 at 23:44
  • @Mardia If you can imagine a statement, I can make $P(n)$ that statement. It's perhaps easier to think of $P(n)$ as a function from the natural numbers to the collection of propositions $P(n)$ to see why you can pick and choose which statements go where. – CyclotomicField Mar 22 '22 at 00:21