The principle of mathematical induction works basically because of the following:

If we have a predicate $P(n)$, then if we have:

P(0) is true, and

P(n) $\implies$ P(n+1) for all nonnegative integers n,


P(m) is true for all nonnegative integers, m.

However, I was concerned what if $P(n)$ is false or what if the base cases were false, what that would mean about $P(n+1)$.

I think I know the answer to my question but I wanted to check the answer with the community.

Say that we know $P(n_0)$ is false. Thus, the implication $P(n_0) \implies P(n_0+1)$ is true vacuously (from the definition of implies). However, our goal in induction is to establish the truth about some implication $P(n)$ for as many n as we can. However, since $P(n_0)$ is false, then from the truth table of an implication $P(n_0) \implies P(n_0 + 1)$ remains true regardless of the truth value of $P(n_0 + 1)$. Thus, even though the implication is true, we are unable to conclude anything about $P(n_0+1)$ (or any other value of n > n_0) because $P(n_0 + 1)$ might be either true or false. The goal was to establish the truth about $P(n_0+1)$, but since we can't establish which fact is true it (about $P(n_0+1)$), then we cannot continue reasoning about values of $n \geq n_0 + 1$.

Is this correct? Is this why if we have a false statement we cannot have a proof by induction that proceeds correctly? Related to this question, if we find a case were $P(n_0) = False$ but then $P(n_1) = True$ for $n_1 > n_0$, can we still hope to show $P(n_0)$ for $n > n_0$? What about the other way round, if $P(n_0) = True$ but then $P(n_1) = False$, $n_1 > n_0$? Can we still hope induction to help us deduce anything meaningful?


If $P(n_0)$ is false then the implication $P(n_0) \implies P(n_0+1)$ is true (vacuously, as you've pointed out), however, we cannot deduce the truth or otherwise of $P(n_0+1)$ in this case.

The hardest part in a proof by induction is proving $P(n) \implies P(n+1).$ If you've proved this, then all you have to do is find a suitable $n_0$ such that $P(n_0)$ is true. If $P(n_0)$ is false, but $P(n) \implies P(n+1)$, then pick a different value for $n_0$ (assuming that one exists). If you can find just one such $n_0$ (and if you've proved that $ \forall n, P(n) \implies P(n+1)$, then $P(n)$ is true for all $n \geqslant n_0$, by the principle of mathematical induction.

In this case, pick a different value of $n_0$ until $P(n_0)$ is true.

Regarding your last sentence, if $P(n_0)$ is true and $P(n) \implies P(n+1)$, then, by the property of implication, $P(n_0+1)$ is true, as is $P(n_0+2), P(n_0+3), ...$, so you'd never have $P(n_1)$ being false (for any $n_1 \geqslant n_0$)!

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    $\begingroup$ Regarding my last sentence, you say $P(n_1)$ cannot be false by the principle of mathematical induction. Doesn't that mean we must assume as an axiom (or know that its true) that induction actually works. In that case, then your last sentences makes sense, right? $\endgroup$ Sep 10 '14 at 16:21
  • $\begingroup$ @Pinocchio It's not an axiom. If you've proved $P(n_0)$ and if you've proved that $P(n) \implies P(n+1),$ then $P(n_0+1)$ is true (since $P(n_0)$ is true), as is $P(n_0+2)$ (as $P(n_0+1)$ is true), as is ..., as are all integers $\geq n_0.$ $\endgroup$
    – beep-boop
    Sep 10 '14 at 16:24
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    $\begingroup$ @Pinocchio - the principle of math induction : $(P(0) \land \forall n (P(n) \rightarrow P(n+1))) \rightarrow \forall nP(n)$ is an axiom. We can use it with $n_0 > 0$; in that case, the conclusion will be : $(\forall n \ge n_0)P(n)$. Of course, if we have $P(n_0)=True$ but $P(n_1)=False, n_1>n_0$ we cannot conclude, because assuming $n_1=n_0+1$, $P(n_0) \rightarrow P(n_0+1)=False$ (it is $T \rightarrow F$) and thus it is not true that $\forall n(P(n) \rightarrow P(n+1))$, i.e. the induction step does not hold. $\endgroup$ Sep 10 '14 at 16:37
  • $\begingroup$ I know its a made up situation, but it seems to me that if we are able to show $P(n) \implies P(n+1)$, then we should not exist any $n_1 > n_0$ such that $P(n_1) = False$. If that existed, then either induction doesn't work (but it does if we assume it as an axiom) or the induction step was wrongly done or $P(n_1) = False$ is wrong. It's nearly a prove by contradiction that if the inductive step was done right and induction works, then there does not exist an $n_1$ such that $P(n_1)= False$. (because if it did, it would contradict one or both of the previous assumptions) I think this is right. $\endgroup$ Sep 10 '14 at 22:17
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    $\begingroup$ @MauroALLEGRANZA See How to prove the mathematical induction is true? $\endgroup$ Sep 11 '14 at 13:47

What you've said is correct, but it's not the usual way that we deal with "false cases".

Consider the statement $P(n)$ that says $$ 3\ln(1 + n) < n. $$ You might try proving this is true for every $n$, but you'd fail: it's false for $n = 0$ and $n = 1$. This is fairly common: namely, it's often the case that for finitely many values of $n$, $P(n)$ is false. Clearly you therefore can't prove $P(n)$ true for all $n$. You can, however, often find some $n_0$ with the property that all the "bad" $n$s are less than $n_0$. In the example, $n_0 = 6$ works.

So you typically prove instead that the statement is true for all $n > n_0$.

Alas, the principle of induction doesn't apply directly.

The usual subterfuge is to consider a new $S(n) = P(n + n_0)$. (So $S(0)$ says that $P(n_0)$ is true, and $S(1)$ says that $P(n_0 + 1)$ is true, and so on.)

The principle of induction then applies to the statments $S(n)$, for $n = 0, 1, ...$ and from their truth, you can deduce the truth of $P(n)$ for $n = n_0, n_0 + 1, n_0 + 2, \ldots$.

  • $\begingroup$ What about the possibility of a mathematical formula that is provable by induction for $n_0-1$ till $n =100000$, and then fails at one point $n+1 =100001$. Why it is not possible is the key. $\endgroup$
    – jiten
    Dec 26 '17 at 5:10

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