I'm in high school, learning induction

Apparently, induction requires the satisfaction of two steps

  1. The initial or base case: prove that the statement holds for 0, or 1.
  2. The induction step, inductive step, or step case: prove that for every n, if the statement holds for n, then it holds for n + 1. In other words, assume that the statement holds for some arbitrary natural number n, and prove that the statement holds for n + 1.

Here's what I can't wrap my head around. Consider this...

When the base case is wrong and all the other cases as well, the wrong assumption can lead to a correct inductive step.

E.g. prove n=n+2 for n>0 (natural numbers)

Base case wrong since 1 not equal to 3

However, if assuming right for n=k RTP: k+1=k+3

LHS=k+1 (sub k=k+2 from assumption)



Therefore k+1=k+3

So the inductive step is right

This all emphasizes the importance of the base case

However, I just don’t understand why there isn’t an example where the base case satisfies, but not all the cases, which leads to a wrong assumption, BUT can also lead to a correct inductive step. I just conceptualise why. I can’t find an example where it does, but I don’t understand on a core level, why I’m not able to find one, apart from the fact that I just can’t.

If a wrong assumption, can lead to a correct inductive step, as I’ve given an example of, what’s stopping a wrong assumption, where only the base case is right, from producing a correct inductive step, (since assuming for n=k where k>=0, can ofc be wrong, if it’s only right for the base case e.g. proving n^3=n^2 for n>0, is only right for n=1) which would, thus, prove it by induction but not actually be right. I can't find a why to prove this example for the inductive step, but I don't know why I can't find an example where I can.

With the k=k+2 example

  • It doesn't work for any thing
  • Is a wrong assumption (as a result)
  • But leads to a correct inductive step

With the k^2>=k^3 example

  • It only works for 0 and 1 but not all k
  • Is a wrong assumption (as a result)
  • But can't lead to an incorrect inductive step (for some reason)

What is fundamental difference between these two false assumptions which allows one to lead to a correct inductive step but the other, an incorrect one, and why is it exactly that the former will have a wrong base case, but the latter will have a correct one. It works out conveniently for induction but I just don't understand how we can trust it, if it is possible for you to make a correct inductive step with a wrong assumption. I don't understand why, exactly, you can just say that if the inductive step is right but the statement is wrong, the base case will always be wrong to show you that.

In a way, I’m asking why induction is considered foolproof and reliable. It just seems so arbitrary and vague. Please let me know if I've done something wrong, or if I've breached or am unaware to a an aspect of induction

  • 3
    $\begingroup$ "I don't understand why there isn't an example where the base case satisfies, but not all the cases" Look for the fake proof that "All horses are the same color" which has the base case of $n=1$ is satisfied, and the induction step is satisfied going from $n$ to $n+1$ for all $n>2$, but fails because the induction step doesn't work going from $n=1$ to $n=2$ ($n=2$ secretly needing to also be a base case that was ignored) $\endgroup$
    – JMoravitz
    Commented Sep 1, 2021 at 23:53
  • $\begingroup$ It's impossible for the base case to be true, for the inductive step to be correct, and yet for the statement to not be true for all numbers. The reason is because if the base case is true and the inductive step is true, then the statement must be true for all numbers... by induction. $\endgroup$
    – Jack M
    Commented Sep 1, 2021 at 23:54
  • 1
    $\begingroup$ As for your core question, see What makes induction a valid proof technique? and the questions linked in that $\endgroup$
    – JMoravitz
    Commented Sep 1, 2021 at 23:57
  • $\begingroup$ Related, possibly helpful: matheducators.stackexchange.com/questions/10021/… $\endgroup$ Commented Sep 2, 2021 at 0:51

2 Answers 2


Intuitively, induction says "all natural numbers can be reached by counting from $0$".

Let's supposed we've proved both the base case and the inductive step. So we've proved $P(0)$, and we've also proved that for all $n$, if $P(n)$ then $P(n + 1)$.

Then for any number $n$ you can actually count to, you can prove $P(n)$.

Let's start counting with $1 = 0 + 1$. We know that $P(1)$ is true because we know $P(0)$, and we know that $P(0) \implies P(0 + 1)$.

We then count to 2. We know that $P(2)$ is true because $P(1)$ is true and also $P(1) \implies P(2)$.

We then count to 3. We know that $P(3)$ is true because $P(2)$ is true and also $P(2) \implies P(3)$.

We can keep doing this as many times as we want. Each application of $P(n) \implies P(n + 1)$ allows us to count 1 higher. As long as we can reach a number $m$ by counting, we will eventually be able to prove $P(m)$.

  • $\begingroup$ I'm a little confused as to what's so special about the inductive step though as well. What exactly is "if P(n)" saying. It seems to be assuming that it is true for any integer n. And we normally use that assumption to prove the inductive step. But firstly, if you know it's true for any specified integer, wouldn't you obviously know that it would be true for 1 integer higher, since that is part of all integers? And secondly, how is it possible to prove an inductive step, using an assumption which you know could be wrong? How do you know when you are operating on a false premise or not? $\endgroup$ Commented Sep 2, 2021 at 1:21
  • $\begingroup$ @JamesDaSilvaChen Assume $P_0$ and $P_{n+1}$ have been proven, however it's actually untrue for some $P_k$. After relabeling $k=j+1$, we claim $P_{j+1}$ doesn't satisfy the original postulate. This means either $P_j$ is similarly flawed or the proof for $P_{n+1}$ is flawed. Assuming the latter is sound then for the former, you recursively relabel and eventually hit the base case, indicating the proof for $P_0$ is flawed. However, if $P_0$ isn't flawed and $P_{n+1}$ isn't flawed, then the premise that it breaks down for some $P_k$ is flawed. $\endgroup$ Commented Sep 2, 2021 at 1:46
  • $\begingroup$ @JamesDaSilvaChen We're trying to prove $\forall n (P(n) \implies P(n + 1))$. That is, we're trying to show that for any $n$, if $P(n)$ is true, then $P(n + 1)$ is also true. When you say "how do you know when you are operating on a false premise or not?", this question is totally irrelevant. We only care about the case where $P(n)$ is true when we're proving $P(n) \implies P(n + 1)$. $\endgroup$ Commented Sep 2, 2021 at 16:39

After researching more about induction and thinking about it more, I have a greater understanding of its components. My question didn't have much reasoning apart from just doubting that induction actually worked. I can see now why induction is "reliable and foolproof", and the individual importance behind both the base case and inductive step. The test is needed to test an assumption for the base case and see if it's correct, then the inductive step which has been proved for any positive integer K that satisfies the assumption (which the base case would or wouldn't), will then imply the next would also work, and then that would imply the next would as well (since it keeps referring to the next positive integer).

  • $\begingroup$ Please provide additional details in your answer. As it's currently written, it's hard to understand your solution. $\endgroup$
    – Community Bot
    Commented Sep 5, 2021 at 10:00

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