Mathematical Induction and implication

Question

Preface : (Skip this part if you want to go straight to my question. I am writing this so hopefully it doesn't get a duplicate since I couldn't find any satisfactory answer at mathstackexchange yet). Hello, I would like ask why I am failing to understand the Mathematical Induction is always valid. I am aware that this is a very common question being asked and I know there are many attempts to explain why, but I have been having this question for a long time (Actually coming back here to ask since I did not succeed for months). Trust me I have read many posts and answers related to this topic all over the different websites and none of them still satisfied my understanding (including mathstackexchange). I will try to explain clearly where I am confused so hopefully someone can help me.

So from what I know, Mathematical Induction proves that a statement $P(n)$ is true for all natural number if:

$P(0)$ is true.

(My understanding: Mathematical statement is true when $n = 0$.)

Assuming $P(n)$ is true for all natural numbers and if I can show that the implication $P(n) \rightarrow P(n+1)$ is true.

(My understanding: Meaning we don't know if the statement P(n) it's true or false. But if we show that the implication $P(n) \rightarrow P(n+1)$ holds then that is enough for mathematical induction to hold whether the statement $P(n)$ is actually true or not).

If these conditions (1) and (2), holds then that proves that the statement $P(n)$ is true for all natural numbers.

Here is where I find mathematical Induction being fishy.

I am fine with this logic of mathematical induction whenever the statement $P(n)$ is actually true, and the conditions (1) and (2) satisfies. Then I can agree with the common answers I've seen: If $P(0)$ is true and since $P(n) \rightarrow P(n+1)$ is true, $P(1)$ is true. Since $P(1)$ is true with the same logic $P(2)$ is true and so on...

But from what I know, the logical implication has a truth table of a form:

So it is saying that the implication is only false when the statement is true but the result is false. Meaning it is also possible for implication being true when the starting statement $P$ is false. If I am understanding implication correctly (Please correct me if I am wrong), when the statement $P \rightarrow Q$ is true, it means after applying valid logical steps to the initial statement $P$ (whether the statement $P$ is true or not), you can get to the resulting statement $Q$.

I was okay with false statement implying false result being true and false statement implying true statement being true with the following examples:

ex1) False statement implies False statement can be true:

Let the statement $P$ be $-1 = 1$ and result $Q$ be $0 = 2$. I am assuming $P$ and $Q$ are both false for obvious reason. Now since implication is applying logical steps, I can add $1$ to both sides of the statement $P$. This results in $-1 + (1) = 1 + 1 \rightarrow 0 = 2$ which is the statement $Q$. Therefore the implication is true.

ex2) False statement implies true statement can be true:

Let the statement $P$ be $-1 = 1$ and result $Q$ be $1 = 1$. I am assuming $P$ is false for obvious reason but $Q$ is true. Now since implication is applying logical steps, I can square both sides of the statement $P$. This results in $(-1)^2 = (1)^2 \rightarrow 1 = 1$ which is the statement $Q$. Therefore the implication is true.

Now back to the mathematical induction, we have $P(0)$ being true, we assume the statement $P(n)$ is true for all natural number not just $0$ (meaning it could also be false for some numbers greater than $0$, but we don't know) and we show that the implication $P(n) \rightarrow P(n+1)$ is true. But from the logic that I just said, $P(n)$ could be false and still imply $P(n) \rightarrow P(n+1)$ is true. If $P(n)$ is a true statement, applying logical steps should get me to $P(n+1)$ because that is how mathematics is built (To get logical answer from logical steps). But if we don't know if the statement $P(n)$ is false and the implication turns out to be true then what? It still seems like the logic of If $P(0)$ is true and since $P(n) \rightarrow P(n+1)$ is true, $P(1)$ is true. Since $P(1)$ is true with the same logic $P(2)$ is true and so on... should work. But as I said if it turns out that $P(n)$ is not true for all natural number, this shouldn't be the case and I can't imagine what the result would be.

I am assuming there should exist some statement $P(n)$ being false for some natural number (True for $P(0)$) but still true for $P(n) \rightarrow P(n+1)$ since logic says it is possible for a false statement to imply true statement and there are infinitely many statements that can satisfy this (I am assuming...).

I tried to come up with my own example of this and I couldn't but in case this will help people who reads this can understand better on where I am getting wrong I will still put it:

Let the statement $P(n)$ on natural number be $$\sum_{n=0}^k (n+1)^{3n} = k!3^k$$.

We can show the base case $P(0)$ is true.

Assuming that this statement is true for all natural number and applying logical steps, I get a contradiction and fail to show that implication is false. (Maybe it is possible to show that the implication is true by doing clever tricks?). In this case the example turns out to fail, but I believe there should be such a statement $P(n)$ that would satisfy my reasoning.

So where am I getting wrong? Why am I thinking that such a $P(n)$ should exist and why shouldn't it exist?

I hope I was clear and if there is any ambiguity or questions please ask.

Regarding the example... We have that it holds for $n=0$. Presumably (if I interpret it correctly) it does not hold for $n=1$. But this means that we cannot prove that $P(n) \to P(n+1)$ holds for every $n$, because for $n=0$ we have that $P(0) \to P(1)$ is false. — Mauro ALLEGRANZA, Sep 17 '20 at 09:06
@Mauro ALLEGRANZA Yeah the example doesn't work I agree and in this case P(1) does not hold so it is obvious that it fails. But I am arguing that there could be a statement that could be correct for many natural number so it isn't obvious to check that P(k) fails for some number also it will give $P(n)→P(n+1)$ true. — Mardia, Sep 17 '20 at 12:41
Not clear... If you prove (with a correct proof) that $P(0)$ holds and $P(n) \to P(n+1)$ holds for every $n$, then there is no possibility to find a $k$ such that $P(k)$ fails, because the inductive proof shows that $P(n)$ holds for every $n$. — Mauro ALLEGRANZA, Sep 17 '20 at 12:46
@Mauro ALLEGRANZA Let's say a false statement satisfies $P(0)$ to be true and in logic it is possible for a false statement $P(n)$ to imply $P(n)→P(n+1)$ is true if I am understanding the idea of implication correctly. So why is this case not possible? — Mardia, Sep 17 '20 at 12:56
A false statement satisfies P(0) to be true is simply no sense. Either it is true or false. — Mauro ALLEGRANZA, Sep 17 '20 at 13:05
@MauroALLEGRANZA Maybe this is where I am confused. If we say a statement $P(n)$ to be true for all $n$ but if we have a number $k$ such that $p(k)$ is false then isn't the statement false? — Mardia, Sep 17 '20 at 13:09
Notice the only way $ P(0) \land [P(0)\implies P(1)]$ is true is that both P(0) and P(1) are true. Procceed by induction and for every n must hold. — Garmekain, Sep 17 '20 at 13:38
@Garmekain That seem to make sense. So where did I go wrong when I thought $P(n)→P(n+1)$ is true while $P(n)$ is false but $P(n+1)$ is true? Am I understanding the idea of logical implication wrong? — Mardia, Sep 17 '20 at 13:52
You are just not applying the base case. It is true that "for all n: P(n)->P(n+1)" does not imply that P(n) is true(it can be false in every case), but with the base case the only way that can be true is that for all n is true. — Garmekain, Sep 17 '20 at 14:15
@Garmekain I think I am almost there. If the base case is true, the only way for $P(n)→P(n+1)$ to be true is for when it applies all $n$ because of the nature of mathematics? I'm thinking of an equation $x(n)$ that is not true for all $n$ can't get to $x(n+1)$ by proper mathematical steps. — Mardia, Sep 17 '20 at 14:36
If you prove P(0) and assuming P(n), prove P(n+1), the only way this can be true is for every n is true by nature of implication. — Garmekain, Sep 17 '20 at 15:16

score 3 · Accepted Answer · answered Sep 18 '20 at 05:08

The Principle of Mathematical Induction

For any unary predicate $P$ we have:

$$P(0) \land \forall a\in N:[P(a) \implies P(a+1)] \implies \forall b\in N: P(b)~~~~(1)$$

Or equivalently:

$$\exists b\in N: \neg P(b) \implies [P(0) \implies \exists a\in N:[P(a) \land \neg P(a+1)]]~~~~(2)$$

Suppose you want to prove the hypothesis that $\forall b\in N: P(b)$ and that you have already shown that $P(0)$ is true.

Let $k$ be any element of $N$.

We can prove that $P(k)\implies P(k+1)$ in the usual way by first assuming that $P(k)$ is true and then proving that $P(k+1)$ must also be true.

Technically, we could also do either of the following (not necessarily the same thing for each value of $k$):

Prove both $P(k)$ and $P(k+1)$ are true (corresponding to line 1 of the truth table)
Assume $P(k)$ is true and $P(k+1)$ is false. Then obtain a contradiction (corresponding to line 2 of the truth table)

What about simply proving $P(k)$ is false as you suggested (corresponding to lines 3 and 4 of the truth table)? As we see from (2) above, doing so would falsify the initial hypothesis.

Can you explain why the technique you use in inductive step proves the statement $P(k)⟹P(k+1) $? — Mardia, Sep 20 '20 at 04:14
@Mardia From the truth table for $A \implies B$, we see, in line 1, that if $A$ is true and $B$ is true, then '$A\implies B$' is true. On line 2, we see that if it is false that both $A$ true and $B$ is false, then '$A\implies B$' is true. — Dan Christensen, Sep 21 '20 at 04:24

score 1 · Answer 2 · answered Sep 17 '20 at 08:59

1

Long comment

Step 2), the so-called Inductive step is a little bit different...

Let assume that $P(n)$ is true and prove $P(n+1)$, that amounts to proving the conditional: $P(n) → P(n+1)$.

In terms of proof, we assume that $P(n)$ holds for an "unspecified" $n$ and we prove $P(n) \to P(n+1)$. Due to the fact that $n$ is whatever, we can "generalize" it to $\forall n (P(n) \to P(n+1)$.

Now it works...

We start with $P(0)$: it is true. Then use $\forall n (P(n) \to P(n+1)$ and instantiate it with $n=0$, that means: $P(0) \to P(0+1)$.

Use $P(0)$ and Modus Ponens to get: $P(1)$.

And so on... The structure of the inductive proof will guarantee us that the "chain" of inference will never break.

answered Sep 17 '20 at 08:59

Mauro ALLEGRANZA

99,247

So why can’t there be a false statement such that P(n) is false but P(n+1) is true? Is there a proof for showing that there is no such P(n)? – Mardia Sep 17 '20 at 12:36
@Obviously there can be... Your example fails for $n=0$. – Mauro ALLEGRANZA Sep 17 '20 at 12:44
Sorry I don't understand... I am aware that the example I provided is wrong but I still gave the example to better show where I am confused. So you are saying there is such a statement P(n) that satisfies the condition of the mathematical induction? – Mardia Sep 17 '20 at 12:49
1

@Mardia If $p(N)$ is false for some particular $N$, then pick the least such $N$ (can do by well ordering of non-negative integers). Then $N$ can't be $0$ since $P(0)$ holds, and so $P(N-1)$ holds by choice of $N$, and hence $P(N)$ holds because the induction step (2) is for ALL $n$, including $n = N-1$. Contradiction. – Ned Sep 17 '20 at 13:06
@Ned I am guessing here that induction step is already proven to be true. But what if that induction step was proven using a false statement? – Mardia Sep 17 '20 at 13:14
@Mardia Want do mean "using" a false statement"? You prove "For any $n$,if $P(n)$ then $P(n+1)$". That WHOLE sentence is true, with the quantifier and the implication. Now follow Mauro's argument which shows why you can then conclude, step by step, that each of the following are true: $P(0), P(1), P(2), P(3), ...., P(n), P(n+1), ...$, so there is no first counterexample -- and hence no counterexamples, period. – Ned Sep 17 '20 at 20:52
Last try... if P(n) is false for e g n=100 and true for n=99, as already explained, $P(99) \to P(100)$ will be false and this the formula $\forall n (P(n) \to P(n+1))$ false and thus the proof fails. – Mauro ALLEGRANZA Sep 18 '20 at 08:51

Mathematical Induction and implication

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