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While proving a statement of $f(n)$ using mathematical induction we do the following-

  • we prove it for some natural number which satisfies the condition of $n$.
  • We assume it true for some $k$.
  • Then we try to prove it for $k+1$ by using the statement obtained by assuming it to be true for $k$.

Now,my question is that we see that the proof using $k+1$ is totally dependent on proof using $k$.But there is no guarantee that the statement is true for k.We have only assumed it.Our assumption maybe right or wrong.The how can be prove $k+1$ on the basis of k if it is not certain that k is true. Please help.Thank You.

Vincenzo Oliva
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Soham
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  • Think of induction like climbing a ladder. The base case is "Prove you can reach rung number $1$." The inductive step is "Prove that, for any $k$, if you can reach rung $k$ then you can reach rung $k+1$." Once you've proven these, what does this mean? It means you can reach rung $1$. Now, you can also reach rung $2$ — just apply the inductive step to $k=1$. Now you can reach rung $3$ — inductive step with $k=2$ — and you can reach rung $4$ and you can reach rung $5$… (Why did I need to prove you can reach rung $2$ before I proved that you could reach rung $3$?) – Akiva Weinberger Aug 18 '15 at 09:33

3 Answers3

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Let's say you have proved that if $f(k)$ is true, then $f(k + 1)$ is true. In your base case, you have also proved that $f(n)$ is true for some $n$. Now, with both these in mind, we can conclude that $f(n + 1), f(n + 2), f(n + 3), \dots$ are all true.

In other words, the assumption that $f(k)$ is true may be right or wrong for arbitrary $k$, but there exists at least one $k$ for which it is true, namely $k = n$ which was shown in the base case.

mrp
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Indeed what you described in the second part is not complete. As it stands, it is a proof of a conditional : if $P(k)$ is true, then so is $P(k+1)$. What you're missing is that you have also a base case. You showed there is an $n$ such that $P(n)$ holds, and thus your proposition is true for all $k\ge n$.

Vincenzo Oliva
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  • If by showing $k_0$ is true for P(n) it verifies that the proposition is true for all $k>=n$ , then by showing the proof holds for any one number we can confirm that it holds for all other numbers.Then there is no need to assume any k or (k+1) – Soham Aug 18 '15 at 07:58
  • @tatan No, hold on. There is no $n$ with such a property. The " for all $k\ge n$ " step follows exactly from having proved that if the proposition holds for any one number, then it does for its successor also. Like dominoes. – Vincenzo Oliva Aug 18 '15 at 08:01
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Let' clarify the concept on a specific example.

We know that $$1+2+\cdots +n=\frac{n(n+1)}2.\tag 1$$

Prove this statement a fake naive way:

  1. Observe that $(1)$ is true for $n=2$. Indeed, $$3=1+2=\frac {2\times 3}2.\tag 2$$
  2. Observe that you can prove the same for $n=3$ based on $(2)$ now: $$6=1+2+3=\frac{2\times3}2+3=\frac{2\times3+2\times3}2=\frac{3\times 4}{2}.\tag 3$$
  3. You do it once again for $1+2+3+4$ based on $(3)$: $$1+2+3+4=(1+2+3)+4=$$$$=\frac{3\times 4}2+4=\frac{3\times4+2\times 4}2=\frac{4\times5}2.$$
  4. You notice the pattern now and try to do it in general:

$$1+2+\cdots n+n+1=\frac{n(n+1)}2+n+1=$$ $$=\frac{n(n+1)+2(n+1)}2=\frac{(n+1)(n+2)}2.$$ You generalize the idea now: If I could do it for $2$ and if I could do it for $n+1$ knowing that it is true for $n$ then it will be true for any natural number.

One more step ahead and you say that if this principle worked for the property $$1+2+\cdots +n=\frac{n(n+1)}{2}$$ then it will work for any property $P(n)$.

We've seen that every proof based on the principle of mathematical induction could be done step by step for any $n$. In that case you do not assume that the statement is true for an $n$ but you directly prove that. So, mathematical induction is a practical simplification of a special kind of inductive proof as long as we talk about the truth of a statement for finitely many $n$'s. The principle of mathematical induction becomes an unproven axiom if it comes to the statement that something is true for all $n$.

zoli
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  • Is it always that if a statement it true for $n$,it will be true for the next $(n+x)$ for some $x$ which satisfies the condition for $n$ as specified in the problem? – Soham Aug 18 '15 at 09:04
  • @tatan: I don't get that. The second step: If something is assumed (or proved; step by step) to be true for an $n$ and it can be proved that if it is true for an $n$ then it is true for $n+1$ then it is true for all $n$. – zoli Aug 18 '15 at 09:11