2

I know the principle of mathematical induction. The only thing that causes my confusion is that we suppose a statement is true for $n=k$ then we prove the statement is also true for $n=k+1$ but how can we suppose $n=k$ to be true? What if a statement is true for $n=k+1$ and is not true for $n=k$? Does $k$ mean to be starting from $1$ or $2$ if in the base case we prove the statement to be valid for $n=1$? Please help me with this confusion.

pjs36
  • 17,979
Waqar
  • 267
  • 2
    We use the base case to prove the inductive case. Think of it as a ladder: If we prove it holds for $n=1$, then it also holds for $n=2$. Since it holds for $n=2$, it holds for $n=3$, and so on. An inductive proof is valid if and only if both the base case and the inductive case are valid. – Edward Jiang Apr 25 '16 at 19:54
  • 1
    "What if a statement is not true for n=k?" - then we cannot use it as the first step (aka the assumption) of the proof. – barak manos Apr 25 '16 at 19:54
  • 2
    We prove that: if it is true for $n=k$, then it is true for $n=k+1$. This allows us to start from $n=0$ (we already have proved it true) and apply the previous reault, to deduce it for $n=1$, and so on... – Mauro ALLEGRANZA Apr 25 '16 at 19:56
  • The base case depends on what you are trying to prove. If the problem says to prove it for all n, then your base case is n = 0 or n = 1 depending on your definition of the natural numbers. Sometimes you will be asked to prove that the statement is true for n >= k. Suppose (for example k =5). Then your base case will be n = 5. – user137481 Apr 25 '16 at 19:59
  • Keep in mind that a proof by induction is two steps- proving that the statement holds for $n=1$ (your starting point can be anything, though typically an integer, but it's usually $1$), the base step, and proving that if the statement holds for $n=k$, then it holds for $n=k+1$, the inductive step. Neither step by itself is sufficient, but if we have both, then we have an inductive proof, because we have that it holds for $n=1$ by the base step, and so it holds for $n=2$ by the inductive step, and for $n=3$ similarly, and $n=4$, and $n=5$, and... – Kevin Long Apr 25 '16 at 20:01
  • 6
    See Mathematical induction: why can we assume that $P(k)$ holds. – Mauro ALLEGRANZA Apr 25 '16 at 20:16
  • Imagine you are holding a bucket of natural numbers. Suppose you can show that the number 1 is in your bucket. Suppose further that if you find any number (k) in your bucket, you will somehow be able to find the next number (k+1) in there. Since 1 is in there, 2 must also be in there. Since 2 is there, 3 must also be in there. And so on. It stands to reason that you must have ALL natural numbers in your bucket – Dan Christensen May 01 '16 at 20:01

2 Answers2

1

Imagine the theorem you are trying to prove is a set of stairs and proving the theorem amounts to getting to the top of the stairs in this analogy. Proving the base case amounts to climbing the first step. Think of the inductive step as follows: if you manage to get yourself to the $n^{th}$ step, then you can get to the $(n +1)^{th}$ step. So if you can get to the $1^{st}$ step (base case), you can get to the next step, i.e. the $2^{nd}$ step. But then you can get to the $3^{rd}$, and then to the $4^{th}$, and so on until you eventually reach the top of the stairs and yell "Huzzah!" because you have proved your theorem.

Oiler
  • 3,413
0

Suppose we show, for a range of values of $k$, that $S(k)\implies S(k+1)$ for any $ k$ in the range.. This does not imply that $S(k)$ is true for any $k$. For example,when $S(k)$ is $k=k+1$ then we can prove that $S(k)\implies S(k+1).$

If you can exhibit some $k_0\in N$ such that $S(k_0)$ is true, and if you can show that $S(k)\implies S(k+1)$ for any natural number $k\geq k_0,$ then you have proved $\forall k\in N\;(k\geq k_0\implies S(k).$

Explanation : We can show that if $\exists k\geq k_0\; (\;\neg S(k)\;)$ then we have a paradox: There would be a least such $k$, which we can call $k_1$ . Now $k_1>k_0$ because we know $S(k_0)$ is true. Then $k_1-1\geq k_0,$ and $S(k_1-1)$ is true, otherwise $k_1$ would not be the LEAST $k\geq k_0$ such that $\neg S(k).$ But then we have $S(k_1-1)$ and also $S(k_1-1)\implies S(k_1)$ , so we have $S(k_1),$ contradicting $\neg S(k_1)$.

DanielWainfleet
  • 57,985