Section 6.2 Basic induction
Suppose we want to show that n! is at least 2^n-2\text{,} for every n \ge 1 (where n must be an integer). We could start verifying this fact for each of the possible values for n\text{:}Theorem 6.2.1 (Principle of Mathematical Induction).
Let P(n) be an assertion about the integer n\text{.} If we know that
the assertion P(n_0) is true for some particular integer n_0\text{;} and
for any integer k \ge n_0\text{,} if P(k) is true then P(k+1) must also be true,
then P(n) is true for every integer n \ge n_0\text{.}
Definition 6.2.2.
In a proof by induction, determining that P(n_0) is true for some particular integer n_0 is called the base case. Proving the conditional statement that P(k) \Rightarrow P(k+1) for every k \ge n_0 is called the inductive step. The assumption we make in the inductive step, that P(k) is true for some arbitrary k \ge n_0\text{,} is called the inductive hypothesis, and can be referred to by (IH) when it is being used in the proof.
Example 6.2.3.
Prove by induction that \(n!\ge 2^n-2\text{,}\) for every integer \(n \ge 2\text{.}\) (This inequality is actually true for every \(n \ge 0\text{,}\) but the proof is considerably simpler if we restrict our attention to \(n \ge 2\text{.}\))
Base case: \(n=2\text{.}\) We have \(n!=2!=2\text{,}\) and
Certainly \(2 \ge 2\text{,}\) so the inequality holds for \(n=2\text{.}\) This completes the proof of the base case.
Inductive step: We begin with the inductive hypothesis. Let \(k \ge 2\) be arbitrary, and suppose that the inequality holds for \(n=k\text{;}\) that is, assume that \(k! \ge 2^k-2\text{.}\)
Now we want to deduce that
Let's start from the left-hand side of this inequality. By the definition of factorial, we know that
Now that we have \(k!\) in the expression, we're in a position to apply the inductive hypothesis; that is,
Since \(k \ge 2\text{,}\) we have \(k+1 \ge 3\text{,}\) so
Again, since \(k \ge 2\text{,}\) we have \(2^k \ge 4\text{,}\) so \(2^k-6 \ge -2\text{.}\) Hence
which is what we wanted to deduce. This completes the proof of the inductive step.
By the Principle of Mathematical Induction, \(n! \ge 2^n-2\) for every integer \(n \ge 2\text{.}\)
Example 6.2.4.
Consider the sum of the first \(n\) integers. We can think about this as a recursively-defined sequence, by defining \(s_1=1\text{,}\) and \(s_{n}=s_{n-1}+n\text{,}\) for every \(n \ge 2\text{.}\) Thus, \(s_2=1+2\text{;}\)
and so on. Prove by induction that \(s_n=n(n+1)/2\text{,}\) for every \(n \ge 1\text{.}\)
Base case: \(n=1\text{.}\) We have \(s_n=s_1=1\text{,}\) and
so the equality holds for \(n=1\text{.}\) This completes the proof of the base case.
Inductive step: We begin with the inductive hypothesis. Let \(k \ge 1\) be arbitrary, and suppose that the equality holds for \(n=k\text{;}\) that is, assume that \(s_k=k(k+1)/2\text{.}\)
Now we want to deduce that
Using the recursive relation, we have \(s_{k+1}=s_k+(k+1)\) since \(k+1 \ge 2\text{,}\) and using the inductive hypothesis, we have \(s_k=k(k+1)/2\text{,}\) so putting these together, we see that
Taking out a common factor of \(k+1\) gives
which is what we wanted to deduce. This completes the proof of the inductive step.
By the Principle of Mathematical Induction, \(s_n=n(n+1)/2\) for every \(n \ge 1\text{.}\)
Example 6.2.5.
Here is a βproof by inductionβ (without a base case) that every integer \(n\) is at least \(1000\text{.}\)
Inductive step: We begin with the inductive hypothesis. Let \(k\) be arbitrary, and suppose that \(k \ge 1000\text{.}\)
Now we want to deduce that \(k+1 \ge 1000\text{.}\) But clearly,
(by our inductive hypothesis), which is what we wanted to deduce. This completes the proof of the inductive step.
By the Principle of Mathematical Induction, \(n \ge 1000\) for every integer \(n\text{.}\)
Exercises 6.2.6.
Use the Principle of Mathematical Induction to prove the following:
For the recursively-defined sequence given by \(b_1=5\) and \(b_n=b_{n-1}+4\) for all \(n \ge 2\text{,}\) prove that for every integer \(n \ge 1\text{,}\) \(b_n=5+4(n-1)\text{.}\)
For the recursively-defined sequence given by \(c_1=3\) and \(c_n=c_{n-1}+3 \cdot 2^{n-1}\) for all \(n \ge 2\text{,}\) prove that for every integer \(n \ge 1\text{,}\) \(c_n=3(2^n-1)\text{.}\)
Prove that for every integer \(n \ge 0\text{,}\) \(n! \ge n\text{.}\)
Prove that for every integer \(n \ge 0\text{,}\) \(4^{n}-1\) is divisible by 3.
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Starting with \(n=2\) and increasing \(n\) from there, calculate the first few values for the product
\begin{equation*} t_n=\prod_{j=2}^n (1-\frac{1}{j})\text{.} \end{equation*}Conjecture a closed formula for \(t_n\) based on the values you have calculated, and use induction to prove that your formula is correct.
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Prove that for every integer \(n \ge 1\text{,}\)
\begin{equation*} \sum_{j=1}^{n}j! \le \frac{1}{2}(n+1)! \end{equation*} -
Define \(c_0=1\) and for \(n \ge 1\text{,}\) define \(c_n=n c_{n-1}+1\text{.}\) Prove by induction: for \(n \ge 0\text{,}\)
\begin{equation*} c_n = \sum_{j=0}^n\frac{n!}{(n-j)!}\text{.} \end{equation*}