Chapter 5 Rosen

• mathematical induction for mathematical equalities and inequalities
• strong induction

First of all, mathematical induction is not a way to derive useful equations.

Must come up with the possible equation first, then prove it is true.

Like, what is the sum of $1 + 2 + 3 + \cdots + n$?

Try it on the board.

First of all, mathematical induction is not a way to derive useful equations.

Must come up with the possible equation first, then prove it is true.

Like, what is the sum of $1 + 2 + 3 + \cdots + n$?

We came up with $$\begin{align*} 1 + 2 + 3 + \cdots + n &= \frac{n^2}{2} + \frac{n}{2} = \frac{n(n+1)}{2} \end{align*}$$

Are you sure?

Seems to work for a few small values of $n$ that we try.

How to be sure it is true for all $n \ge 1$?

Yep, mathematical induction.

Prove $P(n)$: $1+2+\cdots+n = \frac{n(n+1)}{2}$ for $n \ge 1$.

Basis:

Inductive Hypothesis:

Inductive Step:

Prove $P(n)$: $1+2+\cdots +n = \frac{n(n+1)}{2}$ for $n \ge 1$.

Basis: $P(n=1)$: $1 = \frac{1(1+1)}{2} = \frac{2}{2} = 1$. True!

Inductive Hypothesis:

Inductive Step:

Prove $P(n)$: $1+2+\cdots +n = \frac{n(n+1)}{2}$ for $n \ge 1$.

Basis: $P(n=1)$: $1 = \frac{1(1+1)}{2} = \frac{2}{2} = 1$. True!

Inductive Hypothesis: $P(n=k)$: Assume $1+2+\cdots +k = \frac{k(k+1)}{2}$ is True.

Inductive Step:

Prove $P(n)$: $1+2+\cdots +n = \frac{n(n+1)}{2}$ for $n \ge 1$.

Basis: $P(n=1)$: $1 = \frac{1(1+1)}{2} = \frac{2}{2} = 1$. True!

Inductive Hypothesis: $P(n=k)$: Assume $1+2+\cdots +k = \frac{k(k+1)}{2}$ is True.

Inductive Step: Prove $P(n=k+1)$: $1+2+\cdots+(k+1) = \frac{(k+1)((k+1)+1)}{2}$

Let's start with the statement of $P(n=k+1)$, then find part of expression that matches part of inductive hypothesis.

$$\begin{align*} 1+2+\cdots +k + (k+1) &= \frac{(k+1)((k+1)+1)}{2},\;\;\;P(n=k+1)\\ \frac{k(k+1)}{2} + (k+1) &= \frac{(k+1)((k+1)+1)}{2},\;\;\;\mbox{by Ind. Hyp.}\\ \frac{k(k+1)}{2} + \frac{2(k+1)}{2} &= \frac{(k+1)((k+1)+1)}{2}\\ \frac{k(k+1)+2(k+1)}{2} &= \frac{(k+1)((k+1)+1)}{2}\\ \frac{k^2 + k + 2k + 2}{2} &= \frac{(k+1)((k+1)+1)}{2}\\ \frac{k(k+1) + 2k + 2}{2} &= \frac{(k+1)((k+1)+1)}{2}\\ \frac{k(k+1) + 2(k+1)}{2} &= \frac{(k+1)((k+1)+1)}{2}\\ \frac{(k+2)(k+1)}{2} &= \frac{(k+1)(k+2)}{2}\\ \end{align*}$$

Summarizing, prove $P(n)$: $1+2+\cdots +n = \frac{n(n+1)}{2}$ for $n \ge 1$.

Basis: $P(n=1)$: $1 = \frac{1(1+1)}{2} = \frac{2}{2} = 1$. True!

Inductive Hypothesis: $P(n=k)$: Assume $1+2+\cdots +k = \frac{k(k+1)}{2}$ is True.

Inductive Step: Prove $P(n=k+1)$: $1+2+\cdots+(k+1) = \frac{(k+1)((k+1)+1)}{2}$ $$\begin{align*} 1+2+\cdots +k + (k+1) &= \frac{(k+1)((k+1)+1)}{2},\;\;\;P(n=k+1)\\ \frac{k(k+1)}{2} + (k+1) &= \frac{(k+1)((k+1)+1)}{2},\;\;\;\mbox{by Ind. Hyp.}\\ \frac{(k+2)(k+1)}{2} &= \frac{(k+1)(k+2)}{2}\\ \end{align*}$$

The inductive step can go in the reverse direction. Start with the statement of the inductive hypothesis, for $P(n=k)$, that we assume is true. Manipulate both sides of the assumed equation until you get the statement of $P(n=k+1)$.

Inductive Step: Prove $P(n=k+1)$: $1+2+\cdots+(k+1) = \frac{(k+1)((k+1)+1)}{2}$ $$\begin{align*} 1+2+\cdots +k &= \frac{k(k+1)}{2},\;\;\;P(n=k)\\ 1+2+\cdots +k + (k+1) &= \frac{k(k+1)}{2} + k+1\\ 1+2+\cdots +k + (k+1) &= \frac{k(k+1)}{2} + \frac{2(k+1)}{2}\\ 1+2+\cdots +k + (k+1) &= \frac{(k+1)((k+1)+1)}{2} \end{align*}$$

Just to be absurd, what if you thought that $1 + 2 + \cdots + n = n^2$?

Try to prove with mathematical induction.

Basis: $P(n=1)$: $1 = 1^2 = 1$. True!

Inductive Hypothesis: $P(n=k)$: $1+2+\cdots +k = k^2$.

Inductive Step: $P(n=k+1)$: $1+2+\cdots + (k+1) = (k+1)^2$. $$\begin{align*} 1+2+\cdots+k + (k+1) &= (k+1)^2\\ k^2 + (k+1) &= (k+1)^2, \;\;\;\mbox{by Ind. Hyp.}\\ k^2 + (k+1) &= k^2 + 2k + 1\\ k^2 + k+1 &= k^2 + 2k + 1 \;\;\;\mbox{NOT TRUE} \end{align*}$$ We were able to derive something that we know is not true. What went wrong? Our assumption, the Inductive Hypothesis, must not be true!

Let's try another one. What is the sum $1 + 3 + 5 + 7 + \cdots + (2n-1)$?

First, try to find a formula for the sum. Do it on the white board.

Prove $P(n)$: $1+3+5+\cdots + (2n-1) = n^2$

Basis: $P(n=1)$: $1 = 1^2 = 1$. True!

Inductive Hypothesis: $P(n=k)$: $1+3+\cdots +(2k-1) = k^2$.

Inductive Step: $P(n=k+1)$: $1+3+\cdots + (2(k+1)-1) = (k+1)^2$. $$\begin{align*} 1+3+\cdots+(2k-1) + (2(k+1)-1) &= (k+1)^2\\ k^2 + (2(k+1)-1) &= (k+1)^2, \;\;\;\mbox{by Ind. Hyp.}\\ k^2 + 2k + 1 &= (k+1)^2\\ k^2 + 2k +1 &= k^2 + 2k + 1 \;\;\;\mbox{TRUE} \end{align*}$$

How about the geometric series $a + ar + ar^2 + \cdots + ar^n$?

What does it equal?

Remember this trick: $$\begin{align*} a + ar + ar^2 + \cdots + ar^n &= f(n)\\ r(a + ar + ar^2 + \cdots + ar^n) &= r f(n)\\ ar + ar^2 + \cdots + ar^n + ar^{n+1} &= r f(n)\\ \end{align*}$$ Subtract the first line from the third line to get $$\begin{align*} ar^{n+1} - a &= r f(n) - f(n)\\ \end{align*}$$ and solve for $f(n)$. $$\begin{align*} ar^{n+1} - a &= r f(n) - f(n)\\ ar^{n+1} - a &= (r-1) f(n)\\ f(n) &= \frac{ar^{n+1} - a}{r-1} \end{align*}$$

Prove $P(n)$: $a+ar+ar^2+\cdots+ar^n = \frac{ar^{n+1} - a}{r-1}$

Basis: $P(n=0)$: $a = \frac{ar^{0+1} - a}{r-1} = \frac{ar-a}{r-1} = \frac{a(r-1)}{r-1} = a$ TRUE!

Inductive Hypothesis: $P(n=k)$: $a+ar+\cdots+ar^k = \frac{ar^{k+1}-a}{r-1}$

Inductive Step:

Prove $P(n)$: $a+ar+ar^2+\cdots+ar^n = \frac{ar^{n+1} - a}{r-1}$

Basis: $P(n=0)$: $a = \frac{ar^{0+1} - a}{r-1} = \frac{ar-a}{r-1} = \frac{a(r-1)}{r-1} = a$ TRUE!

Inductive Hypothesis: $P(n=k)$: $a+ar+\cdots+ar^k = \frac{ar^{k+1}-a}{r-1}$

Inductive Step: $P(n=k+1)$: $a+ar+\cdots+ar^{k+1} = \frac{ar^{k+2}-a}{r-1}$ $$\begin{align*} a+ar+\cdots+ar^k+ar^{k+1} &= \frac{ar^{k+2}-a}{r-1}\\ \frac{ar^{k+1}-a}{r-1} + ar^{k+1} &= \frac{ar^{k+2}-a}{r-1}, \;\;\mbox{by Ind. Hyp.}\\ ar^{k+1}-a + (r-1)ar^{k+1} &= ar^{k+2}-a\\ ar^{k+1}-a + ar^{k+2} - ar^{k+1} &= ar^{k+2}-a\\ ar^{k+2}-a &= ar^{k+2}-a, \;\;\;\mbox{TRUE} \end{align*}$$

Which is bigger, $2^n$ or $n!$?

Which is bigger, $2^n$ or $n!$?

When $n \ge 4$, $n!$ is bigger.

Prove it.

Prove $P(n)$: $2^n < n!$, for $n \ge 4$.

Basis: $P(n=4)$: $2^4 < 4!$, $16 < 24$, TRUE

Inductive Hypothesis:

Inductive Step:

Prove $P(n)$: $2^n < n!$, for $n \ge 4$.

Basis: $P(n=4)$: $2^4 < 4!$, $16 < 24$, TRUE

Inductive Hypothesis: $P(n=k)$: $2^k < k!$.

Inductive Step: $P(n=k+1)$: $2^{k+1} < (k+1)!$.

$$\begin{align*} 2^{k+1} &< (k+1)!, \;\;P(n=k+1)\\ 2^{k+1} &< (k+1) k!,\;\;\mbox{definition of !}\\ 2^{k+1} &< (k+1) 2^k < (k+1) k!, \;\;\mbox{by Ind. Hyp.}\\ 2\; 2^k &< (k+1) 2^k, \;\;\mbox{TRUE, for } k > 1, \mbox{ which it is.}\\ \end{align*}$$

Looking at the sequence of $n^3-n$ values $$0^3-0, 1^3-1, 2^3-2, 3^3-3, 4^3-4,\ldots = 0, 0, 6, 24, 60, \ldots$$ it looks like they are all divisible by 3.

Prove it.

Prove $P(n)$: $n^3-n$ is divisible by 3, for $n > 0$.

Basis: $P(n=1)$: $1^3-1 = 1-1 = 0$, which is divisible by 3.

Inductive Hypothesis: $P(n=k)$: $k^3-k$ is divisible by 3.

Inductive Step: $P(n=k+1)$: $(k+1)^3-(k+1)$ is divisible by 3.

The trick here is to manipulate the expression to a form that looks like the inductive hypothesis.

Inductive Hypothesis: $P(n=k)$: $k^3-k$ is divisible by 3.

Inductive Step: $P(n=k+1)$: $(k+1)^3-(k+1)$ is divisible by 3.

The trick here is to manipulate the expression to a form that looks like the inductive hypothesis.

$$\begin{align*} (k+1)^3-(k+1) &= (k+1)(k^2+2k+1)-k-1\\ &= k^3 +2k^2 + k + k^2+2k+1 -k -1\\ &= k^3 -k +3k^2 + 3k\\ &= (k^3 -k) +3(k^2 + k) \end{align*}$$ By the Inductive Hypothesis, the first term in the last equation is divisible by 3. The last term is obviously divisible by 3, so the last line is divisible by 3, meaning $$\frac{(k^3 -k) +3(k^2 + k)}{3}$$ is an integer.

Sometimes an inductive proof requies the Inductive Hypothesis to include all values of $n$ from a minimum value up to $k$. This is strong induction.

Show that $n$ can be written as a product of primes, for $n>1$.

Basis: $P(n=2)$: $2 = 2\cdot 1$, two primes.

Inductive Hypothesis: $P(n=2,\ldots,k)$: $k$ can be written as product of two primes

Inductive Step: $P(n=k+1)$: $k+1$ is the product of two primes.

If $k+1$ is prime, then statement is true.

If $k+1$ is not prime, then $k+1 = a\,b$ where $2 \le a \le b \le k+1$. Using the Inductive Hypothesis, $a$ and $b$ can be written as the product of primes, so $a\,b$ is the product of primes, so $k+1=a\,b$ is a product of primes.

Exercise 30 in Rosen. What is wrong with the following proof that $a^n = 1$ for all $n\ge0$?

Basis: $P(n=0)$: $a^0 = 1 = 1$. True!

Inductive Hypothesis: $P(n=0,\ldots,k)$: $a^k = 1$.

Inductive Step: $P(n=k+1)$: $$\begin{align*} a^{k+1} &= \frac{a^k a^k}{a^{k-1}}\\ a^{k+1}&= \frac{1 \cdot 1}{1}\\ a^{k+1}& = 1 \end{align*}$$