 # PreTeXt Sample Book: Abstract Algebra (SAMPLE ONLY)

## Section2.1Mathematical Induction and Math in a Title $$A\notsubset B$$

Suppose we wish to show that
\begin{equation*} 1 + 2 + \cdots + n = \frac{n(n + 1)}{2} \end{equation*}
for any natural number $$n\text{.}$$ This formula is easily verified for small numbers such as $$n = 1\text{,}$$ 2, 3, or 4, but it is impossible to verify for all natural numbers on a case-by-case basis. To prove the formula true in general, a more generic method is required.
Suppose we have verified the equation for the first $$n$$ cases. We will attempt to show that we can generate the formula for the $$(n + 1)$$th case from this knowledge. The formula is true for $$n = 1$$ since
\begin{equation*} 1 = \frac{1(1 + 1)}{2}\text{.} \end{equation*}
If we have verified the first $$n$$ cases, then
\begin{align*} 1 + 2 + \cdots + n + (n + 1) & = \frac{n(n + 1)}{2} + n + 1\\ & = \frac{n^2 + 3n + 2}{2}\\ & = \frac{(n + 1)[(n + 1) + 1]}{2}\text{.} \end{align*}
This is exactly the formula for the $$(n + 1)$$th case.
This method of proof is known as mathematical induction. Instead of attempting to verify a statement about some subset $$S$$ of the positive integers $${\mathbb N}$$ on a case-by-case basis, an impossible task if $$S$$ is an infinite set, we give a specific proof for the smallest integer being considered, followed by a generic argument showing that if the statement holds for a given case, then it must also hold for the next case in the sequence. We summarize mathematical induction in the following axiom.
For all integers $$n \geq 3\text{,}$$ $$2^n \gt n + 4\text{.}$$ Since
\begin{equation*} 8 = 2^3 \gt 3 + 4 = 7\text{,} \end{equation*}
the statement is true for $$n_0 = 3\text{.}$$ Assume that $$2^k \gt k + 4$$ for $$k \geq 3\text{.}$$ Then $$2^{k + 1} = 2 \cdot 2^{k} \gt 2(k + 4)\text{.}$$ But
\begin{equation*} 2(k + 4) = 2k + 8 \gt k + 5 = (k + 1) + 4 \end{equation*}
since $$k$$ is positive. Hence, by induction, the statement holds for all integers $$n \geq 3\text{.}$$
Every integer $$10^{n + 1} + 3 \cdot 10^n + 5$$ is divisible by 9 for $$n \in {\mathbb N}\text{.}$$ For $$n = 1\text{,}$$
\begin{equation*} 10^{1 + 1} + 3 \cdot 10 + 5 = 135 = 9 \cdot 15 \end{equation*}
is divisible by 9. Suppose that $$10^{k + 1} + 3 \cdot 10^k + 5$$ is divisible by 9 for $$k \geq 1\text{.}$$ Then
\begin{align*} 10^{(k + 1) + 1} + 3 \cdot 10^{k + 1} + 5& = 10^{k + 2} + 3 \cdot 10^{k + 1} + 50 - 45\\ & = 10 (10^{k + 1} + 3 \cdot 10^{k} + 5) - 45 \end{align*}
is divisible by 9.
We will prove the binomial theorem using mathematical induction; that is,
\begin{equation*} (a + b)^n = \sum_{k = 0}^{n} \binom{n}{k} a^k b^{n - k}\text{,} \end{equation*}
where $$a$$ and $$b$$ are real numbers, $$n \in \mathbb{N}\text{,}$$ and
\begin{equation*} \binom{n}{k} = \frac{n!}{k! (n - k)!} \end{equation*}
is the binomial coefficient. We first show that
\begin{equation*} \binom{n + 1}{k} = \binom{n}{k} + \binom{n}{k - 1}\text{.} \end{equation*}
This result follows from
\begin{align*} \binom{n}{k} + \binom{n}{k - 1} & = \frac{n!}{k!(n - k)!} +\frac{n!}{(k-1)!(n - k + 1)!}\\ & = \frac{(n + 1)!}{k!(n + 1 - k)!}\\ & =\binom{n + 1}{k}\text{.} \end{align*}
If $$n = 1\text{,}$$ the binomial theorem is easy to verify. Now assume that the result is true for $$n$$ greater than or equal to 1. Then
\begin{align*} (a + b)^{n + 1} & = (a + b)(a + b)^n\\ & = (a + b) \left( \sum_{k = 0}^{n} \binom{n}{k} a^k b^{n - k}\right)\\ & = \sum_{k = 0}^{n} \binom{n}{k} a^{k + 1} b^{n - k} + \sum_{k = 0}^{n} \binom{n}{k} a^k b^{n + 1 - k}\\ & = a^{n + 1} + \sum_{k = 1}^{n} \binom{n}{k - 1} a^{k} b^{n + 1 - k} + \sum_{k = 1}^{n} \binom{n}{k} a^k b^{n + 1 - k} + b^{n + 1}\\ & = a^{n + 1} + \sum_{k = 1}^{n} \left[ \binom{n}{k - 1} + \binom{n}{k} \right]a^k b^{n + 1 - k} + b^{n + 1}\\ & = \sum_{k = 0}^{n + 1} \binom{n + 1}{k} a^k b^{n + 1- k}\text{.} \end{align*}
We have an equivalent statement of the Principle of Mathematical Induction that is often very useful.
A nonempty subset $$S$$ of $${\mathbb Z}$$ is well-ordered if $$S$$ contains a least element. Notice that the set $${\mathbb Z}$$ is not well-ordered since it does not contain a smallest element. However, the natural numbers are well-ordered.
The Principle of Well-Ordering is equivalent to the Principle of Mathematical Induction.
Let $$S = \{ n \in {\mathbb N} : n \geq 1 \}\text{.}$$ Then $$1 \in S\text{.}$$ Now assume that $$n \in S\text{;}$$ that is, $$n \geq 1\text{.}$$ Since $$n+1 \geq 1\text{,}$$ $$n+ 1 \in S\text{;}$$ hence, by induction, every natural number is greater than or equal to 1.
We must show that if $$S$$ is a nonempty subset of the natural numbers, then $$S$$ contains a least element. If $$S$$ contains 1, then the theorem is true by Lemma 2.1.7. Assume that if $$S$$ contains an integer $$k$$ such that $$1 \leq k \leq n\text{,}$$ then $$S$$ contains a least element. We will show that if a set $$S$$ contains an integer less than or equal to $$n + 1\text{,}$$ then $$S$$ has a least element. If $$S$$ does not contain an integer less than $$n+1\text{,}$$ then $$n+1$$ is the smallest integer in $$S\text{.}$$ Otherwise, since $$S$$ is nonempty, $$S$$ must contain an integer less than or equal to $$n\text{.}$$ In this case, by induction, $$S$$ contains a least element.
Induction can also be very useful in formulating definitions. For instance, there are two ways to define $$n!\text{,}$$ the factorial of a positive integer $$n\text{.}$$
• The explicit definition: $$n! = 1 \cdot 2 \cdot 3 \cdots (n - 1) \cdot n\text{.}$$
• The inductive or recursive definition: $$1! = 1$$ and $$n! = n(n - 1)!$$ for $$n \gt 1\text{.}$$
Every good mathematician or computer scientist knows that looking at problems recursively, as opposed to explicitly, often results in better understanding of complex issues.