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Chapter 4 Principle of Mathematical Induction (Concepts)
Welcome to this fascinating chapter that introduces a unique and remarkably powerful proof technique known as the Principle of Mathematical Induction, often abbreviated as PMI. While other proof methods like direct proof or proof by contradiction rely on logical deduction from axioms or established theorems, mathematical induction provides a specific framework for proving statements, propositions, or formulas that are claimed to be true for an infinite sequence of cases – typically, for all positive integers ($n = 1, 2, 3, \dots$), or sometimes for all integers greater than or equal to some starting integer.
The underlying logic of PMI is analogous to the domino effect. Imagine an infinite line of dominoes. To ensure all dominoes fall, you need to satisfy two conditions:
- The first domino must be knocked over.
- If any domino falls, it must knock over the next domino in line.
- Condition 1 (Base Case): The statement $P(n)$ is true for the first possible integer, usually $n=1$.
- Condition 2 (Inductive Link): If we assume the statement $P(n)$ is true for some arbitrary positive integer $k$, then it must also be true for the next integer, $k+1$.
The process of constructing a proof using the Principle of Mathematical Induction involves three distinct, mandatory steps:
- Base Step (or Basis Step): Show that the statement $P(n)$ holds true for the initial value specified (commonly $n=1$). This involves substituting $n=1$ into the statement $P(n)$ and verifying its truth directly. This corresponds to knocking over the first domino.
- Inductive Hypothesis: Assume that the statement $P(n)$ is true for some arbitrary, but fixed, positive integer $k$. That is, we explicitly assume $P(k)$ is true. This assumption is the crucial link in the inductive argument – it's like assuming the $k^{th}$ domino falls.
- Inductive Step: Using the Inductive Hypothesis (the assumption that $P(k)$ is true) as a premise, rigorously prove that the statement must also be true for the next integer, $n = k + 1$. That is, we must demonstrate logically that $\mathbf{P(k) \implies P(k+1)}$ (if $P(k)$ is true, then $P(k+1)$ is true). This step involves algebraic manipulation, starting with one side of the $P(k+1)$ statement and using the $P(k)$ assumption to transform it into the other side. This corresponds to showing that if the $k^{th}$ domino falls, it will necessarily cause the $(k+1)^{th}$ domino to fall.
If all three steps – establishing the Base Step, stating the Inductive Hypothesis, and successfully proving the Inductive Step – are completed, then we can invoke the Principle of Mathematical Induction to conclude that the statement $P(n)$ is true for all positive integers $n$ (or all integers from the base case onwards).
We will practice applying this structured proof technique to establish the validity of various mathematical results, including:
- Formulas for sums of series, such as the sum of the first $n$ natural numbers ($\sum n = \frac{n(n+1)}{2}$), the sum of the first $n$ squares ($\sum n^2$), or the sum of the first $n$ cubes ($\sum n^3$).
- Formulas related to terms or sums in arithmetic or geometric progressions.
- Statements concerning divisibility (e.g., proving that $7^n - 3^n$ is divisible by $4$ for all positive integers $n$).
- Proving various mathematical inequalities that depend on $n$.
Mathematical Statement
The Principle of Mathematical Induction is a powerful technique used to prove statements or propositions that are claimed to be true for all natural numbers, or for all natural numbers greater than or equal to some initial natural number. Before we delve into the principle itself, it's important to clearly understand what kind of statements we can prove using this method.
In the context of mathematical induction, we work with assertions or claims that depend on a natural number. These statements typically describe a formula, an equation, an inequality, or some property related to a natural number $n$. Our goal is to prove that such a statement holds true for an infinite sequence of natural numbers.
What is a Mathematical Statement?
A mathematical statement (also called a proposition or an assertion) is a sentence that is precise enough to be definitively classified as either true or false. It cannot be both true and false simultaneously. For example, the sentence "5 is an odd number" is a mathematical statement because it is true. The sentence "All squares are circles" is a mathematical statement because it is false.
Sentences that are subjective or whose truth value depends on context are generally not considered mathematical statements in this strict sense (e.g., "This number is large" is not a mathematical statement).
Deduction, Induction, and Conjecture
In mathematics, we use different forms of reasoning to arrive at conclusions. Understanding the distinction between them is key to appreciating the role of mathematical induction.
Deductive Reasoning
Deduction is the process of reasoning from one or more general statements (premises) to reach a logically certain and specific conclusion. If the premises are true, the conclusion must also be true. This is a "top-down" approach.
A classic example is:
- All men are mortal. (General premise)
- Socrates is a man. (Specific premise)
- Therefore, Socrates is mortal. (Specific conclusion)
Most proofs in geometry and algebra are deductive. We start with axioms, definitions, and previously proven theorems to deduce a new result.
Inductive Reasoning
Inductive reasoning (in the everyday sense) is the process of making a generalization based on a set of specific observations. It involves moving from specific instances to a broader, probable conclusion. This is a "bottom-up" approach.
For example, if you observe that the sun has risen every day of your life, you might form the general conclusion: "The sun rises every day." While this conclusion is very likely to be true tomorrow, it is not logically certain. This form of reasoning is central to the scientific method but is not sufficient for a mathematical proof because a single counterexample can invalidate the generalization.
Conjecture
A conjecture is a mathematical statement that is believed to be true based on the evidence gathered from inductive reasoning (i.e., checking many specific cases), but it has not yet been formally proven through deductive methods. A conjecture is essentially an educated guess that awaits a rigorous proof.
For example, after observing that $1 = \frac{1(2)}{2}$ and $1+2 = \frac{2(3)}{2}$, one might form the conjecture that the sum of the first $n$ natural numbers is always $\frac{n(n+1)}{2}$. This remains a conjecture until it is formally proven for all natural numbers.
The "Principle of Mathematical Induction" is a formal, deductive proof technique that allows us to take a conjecture that appears true and prove it with logical certainty for all natural numbers.
Notation for Mathematical Statements
In mathematical induction, we focus on statements whose truth depends on a natural number variable, usually denoted by $n$. To represent such a statement in a concise way, we use a special notation.
We represent the statement by $P(n)$.
- The letter P stands for the "Proposition" or "Property" being described. It is like the name of the statement.
- The variable (n) indicates that the statement's truth value depends on the natural number $n$.
For example, if our statement is "The number $n$ is greater than 3", we would write it as:
$P(n): n > 3$
By substituting a specific natural number for $n$, say $k$, we get a specific statement $P(k)$ which is either true or false.
- $P(2)$ would be the statement "$2 > 3$", which is False.
- $P(5)$ would be the statement "$5 > 3$", which is True.
This notation allows us to refer to the general statement as $P(n)$ and to specific instances of the statement as $P(1)$, $P(2)$, $P(k)$, and so on.
Examples of Mathematical Statements $P(n)$ involving Natural Numbers:
Let's look at a few examples of statements $P(n)$ that involve a natural number $n$ and see how we can evaluate their truth value for specific values of $n$. Remember that the set of natural numbers is $\mathbb{N} = \{1, 2, 3, 4, ...\}$.
1. Consider the statement $P(n)$: "The sum of the first $n$ natural numbers is equal to $\frac{n(n+1)}{2}$."
Let's check this statement for the first few natural numbers:
-
For $n=1$: $P(1)$ is "The sum of the first 1 natural number is $\frac{1(1+1)}{2}$."
The sum of the first 1 natural number is just 1. The formula gives $\frac{1(2)}{2} = 1$.
So, $P(1)$ is "$1 = 1$", which is True.
-
For $n=2$: $P(2)$ is "The sum of the first 2 natural numbers is $\frac{2(2+1)}{2}$."
The sum of the first 2 natural numbers is $1 + 2 = 3$. The formula gives $\frac{2(3)}{2} = 3$.
So, $P(2)$ is "$3 = 3$", which is True.
-
For $n=3$: $P(3)$ is "The sum of the first 3 natural numbers is $\frac{3(3+1)}{2}$."
The sum of the first 3 natural numbers is $1 + 2 + 3 = 6$. The formula gives $\frac{3(4)}{2} = 6$.
So, $P(3)$ is "$6 = 6$", which is True.
Based on these initial checks, the statement $P(n)$ appears to be true for $n=1, 2, 3$. Mathematical induction is the method used to prove if it is indeed true for all natural numbers $n \ge 1$.
2. Consider the statement $P(n)$: "$2^n > n$."
Let's check this statement for some initial values of $n$:
-
For $n=1$: $P(1)$ is "$2^1 > 1$", which is $2 > 1$. This is True.
-
For $n=2$: $P(2)$ is "$2^2 > 2$", which is $4 > 2$. This is True.
-
For $n=3$: $P(3)$ is "$2^3 > 3$", which is $8 > 3$. This is True.
-
For $n=4$: $P(4)$ is "$2^4 > 4$", which is $16 > 4$. This is True.
The statement $P(n)$ seems to hold for these values. Mathematical induction provides a way to prove this inequality holds for all natural numbers $n \ge 1$.
3. Consider the statement $P(n)$: "The number $n^2 + n + 41$ is a prime number."
Let's check this statement for a few values of $n$:
-
For $n=1$: $P(1)$ is "$1^2 + 1 + 41$ is a prime number". $1 + 1 + 41 = 43$. 43 is a prime number. This is True.
-
For $n=2$: $P(2)$ is "$2^2 + 2 + 41$ is a prime number". $4 + 2 + 41 = 47$. 47 is a prime number. This is True.
-
For $n=3$: $P(3)$ is "$3^2 + 3 + 41$ is a prime number". $9 + 3 + 41 = 53$. 53 is a prime number. This is True.
-
... (If you continue, you will find it's true for $n=1, 2, ..., 39$)
-
Let's jump to $n=40$: $P(40)$ is "$40^2 + 40 + 41$ is a prime number".
$40^2 + 40 + 41 = 1600 + 40 + 41 = 1681$.
Is 1681 a prime number? We can check for factors. Notice that $1681 = 41 \times 41$.
So, 1681 is not prime (it is the square of 41). $P(40)$ is "$1681$ is a prime number", which is False.
This third example is very important. It clearly demonstrates that even if a statement $P(n)$ is true for many initial natural numbers, it is not sufficient to conclude that it is true for all natural numbers. Checking cases only provides evidence, not a proof. This highlights the necessity of a rigorous proof method like the Principle of Mathematical Induction to establish the truth of a statement for an infinite sequence of natural numbers.
The Principle of Mathematical Induction provides a formal framework to prove that a statement $P(n)$ holds for all natural numbers $n \ge n_0$, where $n_0$ is a specified starting natural number (often $n_0=1$). The method ensures that if the statement holds for the first case, and if assuming it holds for any arbitrary case implies it holds for the next case, then it must hold for all subsequent cases indefinitely.
Principle of Mathematical Induction
The Principle of Mathematical Induction (PMI) is a formal method of proof used to show that a given mathematical statement is true for all natural numbers (or an infinite sequence of natural numbers starting from a specific integer).
It's a way to handle proofs that would be impossible to do by checking every single case, because there are infinitely many cases to check.
The Domino Analogy
A helpful way to visualize the principle is to think of an infinitely long row of dominoes, numbered 1, 2, 3, and so on.
How can you be absolutely sure that every single domino will fall?
- You must knock over the first domino. If the first one isn't pushed, the chain reaction will never start. This is the Base Case.
- You must ensure the dominoes are close enough together. Specifically, you need to be certain that if any domino falls, its fall will definitely knock over the next one. This is the Inductive Step.
If you can guarantee these two conditions, the logic is inescapable: The first domino falls (by condition 1). This causes the second to fall (by condition 2). The second falling causes the third to fall (again, by condition 2), and so on, forever. Every domino is guaranteed to fall.
Mathematical Induction applies this same powerful, step-by-step logic to mathematical statements.
Formal Statement of the Principle
Let $P(n)$ be a mathematical statement that depends on a natural number $n$. To prove that $P(n)$ is true for all natural numbers $n \ge 1$, we must satisfy two conditions:
Condition 1: The Base Case
The statement must be true for the very first case, usually $n=1$. We must prove that $P(1)$ is true.
Condition 2: The Inductive Step
We must prove that if the statement is true for any one case, then it must also be true for the very next case. To do this, we:
(a) Assume the statement is true for an arbitrary positive integer $k$. This assumption, "$P(k)$ is true," is called the Inductive Hypothesis.
(b) Use this assumption to prove that the statement is also true for the next integer, $k+1$. In other words, we must prove that $P(k+1)$ is true.
If we successfully prove both the Base Case and the Inductive Step, we can conclude by the Principle of Mathematical Induction that the statement $P(n)$ is true for all natural numbers $n \ge 1$.
Steps for a Proof by Induction
To write a clear and correct proof using mathematical induction, follow this structured four-step process.
Step 1: Base Case (Verification)
State the proposition $P(n)$ you are trying to prove. Then, show it is true for the first value of $n$ (usually $n=1$). Substitute this value into the statement and verify that the equation or inequality holds.
Step 2: Inductive Hypothesis (Assumption)
Assume that the statement $P(n)$ is true for some arbitrary positive integer $k$. Write down this assumption explicitly as the equation or inequality $P(k)$. This is the crucial tool you will use in the next step.
Step 3: Inductive Step (Proof)
Your goal is to prove that $P(k+1)$ is true. Start by writing down what the statement $P(k+1)$ looks like. Then, begin with one side of the $P(k+1)$ statement (usually the more complex side) and use algebraic steps to transform it into the other side. The key move in this step is to find a way to use your Inductive Hypothesis ($P(k)$) from Step 2.
Step 4: Conclusion
After completing the first three steps, state your conclusion formally. For example: "Since the base case is true and the inductive step has been proven, by the Principle of Mathematical Induction, the statement $P(n)$ is true for all natural numbers $n \ge 1$."
Example 1. Prove by induction that for all $n \in \mathbb{N}$, the sum of the first $n$ odd numbers is $n^2$.
Answer:
Let $P(n)$ be the statement: $1 + 3 + 5 + ... + (2n - 1) = n^2$.
Step 1: Base Case (n=1)
We need to show that $P(1)$ is true.
For $n=1$, the Left Hand Side (LHS) is the sum of the first odd number, which is simply 1.
LHS = 1
The Right Hand Side (RHS) is $1^2$.
RHS = $1^2 = 1$
Since LHS = RHS, $P(1)$ is true.
Step 2: Inductive Hypothesis
Assume that $P(k)$ is true for some arbitrary positive integer $k$.
$P(k): 1 + 3 + 5 + ... + (2k - 1) = k^2$
... (i)
Step 3: Inductive Step
We need to prove that $P(k+1)$ is true. The statement $P(k+1)$ is:
$P(k+1): 1 + 3 + 5 + ... + (2(k+1) - 1) = (k+1)^2$
Let's simplify the last term: $2(k+1) - 1 = 2k + 2 - 1 = 2k+1$. So, we want to prove:
$P(k+1): \underbrace{1 + 3 + 5 + ... + (2k - 1)}_{\text{Sum of first k terms}} + (2k + 1) = (k+1)^2$
Now, let's start with the LHS of $P(k+1)$ and use our assumption.
LHS = $1 + 3 + 5 + ... + (2k - 1) + (2k + 1)$
By the inductive hypothesis (i), the sum of the first $k$ terms is $k^2$.
LHS = $k^2 + (2k + 1)$
[Using (i)]
Factoring the quadratic expression:
LHS = $(k+1)^2$
This is equal to the RHS of $P(k+1)$. Thus, we have shown that if $P(k)$ is true, then $P(k+1)$ is also true.
Step 4: Conclusion
Since the base case ($n=1$) is true and we have shown that $P(k) \implies P(k+1)$, by the Principle of Mathematical Induction, the statement $P(n)$ is true for all natural numbers $n \ge 1$.
Example 2. Prove by induction that $7^n - 1$ is divisible by 6 for all $n \in \mathbb{N}$.
Answer:
Let $P(n)$ be the statement: "$7^n - 1$ is divisible by 6."
Step 1: Base Case (n=1)
For $n=1$, the statement is "$7^1 - 1$ is divisible by 6".
$7^1 - 1 = 6$
Since 6 is divisible by 6, $P(1)$ is true.
Step 2: Inductive Hypothesis
Assume that $P(k)$ is true for some arbitrary positive integer $k$. This means we assume "$7^k - 1$ is divisible by 6".
If a number is divisible by 6, it can be written as $6m$ for some integer $m$. So, our assumption is:
$7^k - 1 = 6m$, for some integer $m$.
... (i)
Step 3: Inductive Step
We need to prove that $P(k+1)$ is true, which is the statement "$7^{k+1} - 1$ is divisible by 6".
Let's examine the expression $7^{k+1} - 1$ and try to use our assumption.
$7^{k+1} - 1 = 7 \cdot 7^k - 1$
We want to use the term $7^k - 1$. We can rewrite the expression:
$ = 7 \cdot 7^k - 7 + 6$
$ = 7(7^k - 1) + 6$
From our inductive hypothesis (i), we know $7^k - 1 = 6m$.
$ = 7(6m) + 6$
[Using (i)]
Now, factor out 6:
$ = 6(7m + 1)$
Since $m$ is an integer, $7m+1$ is also an integer. Therefore, $6(7m+1)$ is a multiple of 6, which means it is divisible by 6. Thus, $P(k+1)$ is true.
Step 4: Conclusion
Since the base case is true and the inductive step has been proven, by the Principle of Mathematical Induction, $7^n - 1$ is divisible by 6 for all natural numbers $n \ge 1$.
Variants of Induction
The starting point of an induction proof does not have to be $n=1$. Sometimes a statement is only true for integers greater than a certain value.
The principle can be modified: if you can prove a base case $P(n_0)$ and then prove that $P(k) \implies P(k+1)$ for all $k \ge n_0$, then the statement is true for all integers $n \ge n_0$.
Example 3. Prove that $2^n > n^2$ for all integers $n \ge 5$.
Answer:
Let $P(n)$ be the statement: "$2^n > n^2$". We want to prove this for $n \ge 5$.
Step 1: Base Case (n=5)
Our starting point is $n_0 = 5$. We check if $P(5)$ is true.
$P(5): 2^5 > 5^2$
$32 > 25$. This is true. So, the base case holds.
Step 2: Inductive Hypothesis
Assume $P(k)$ is true for some arbitrary integer $k \ge 5$.
$P(k): 2^k > k^2$
... (i)
Step 3: Inductive Step
We want to prove $P(k+1)$ is true, i.e., $2^{k+1} > (k+1)^2$.
Start with the LHS of $P(k+1)$:
LHS = $2^{k+1} = 2 \cdot 2^k$
Using our inductive hypothesis (i), we know $2^k > k^2$. So we can say:
$2 \cdot 2^k > 2 \cdot k^2$
Now our goal is to show that this expression, $2k^2$, is greater than the RHS of $P(k+1)$, which is $(k+1)^2$. Let's check if $2k^2 > (k+1)^2$ is true for $k \ge 5$.
$2k^2 > k^2 + 2k + 1$
$k^2 - 2k - 1 > 0$
The roots of $k^2-2k-1=0$ are $k = \frac{2 \pm \sqrt{4-4(1)(-1)}}{2} = 1 \pm \sqrt{2}$. So the parabola opens up and is positive for $k > 1+\sqrt{2} \approx 2.414$. Since our assumption is for $k \ge 5$, the inequality $k^2 - 2k - 1 > 0$ is definitely true.
So, we have the chain of inequalities:
$2^{k+1} = 2 \cdot 2^k > 2k^2 > (k+1)^2$
Therefore, $2^{k+1} > (k+1)^2$. This proves that $P(k+1)$ is true.
Step 4: Conclusion
Since the base case ($n=5$) is true and the inductive step holds for all $k \ge 5$, by PMI, $2^n > n^2$ is true for all integers $n \ge 5$.