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Principle of Mathematical Induction
Mathematical Statement (for Induction)
Statements Dependent on Natural Numbers
In mathematics, we frequently encounter propositions or assertions whose truth depends on the value of a positive integer, also known as a natural number. For instance, we might make a claim about the sum of the first $n$ natural numbers, or about a property that holds for all numbers $n$ greater than or equal to a specific integer. These are examples of mathematical statements that are parameterized by a natural number.
A mathematical statement (or proposition) for induction is essentially a function or formula $P(n)$ that produces a statement (which is either true or false) for each natural number $n$. We are interested in proving that such a statement $P(n)$ is true for *all* natural numbers $n \in \{1, 2, 3, \ldots\}$, or sometimes for all natural numbers $n$ greater than or equal to a specific starting integer (like $n \ge 5$).
We typically represent such a statement using the notation $P(n)$ to explicitly show its dependence on the natural number variable $n$. The goal is to show that $P(n)$ holds true for every integer in the specified range.
Examples of Mathematical Statements for Induction
Here are several examples of types of mathematical statements $P(n)$ that are commonly proven using the Principle of Mathematical Induction:
Statements about sums of series:
$P(n)$: The sum of the first $n$ positive integers is given by the formula $\frac{n(n+1)}{2}$.
Let's see what this statement means for specific values of $n$:
- For $n=1$, $P(1)$ is the statement: "The sum of the first 1 positive integer is $\frac{1(1+1)}{2}$." This simplifies to $1 = \frac{1 \times 2}{2} = 1$, which is a true statement.
- For $n=2$, $P(2)$ is the statement: "The sum of the first 2 positive integers is $\frac{2(2+1)}{2}$." This simplifies to $1+2 = \frac{2 \times 3}{2} = 3$, which is a true statement ($3=3$).
- For $n=3$, $P(3)$ is the statement: "The sum of the first 3 positive integers is $\frac{3(3+1)}{2}$." This simplifies to $1+2+3 = \frac{3 \times 4}{2} = 6$, which is a true statement ($6=6$).
The statement $P(n)$ asserts that this formula $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ is valid for every natural number $n$.
Statements about sums of other sequences:
$P(n)$: The sum of the first $n$ positive odd numbers is $n^2$. (The $n$-th positive odd number is $2n-1$). The sum is $1 + 3 + 5 + \ldots + (2n-1)$.
- For $n=1$, $P(1)$: "The sum of the first 1 positive odd number is $1^2$." This is $1 = 1$, true.
- For $n=2$, $P(2)$: "The sum of the first 2 positive odd numbers is $2^2$." This is $1 + 3 = 4$, true.
- For $n=3$, $P(3)$: "The sum of the first 3 positive odd numbers is $3^2$." This is $1 + 3 + 5 = 9$, true.
Statements about inequalities:
$P(n)$: For all natural numbers $n \ge 1$, $2^n > n$.
- For $n=1$, $P(1)$: "$2^1 > 1$." This is $2 > 1$, true.
- For $n=2$, $P(2)$: "$2^2 > 2$." This is $4 > 2$, true.
- For $n=3$, $P(3)$: "$2^3 > 3$." This is $8 > 3$, true.
Statements about divisibility:
$P(n)$: For all natural numbers $n \ge 1$, the expression $7^n - 3^n$ is divisible by $4$.
- For $n=1$, $P(1)$: "$7^1 - 3^1$ is divisible by $4$." This is $7 - 3 = 4$, which is divisible by 4. True.
- For $n=2$, $P(2)$: "$7^2 - 3^2$ is divisible by $4$." This is $49 - 9 = 40$, which is divisible by 4 ($40 = 4 \times 10$). True.
Statements with a starting point other than 1:
$P(n)$: For all natural numbers $n \ge 5$, $2^n > n^2$.
Note the condition $n \ge 5$. The statement is being claimed only for natural numbers starting from 5.
- Let's check values less than 5:
- For $n=1$: $2^1 = 2$, $1^2 = 1$. $2 > 1$, True. ($P(1)$ is true, but not part of the claim for $n \ge 5$).
- For $n=2$: $2^2 = 4$, $2^2 = 4$. $4 \not> 4$, False.
- For $n=3$: $2^3 = 8$, $3^2 = 9$. $8 \not> 9$, False.
- For $n=4$: $2^4 = 16$, $4^2 = 16$. $16 \not> 16$, False.
- Now check the starting point for the claim:
- For $n=5$, $P(5)$: "$2^5 > 5^2$." This is $32 > 25$, True. The base case for the induction proof starting from $n=5$ would be checking $P(5)$.
- Let's check values less than 5:
The Principle of Mathematical Induction is a powerful technique used to rigorously prove that such statements $P(n)$ are true for all natural numbers (or for all natural numbers greater than or equal to a specified starting integer). It provides a step-by-step logical process to establish the truth of these infinite number of statements.
Principle of Mathematical Induction: Statement and Steps
The Principle
The Principle of Mathematical Induction (PMI) is one of the most fundamental and powerful proof techniques in mathematics, particularly useful for proving statements that assert a property holds for all natural numbers (or all natural numbers starting from a certain point). It provides a rigorous way to move from showing a statement is true for a specific initial case to concluding its truth for an infinite sequence of cases.
To intuitively understand the Principle of Mathematical Induction, consider the analogy of an infinitely long line of dominoes. Suppose these dominoes are arranged such that if any single domino falls, it will knock over the very next domino in the line. To ensure that *all* the dominoes in this infinite line will eventually fall, only two conditions are necessary:
- You must push over the first domino. (This starts the chain reaction).
- You must ensure the mechanism works: if any arbitrary domino falls, the design guarantees that the next one will also fall. (This ensures the reaction continues indefinitely).
If both these conditions are met, the entire line of dominoes will fall, one after the other.
The Principle of Mathematical Induction applies this same logic to prove statements $P(n)$ about natural numbers $n$. If we can show that the statement holds for the first relevant natural number (like knocking over the first domino) and that the truth of the statement for any natural number implies its truth for the next natural number (like ensuring a falling domino knocks over the next), then the statement must be true for all natural numbers (like all dominoes falling).
Statement of the Principle of Mathematical Induction
Let $P(n)$ be a mathematical statement or proposition that depends on a natural number $n$. Suppose we want to prove that $P(n)$ is true for all natural numbers $n$ greater than or equal to a fixed integer $n_0$ (usually $n_0 = 1$, but it can be any starting integer like 0, 2, 5, etc.).
The Principle of Mathematical Induction states that $P(n)$ is true for all natural numbers $n \ge n_0$, if the following two conditions are met:
Basis Step (Base Case): The statement $P(n)$ is true for the initial value $n_0$. That is, $P(n_0)$ is true.
(This is equivalent to knocking over the first domino).
Inductive Step: If the statement $P(n)$ is true for some arbitrary natural number $k \ge n_0$ (this assumption is called the inductive hypothesis), then it must follow that the statement is also true for the next natural number, $k+1$. That is, the implication $P(k) \implies P(k+1)$ is true for all $k \ge n_0$.
(This is equivalent to showing that if any domino $k$ falls, domino $k+1$ will also fall).
If both the Basis Step and the Inductive Step are successfully proven, then the Principle of Mathematical Induction allows us to conclude that the statement $P(n)$ is true for all natural numbers $n$ in the specified range, i.e., for all $n \ge n_0$.
In most common applications where the statement is claimed for all natural numbers starting from 1 ($n \ge 1$), the Basis Step involves proving $P(1)$ is true, and the Inductive Step involves proving that for any $k \ge 1$, if $P(k)$ is true, then $P(k+1)$ is true. Once these are established, the chain of implications $P(1) \implies P(2) \implies P(3) \implies \ldots$ demonstrates the truth of $P(n)$ for all $n \in \{1, 2, 3, \ldots\}$.
Steps of a Proof by Mathematical Induction
To construct a formal proof using the Principle of Mathematical Induction to show that a statement $P(n)$ is true for all natural numbers $n \ge n_0$, follow these standard steps:
Formulate the Statement: Clearly write down the statement $P(n)$ that you intend to prove, explicitly mentioning the variable $n$ (representing a natural number) and the range of $n$ (e.g., "for all natural numbers $n \ge 1$" or "for all integers $n \ge 5$").
Basis Step (Base Case):
- Identify the smallest natural number $n_0$ for which the statement is claimed to be true. This is your starting point.
- Show, by direct substitution or calculation, that the statement $P(n_0)$ is true. This is the first crucial step to start the inductive process.
Inductive Hypothesis:
- Clearly state the assumption for the inductive step. Assume that the statement $P(k)$ is true for some arbitrary natural number $k \ge n_0$. This assumption is the "inductive hypothesis". Write out what $P(k)$ means in terms of the specific mathematical expression or property you are working with.
Inductive Step:
- This is the core of the proof. The goal is to prove that if $P(k)$ is true (as assumed in the inductive hypothesis), then $P(k+1)$ must also be true.
- Start with the expression or condition for $P(k+1)$. Try to manipulate it algebraically or logically, aiming to introduce the expression or condition from $P(k)$ (your inductive hypothesis).
- Use the truth of $P(k)$ (from the inductive hypothesis) at some point in your logical steps to demonstrate the truth of $P(k+1)$. Show a clear logical flow from the assumption $P(k)$ to the conclusion $P(k+1)$.
Conclusion:
- State that because both the Basis Step ($P(n_0)$ is true) and the Inductive Step ($P(k) \implies P(k+1)$ for $k \ge n_0$) have been successfully demonstrated, the Principle of Mathematical Induction guarantees that the statement $P(n)$ is true for all natural numbers $n \ge n_0$.
Visualizing the Chain Reaction
Let's illustrate how the steps connect to prove the statement $P(n)$ for all $n \ge 1$, assuming we've proven $P(1)$ (Basis Step) and $P(k) \implies P(k+1)$ for any $k \ge 1$ (Inductive Step):
- The Basis Step establishes that $P(1)$ is true.
- Now, apply the Inductive Step for $k=1$. It says $P(1) \implies P(1+1)$, which is $P(1) \implies P(2)$. Since we know $P(1)$ is true, the implication means $P(2)$ must also be true.
- Next, apply the Inductive Step for $k=2$. It says $P(2) \implies P(2+1)$, which is $P(2) \implies P(3)$. Since we just found that $P(2)$ is true, this implication means $P(3)$ must also be true.
- Continue this process. For $k=3$, $P(3) \implies P(4)$. Since $P(3)$ is true, $P(4)$ must be true.
- This chain continues indefinitely, covering $n=1, 2, 3, 4, 5, \ldots$, thus proving the statement $P(n)$ for all natural numbers $n \ge 1$.
Mathematical induction is a specific and powerful proof technique for statements indexed by natural numbers. It is not a method for discovering formulas but for proving that a conjectured formula or statement is correct.
Applications of Mathematical Induction
The Principle of Mathematical Induction (PMI) is a versatile and indispensable tool in mathematics for proving statements that are universally true for all natural numbers (or for all natural numbers from a certain starting point). It is not limited to a single type of problem but can be applied to prove a wide array of mathematical propositions indexed by natural numbers.
Types of Statements Proven by Induction
Mathematical induction is commonly used to prove statements falling into categories such as:
- Summation Formulas: Proving that a formula for the sum of the first $n$ terms of a sequence (e.g., an AP, a GP, or sums of powers like $1^2 + 2^2 + \ldots + n^2$) holds true for all natural numbers $n$.
- Divisibility Statements: Proving that an algebraic expression involving $n$ is perfectly divisible by a specific integer for all natural numbers $n$.
- Inequalities: Proving that an inequality involving $n$ is valid for all natural numbers $n$ starting from a particular integer $n_0$.
- Properties of Sequences Defined Recursively: Proving closed-form formulas or other properties for sequences defined by a recursive relation (where a term is defined based on previous terms), such as the Fibonacci sequence.
- Properties in Discrete Mathematics: Proving theorems and properties related to structures like graphs, trees, and algorithms, where the size or complexity of the structure is parameterized by a natural number.
- Formulas for Products: Proving formulas for the product of the first $n$ terms of a sequence.
Examples of Proofs by Induction
Let's illustrate the application of the Principle of Mathematical Induction with detailed examples of common proof types.
Example 1: Proving a Summation Formula (Sum of First n Positive Integers)
Example 1. Prove by mathematical induction that for all natural numbers $n \ge 1$, the sum of the first $n$ positive integers is given by the formula: $1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}$.
Proof:
Let $P(n)$ be the statement: $1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}$. We want to prove $P(n)$ is true for all natural numbers $n \ge 1$.
Step 1: Basis Step (Base Case: $n_0=1$)
We need to show that the statement $P(1)$ is true.
$P(1)$ is the statement "$1 = \frac{1(1+1)}{2}$".
Left Hand Side (LHS) of $P(1)$: The sum of the first 1 positive integer is just $1$.
Right Hand Side (RHS) of $P(1)$: $\frac{1(1+1)}{2} = \frac{1 \times 2}{2} = \frac{2}{2} = 1$.
Since LHS = RHS ($1 = 1$), the statement $P(1)$ is true. This completes the Basis Step.
Step 2: Inductive Hypothesis
Assume that the statement $P(k)$ is true for some arbitrary natural number $k \ge 1$.
That is, we assume that the formula holds for $n=k$:
$1 + 2 + 3 + \ldots + k = \frac{k(k+1)}{2} $
... (Inductive Hypothesis)
Step 3: Inductive Step
We need to prove that the statement $P(k+1)$ is true, using the assumption from the Inductive Hypothesis ($P(k)$ is true). $P(k+1)$ is the statement where $n$ is replaced by $k+1$. The sum on the left side includes one more term, $(k+1)$.
$P(k+1)$ is the statement: $1 + 2 + 3 + \ldots + k + (k+1) = \frac{(k+1)((k+1)+1)}{2} = \frac{(k+1)(k+2)}{2}$.
Consider the Left Hand Side (LHS) of the statement $P(k+1)$:
$$ \text{LHS of } P(k+1) = 1 + 2 + 3 + \ldots + k + (k+1) $$We can group the first $k$ terms together:
$$ \text{LHS} = (1 + 2 + 3 + \ldots + k) + (k+1) $$By the Inductive Hypothesis (Step 2), we know that the sum of the first $k$ terms ($1 + 2 + \ldots + k$) is equal to $\frac{k(k+1)}{2}$. Substitute this into the expression:
$$ \text{LHS} = \left(\frac{k(k+1)}{2}\right) + (k+1) $$Now, we need to manipulate this expression to show that it is equal to the Right Hand Side (RHS) of $P(k+1)$, which is $\frac{(k+1)(k+2)}{2}$.
Find a common denominator for the two terms:
$$ \text{LHS} = \frac{k(k+1)}{2} + \frac{2(k+1)}{2} $$Combine the numerators over the common denominator:
$$ \text{LHS} = \frac{k(k+1) + 2(k+1)}{2} $$Notice that $(k+1)$ is a common factor in the numerator. Factor it out:
$$ \text{LHS} = \frac{(k+1)(k + 2)}{2} $$This result is exactly the Right Hand Side (RHS) of the statement $P(k+1)$.
Thus, we have shown that if $P(k)$ is true, then $P(k+1)$ is also true. This completes the Inductive Step.
Step 4: Conclusion
Since the Basis Step ($P(1)$ is true) and the Inductive Step ($P(k) \implies P(k+1)$ for all $k \ge 1$) have been proven, by the Principle of Mathematical Induction, the statement $P(n)$ is true for all natural numbers $n \ge 1$.
Therefore, the formula $1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}$ is true for all natural numbers $n$.
Example 2: Proving a Divisibility Statement
Example 2. Prove by mathematical induction that for all natural numbers $n \ge 1$, the expression $n^2 + n$ is an even number.
Proof:
Let $P(n)$ be the statement "$n^2 + n$ is an even number". An integer is even if it is divisible by 2, or if it can be written in the form $2m$ for some integer $m$.
Step 1: Basis Step (Base Case: $n_0=1$)
We need to show that the statement $P(1)$ is true.
$P(1)$ is the statement "$1^2 + 1$ is an even number".
Evaluate the expression for $n=1$: $1^2 + 1 = 1 + 1 = 2$.
The number 2 is an even number (since $2 = 2 \times 1$, where 1 is an integer). So, the statement $P(1)$ is true. This completes the Basis Step.
Step 2: Inductive Hypothesis
Assume that the statement $P(k)$ is true for some arbitrary natural number $k \ge 1$.
That is, we assume that $k^2 + k$ is an even number. This means we can write $k^2 + k = 2m$ for some integer $m$.
$k^2 + k = 2m $
... (Inductive Hypothesis)
Step 3: Inductive Step
We need to prove that the statement $P(k+1)$ is true, using the Inductive Hypothesis. $P(k+1)$ is the statement "$(k+1)^2 + (k+1)$ is an even number".
Consider the expression for $P(k+1)$, which is $(k+1)^2 + (k+1)$:
$$ (k+1)^2 + (k+1) $$Expand the term $(k+1)^2$ using the identity $(a+b)^2 = a^2 + 2ab + b^2$:
$$ = (k^2 + 2k + 1) + (k+1) $$Now, rearrange and group the terms in a way that allows us to use the Inductive Hypothesis ($k^2 + k = 2m$). Group the $k^2$ and $k$ terms together:
$$ = (k^2 + k) + (2k + 1 + 1) $$ $$ = (k^2 + k) + 2k + 2 $$By the Inductive Hypothesis (Step 2), we know that $k^2 + k = 2m$ for some integer $m$. Substitute $2m$ for $(k^2 + k)$ in the expression:
$$ = (2m) + 2k + 2 $$ $$ = 2m + 2k + 2 $$Factor out the common factor of 2 from all terms:
$$ = 2(m + k + 1) $$Let $M = m + k + 1$. Since $m$ is an integer (from the inductive hypothesis) and $k$ is a natural number (and therefore an integer), the sum $m + k + 1$ is also an integer. So, $M$ is an integer.
The expression $(k+1)^2 + (k+1)$ can be written as $2M$, where $M$ is an integer. By definition, this means that $(k+1)^2 + (k+1)$ is an even number. Thus, the statement $P(k+1)$ is true.
We have successfully shown that if $P(k)$ is true, then $P(k+1)$ is also true. This completes the Inductive Step.
Step 4: Conclusion
Since the Basis Step ($P(1)$ is true) and the Inductive Step ($P(k) \implies P(k+1)$ for all $k \ge 1$) have been proven, by the Principle of Mathematical Induction, the statement $P(n)$ is true for all natural numbers $n \ge 1$.
Therefore, $n^2 + n$ is an even number for all natural numbers $n$.
Alternative reasoning for $n^2+n$:
Note that $n^2 + n = n(n+1)$. This is the product of two consecutive integers. In any pair of consecutive integers, one must be even and the other must be odd. The product of an even integer and any other integer (even or odd) is always even. Thus, $n(n+1)$ is always even for any integer $n$. Since natural numbers are a subset of integers, $n^2+n$ is always even for any natural number $n$. While this reasoning is simpler, the induction proof demonstrates the formal application of the Principle.
These examples illustrate the power of mathematical induction in proving statements about natural numbers. By establishing the truth for the initial case and proving the implication from $k$ to $k+1$, we can confidently conclude the truth of the statement for an infinite set of numbers.