Mathematical Induction is a fundamental proof technique used to prove that a statement $P(n)$ is true for every natural number $n$ (or for every integer $n \ge n_0$ for some fixed integer $n_0$). It is based on the idea of a "domino effect." If you can show that the first domino falls, and that if any domino falls, the next one will also fall, then all dominoes will eventually fall.

To prove that a statement $P(n)$ is true for all integers $n \ge n_0$:

1. Base Case (Basis Step): Show that $P(n_0)$ is true. This establishes the starting point for the induction.

2. Inductive Step: Assume that $P(k)$ is true for an arbitrary integer $k \ge n_0$ (this is called the Inductive Hypothesis). Then, show that $P(k+1)$ must also be true. This demonstrates that if the statement holds for some integer $k$, it must also hold for the next integer $k+1$.

Once both steps are successfully completed, the Principle of Mathematical Induction guarantees that $P(n)$ is true for all integers $n \ge n_0$.

* Base Case ($n=1$):

$P(1)$: $\sum_{i=1}^{1} i = 1$.

Formula: $\frac{1(1+1)}{2} = \frac{1 \cdot 2}{2} = 1$.

Since $1=1$, $P(1)$ is true.

* Inductive Hypothesis: Assume $P(k)$ is true for an arbitrary integer $k \ge 1$.

That is, assume $\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$.

* Inductive Step: We need to show that $P(k+1)$ is true, i.e., $\sum_{i=1}^{k+1} i = \frac{(k+1)((k+1)+1)}{2} = \frac{(k+1)(k+2)}{2}$.

Start with the left-hand side of $P(k+1)$:

$\sum_{i=1}^{k+1} i = \left(\sum_{i=1}^{k} i\right) + (k+1)$

By the Inductive Hypothesis, we can substitute $\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$:

$= \frac{k(k+1)}{2} + (k+1)$

Factor out $(k+1)$:

$= (k+1)\left(\frac{k}{2} + 1\right)$

$= (k+1)\left(\frac{k+2}{2}\right)$

$= \frac{(k+1)(k+2)}{2}$

This is the right-hand side of $P(k+1)$. Thus, $P(k+1)$ is true.

By the Principle of Mathematical Induction, the formula $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ is true for all integers $n \ge 1$.

* Base Case ($n=1$):

$P(1)$: $1^3 - 1 = 1 - 1 = 0$.

Since $0$ is divisible by 3 ($0 = 3 \cdot 0$), $P(1)$ is true.

* Inductive Hypothesis: Assume $P(k)$ is true for an arbitrary integer $k \ge 1$.

That is, assume $k^3 - k$ is divisible by 3. This means $k^3 - k = 3m$ for some integer $m$.

* Inductive Step: We need to show that $P(k+1)$ is true, i.e., $(k+1)^3 - (k+1)$ is divisible by 3.

Expand $(k+1)^3 - (k+1)$:

$(k+1)^3 - (k+1) = (k^3 + 3k^2 + 3k + 1) - (k+1)$

$= k^3 + 3k^2 + 3k + 1 - k - 1$

$= k^3 - k + 3k^2 + 3k$

$= (k^3 - k) + 3k^2 + 3k$

By the Inductive Hypothesis, $k^3 - k = 3m$ for some integer $m$:

$= 3m + 3k^2 + 3k$

$= 3(m + k^2 + k)$

Since $(m + k^2 + k)$ is an integer, $(k+1)^3 - (k+1)$ is divisible by 3. Thus, $P(k+1)$ is true.

By the Principle of Mathematical Induction, $n^3 - n$ is divisible by 3 for all integers $n \ge 1$.

* Base Case ($n=1$):

$P(1)$: $2^1 > 1$, which is $2 > 1$. This is true.

* Inductive Hypothesis: Assume $P(k)$ is true for an arbitrary integer $k \ge 1$.

That is, assume $2^k > k$.

* Inductive Step: We need to show that $P(k+1)$ is true, i.e., $2^{k+1} > k+1$.

Start with the left-hand side of $P(k+1)$:

$2^{k+1} = 2 \cdot 2^k$

By the Inductive Hypothesis, $2^k > k$:

$2 \cdot 2^k > 2 \cdot k$

So, $2^{k+1} > 2k$.

We want to show $2^{k+1} > k+1$. We know $2k = k+k$.

Since $k \ge 1$, we have $k \ge 1$.

Therefore, $2k = k+k \ge k+1$.

Combining the inequalities: $2^{k+1} > 2k \ge k+1$.

Thus, $2^{k+1} > k+1$. So, $P(k+1)$ is true.

By the Principle of Mathematical Induction, $2^n > n$ is true for all integers $n \ge 1$.

* Base Cases: (Since $F_n$ depends on two previous terms, we might need two base cases for strong induction, or just one for weak induction if the inductive step can handle it. For $F_n < 2^n$, $n=1$ and $n=2$ are good starting points.)

$P(1)$: $F_1 = 1 < 2^1 = 2$. True.

$P(2)$: $F_2 = F_1 + F_0 = 1 + 0 = 1 < 2^2 = 4$. True.

* Inductive Hypothesis: Assume $P(k)$ is true for an arbitrary integer $k \ge 2$.

That is, assume $F_k < 2^k$. (For this type of problem, strong induction is often more natural, assuming $F_i < 2^i$ for all $i \le k$). Let's use strong induction here.

Assume $F_i < 2^i$ for all integers $1 \le i \le k$.

* Inductive Step: We need to show that $P(k+1)$ is true, i.e., $F_{k+1} < 2^{k+1}$.

For $k+1 \ge 2$, we have $F_{k+1} = F_k + F_{k-1}$.

By the Strong Inductive Hypothesis, $F_k < 2^k$ and $F_{k-1} < 2^{k-1}$ (since $k-1 \ge 1$).

So, $F_{k+1} = F_k + F_{k-1} < 2^k + 2^{k-1}$

$= 2^{k-1}(2 + 1)$

$= 3 \cdot 2^{k-1}$

We want to show $F_{k+1} < 2^{k+1}$.

We have $3 \cdot 2^{k-1}$. We know that $3 < 4 = 2^2$.

So, $3 \cdot 2^{k-1} < 2^2 \cdot 2^{k-1} = 2^{2+k-1} = 2^{k+1}$.

Therefore, $F_{k+1} < 2^{k+1}$. Thus, $P(k+1)$ is true.

By the Principle of Strong Mathematical Induction, $F_n < 2^n$ for all integers $n \ge 1$.

Strong Induction is a variation of mathematical induction where the inductive hypothesis is stronger. Instead of assuming $P(k)$ is true to prove $P(k+1)$, strong induction assumes that $P(i)$ is true for all integers $i$ from $n_0$ up to $k$.

To prove that a statement $P(n)$ is true for all integers $n \ge n_0$ using strong induction:

1. Base Case (Basis Step): Show that $P(n_0)$ is true (and possibly $P(n_0+1), \dots, P(m)$ if the inductive step requires multiple preceding values).

2. Inductive Step: Assume that $P(i)$ is true for all integers $i$ such that $n_0 \le i \le k$, for an arbitrary integer $k \ge n_0$ (this is the Strong Inductive Hypothesis). Then, show that $P(k+1)$ must also be true.

Strong induction is particularly useful when the truth of $P(k+1)$ depends not just on $P(k)$, but on several preceding cases (e.g., $P(k-1)$, $P(k-2)$, etc.), as often seen in recurrence relations. It is logically equivalent to weak induction; any proof by strong induction can be converted into a proof by weak induction, and vice-versa.

* Incorrect Base Case: Failing to prove the base case, or proving it for an incorrect starting value $n_0$.

* Not Using the Inductive Hypothesis: The inductive hypothesis is the crucial link. If it's not used to transform the expression for $P(k+1)$, the proof is likely flawed.

* Insufficient Base Cases for Strong Induction: When using strong induction for recurrence relations, ensure enough base cases are proven to cover the smallest values that the recurrence relation refers to. For example, if $P(k+1)$ depends on $P(k-1)$ and $P(k-2)$, then $P(n_0)$ and $P(n_0+1)$ might both need to be proven as base cases.

* Mathematical Induction: A powerful proof technique used to establish that a given statement $P(n)$ is true for all natural numbers $n$ (or all integers $n \ge n_0$). It involves proving a base case and an inductive step.

* Base Case (Basis Step): The first step in an inductive proof, where the statement $P(n_0)$ is shown to be true for the smallest relevant integer $n_0$. This establishes the starting point for the chain of implications.

* Inductive Hypothesis: The assumption made in the inductive step that the statement $P(k)$ is true for an arbitrary integer $k \ge n_0$. This assumption is crucial for proving the next case.

* Inductive Step: The second main step in an inductive proof, where it is shown that if the inductive hypothesis $P(k)$ is true, then the statement $P(k+1)$ must also be true. This demonstrates the "domino effect."

* Natural Numbers ($\mathbb{N}$): The set of positive integers $\{1, 2, 3, \dots\}$. Sometimes, it includes 0, i.e., $\{0, 1, 2, 3, \dots\}$, depending on the context. In this document, it typically refers to $n \ge 1$ unless $n_0$ is specified otherwise.

* Base Case ($n=0$): $ar^0 = a$. Formula: $\frac{a(r^{0+1}-1)}{r-1} = \frac{a(r-1)}{r-1} = a$. True.

* Inductive Hypothesis: Assume $\sum_{i=0}^{k} ar^i = \frac{a(r^{k+1}-1)}{r-1}$.

* Inductive Step: Show $\sum_{i=0}^{k+1} ar^i = \frac{a(r^{k+2}-1)}{r-1}$.

$\sum_{i=0}^{k+1} ar^i = \left(\sum_{i=0}^{k} ar^i\right) + ar^{k+1}$

$= \frac{a(r^{k+1}-1)}{r-1} + ar^{k+1}$ (by IH)

$= \frac{a(r^{k+1}-1) + ar^{k+1}(r-1)}{r-1}$

$= \frac{ar^{k+1}-a + ar^{k+2}-ar^{k+1}}{r-1}$

$= \frac{ar^{k+2}-a}{r-1} = \frac{a(r^{k+2}-1)}{r-1}$. This proves the formula.

* Base Case ($n=1$): $6^1 - 1 = 5$, which is divisible by 5. True.

* Inductive Hypothesis: Assume $6^k - 1$ is divisible by 5, i.e., $6^k - 1 = 5m$ for some integer $m$.

* Inductive Step: Show $6^{k+1} - 1$ is divisible by 5.

$6^{k+1} - 1 = 6 \cdot 6^k - 1$

From IH, $6^k = 5m + 1$. Substitute this:

$= 6(5m + 1) - 1$

$= 30m + 6 - 1$

$= 30m + 5$

$= 5(6m + 1)$. Since $6m+1$ is an integer, $6^{k+1}-1$ is divisible by 5.

* Base Case ($n=4$): $4! = 24$. $2^4 = 16$. Since $24 > 16$, $P(4)$ is true.

* Inductive Hypothesis: Assume $k! > 2^k$ for an arbitrary integer $k \ge 4$.

* Inductive Step: Show $(k+1)! > 2^{k+1}$.

$(k+1)! = (k+1) \cdot k!$

By IH, $k! > 2^k$:

$(k+1) \cdot k! > (k+1) \cdot 2^k$

Since $k \ge 4$, we know $k+1 > 2$.

Therefore, $(k+1) \cdot 2^k > 2 \cdot 2^k = 2^{k+1}$.

Combining, $(k+1)! > (k+1) \cdot 2^k > 2^{k+1}$. Thus, $(k+1)! > 2^{k+1}$.

This document, "Xirius-mathematicalinduction2-COS203.pdf," provides an in-depth exploration of Mathematical Induction, a cornerstone proof technique in discrete mathematics and computer science. It is structured to guide students through the rigorous application of induction to prove statements concerning natural numbers. The core of the document revolves around the Principle of Mathematical Induction (PMI), which asserts that to prove a statement $P(n)$ for all integers $n \ge n_0$, one must successfully complete two steps: the Base Case and the Inductive Step.

The Base Case requires demonstrating that the statement $P(n_0)$ holds true for the initial integer $n_0$. This is the foundational step, establishing the first instance where the property is valid. Following this, the Inductive Step involves assuming the Inductive Hypothesis, which states that $P(k)$ is true for an arbitrary integer $k \ge n_0$. With this assumption, the goal is then to logically deduce that $P(k+1)$ must also be true. This step is crucial as it establishes the chain reaction: if the property holds for any $k$, it must also hold for the subsequent integer $k+1$. The document emphasizes that both steps are indispensable; failure in either invalidates the inductive proof.

The utility of mathematical induction is showcased through its diverse applications. The document meticulously details how to apply induction to prove summation formulas, such as the sum of the first $n$ integers $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$, by breaking down the sum for $k+1$ into the sum for $k$ plus the $(k+1)$-th term, and then applying the inductive hypothesis. It also covers proving divisibility properties, demonstrating how to show that an expression like $n^3 - n$ is always divisible by 3 by algebraically manipulating the $(k+1)$-th case to reveal a multiple of the divisor, leveraging the inductive hypothesis. Furthermore, the document illustrates the use of induction for proving inequalities, such as $2^n > n$, often requiring careful algebraic manipulation and sometimes additional inequalities (e.g., $k+1 > 2$ for $k \ge 1$) to bridge the gap between $P(k)$ and $P(k+1)$. Finally, it addresses proving properties of recurrence relations and sequences, like the Fibonacci sequence, where the definition of a term depends on previous terms, making induction a natural fit.

A significant aspect covered is Strong Mathematical Induction, a variant where the inductive hypothesis is strengthened to assume that $P(i)$ is true for all integers $i$ from $n_0$ up to $k$, not just for $P(k)$. This form is particularly advantageous when the truth of $P(k+1)$ relies on more than just the immediately preceding case, often seen in proofs involving recurrence relations. While logically equivalent to weak induction, strong induction can simplify proofs in certain contexts.

The document also serves as a guide to avoiding common pitfalls, such as failing to establish a correct base case, erroneously assuming $P(k+1)$ instead of deriving it, making algebraic errors, or not effectively utilizing the inductive hypothesis. By providing clear definitions for terms like "Mathematical Induction," "Base Case," "Inductive Hypothesis," and "Inductive Step," the document ensures a solid conceptual foundation. Through its comprehensive examples and detailed explanations, this resource aims to empower students of COS203 to master mathematical induction as a fundamental and indispensable tool for rigorous mathematical and computational reasoning.