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Questions and Answers
In a proof by mathematical induction, what is the purpose of the inductive step?
In a proof by mathematical induction, what is the purpose of the inductive step?
- To prove the statement is true for a specific initial value.
- To assume the statement is false for all natural numbers.
- To show that if the statement is true for _n = k_, it is also true for _n = k + 1_. (correct)
- To define the statement for all real numbers.
When proving a statement using mathematical induction, which of the following is a common error to avoid?
When proving a statement using mathematical induction, which of the following is a common error to avoid?
- Using the inductive hypothesis to prove the base case.
- Assuming the statement is false for _n = k_ + 1.
- Proving the statement for all real numbers instead of integers.
- Omitting the explicit statement of the conclusion. (correct)
Which of the following scenarios would most likely require strong induction instead of weak induction?
Which of the following scenarios would most likely require strong induction instead of weak induction?
- When the base case is difficult to prove.
- Proving a simple inequality for all natural numbers.
- Proving a statement about the sum of the first n natural numbers.
- When the truth of the statement for _n = k + 1_ depends on the truth of the statement for multiple preceding values. (correct)
What is the first step in proving $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ for all positive integers n using mathematical induction?
What is the first step in proving $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ for all positive integers n using mathematical induction?
In the context of mathematical induction, what does the 'inductive hypothesis' involve?
In the context of mathematical induction, what does the 'inductive hypothesis' involve?
When using mathematical induction to prove that $n^3 - n$ is divisible by 3 for all positive integers n, what algebraic manipulation would be most useful in the inductive step?
When using mathematical induction to prove that $n^3 - n$ is divisible by 3 for all positive integers n, what algebraic manipulation would be most useful in the inductive step?
You are using mathematical induction to prove an inequality. After establishing the base case and stating the inductive hypothesis, what is your next goal?
You are using mathematical induction to prove an inequality. After establishing the base case and stating the inductive hypothesis, what is your next goal?
Suppose you want to prove a statement P(n) for all integers n ≥ a using mathematical induction. You have completed the basis step, showing that P(a) is true. What is the next step you should take?
Suppose you want to prove a statement P(n) for all integers n ≥ a using mathematical induction. You have completed the basis step, showing that P(a) is true. What is the next step you should take?
Which of the following is a valid application of mathematical induction?
Which of the following is a valid application of mathematical induction?
Why is it important to prove the base case in a proof by mathematical induction?
Why is it important to prove the base case in a proof by mathematical induction?
Consider a proposition $P(n)$ defined for all positive integers $n$. Under what condition is it most appropriate to use strong induction instead of weak induction to prove $P(n)$?
Consider a proposition $P(n)$ defined for all positive integers $n$. Under what condition is it most appropriate to use strong induction instead of weak induction to prove $P(n)$?
You are attempting to prove a statement $P(n)$ for all positive integers $n$ using mathematical induction, but you find that the inductive step fails for $n = 5$. What can you definitively conclude?
You are attempting to prove a statement $P(n)$ for all positive integers $n$ using mathematical induction, but you find that the inductive step fails for $n = 5$. What can you definitively conclude?
When proving a statement $P(n)$ for $n ≥ 1$ using mathematical induction, you establish the base case and assume $P(k)$ is true. In the inductive step, you derive a new statement $Q(k+1)$ from $P(k)$ but fail to show that $Q(k+1)$ is equivalent to $P(k+1)$. What is the most accurate conclusion?
When proving a statement $P(n)$ for $n ≥ 1$ using mathematical induction, you establish the base case and assume $P(k)$ is true. In the inductive step, you derive a new statement $Q(k+1)$ from $P(k)$ but fail to show that $Q(k+1)$ is equivalent to $P(k+1)$. What is the most accurate conclusion?
Suppose you are using mathematical induction to prove a statement of the form 'For all $n ≥ a$, $P(n)$ is true,' where $a$ is an integer. What is the significance of choosing the smallest possible value for $a$?
Suppose you are using mathematical induction to prove a statement of the form 'For all $n ≥ a$, $P(n)$ is true,' where $a$ is an integer. What is the significance of choosing the smallest possible value for $a$?
How does strong induction differ fundamentally from weak induction in the context of proving a statement $P(n)$ for all positive integers $n$?
How does strong induction differ fundamentally from weak induction in the context of proving a statement $P(n)$ for all positive integers $n$?
Which of the following best describes a scenario where mathematical induction would be most appropriate to use as a proof technique?
Which of the following best describes a scenario where mathematical induction would be most appropriate to use as a proof technique?
When using mathematical induction to prove a statement $P(n)$ for all $n ≥ 1$, the inductive step involves showing that $P(k)$ implies $P(k+1)$. What is the logical basis for this implication?
When using mathematical induction to prove a statement $P(n)$ for all $n ≥ 1$, the inductive step involves showing that $P(k)$ implies $P(k+1)$. What is the logical basis for this implication?
In the context of mathematical induction, what is the primary reason for ensuring the base case is true?
In the context of mathematical induction, what is the primary reason for ensuring the base case is true?
While attempting a proof by mathematical induction, one encounters difficulty in establishing the base case. How might one proceed?
While attempting a proof by mathematical induction, one encounters difficulty in establishing the base case. How might one proceed?
Suppose you are trying to prove a statement about a recursively defined sequence using mathematical induction. After assuming $P(k)$ is true, you find that you need $P(k-1)$ to be true in order to prove $P(k+1)$. Which approach is most suitable?
Suppose you are trying to prove a statement about a recursively defined sequence using mathematical induction. After assuming $P(k)$ is true, you find that you need $P(k-1)$ to be true in order to prove $P(k+1)$. Which approach is most suitable?
A student is trying to prove $\sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2$ using mathematical induction. They have correctly established the base case and inductive hypothesis. In the inductive step, they need to manipulate the expression $\left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3$. What is the most strategic next step to simplify this expression and prove the inductive step?
A student is trying to prove $\sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2$ using mathematical induction. They have correctly established the base case and inductive hypothesis. In the inductive step, they need to manipulate the expression $\left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3$. What is the most strategic next step to simplify this expression and prove the inductive step?
Consider the proposition $P(n): n^2 + 1$ is divisible by 4. A student attempts to prove this by induction. They show a base case holds, assume $P(k)$ is true, and then try to prove $P(k+1)$. Where does the proof most likely break down?
Consider the proposition $P(n): n^2 + 1$ is divisible by 4. A student attempts to prove this by induction. They show a base case holds, assume $P(k)$ is true, and then try to prove $P(k+1)$. Where does the proof most likely break down?
You are using mathematical induction to prove that a certain property holds for all binary trees of height $n$. Suppose the inductive hypothesis assumes the property holds for all trees of height up to $k$. What must be demonstrated in the inductive step?
You are using mathematical induction to prove that a certain property holds for all binary trees of height $n$. Suppose the inductive hypothesis assumes the property holds for all trees of height up to $k$. What must be demonstrated in the inductive step?
A common mistake in proofs by mathematical induction is to assume the very statement one is trying to prove. How can this error manifest in the inductive step?
A common mistake in proofs by mathematical induction is to assume the very statement one is trying to prove. How can this error manifest in the inductive step?
Consider the recursive sequence defined as $a_1 = 1$, $a_2 = 2$, and $a_n = a_{n-1} + a_{n-2}$ for $n > 2$. What is a suitable proposition $P(n)$ to prove using strong induction?
Consider the recursive sequence defined as $a_1 = 1$, $a_2 = 2$, and $a_n = a_{n-1} + a_{n-2}$ for $n > 2$. What is a suitable proposition $P(n)$ to prove using strong induction?
What is the most significant reason for verifying the base case in a proof by mathematical induction?
What is the most significant reason for verifying the base case in a proof by mathematical induction?
In a proof by mathematical induction, suppose you successfully prove that if $P(k)$ is true, then $P(k+1)$ is true. However, further analysis reveals that $P(1)$ is actually false. What can you conclude about the truth of $P(n)$ for all $n > 1$?
In a proof by mathematical induction, suppose you successfully prove that if $P(k)$ is true, then $P(k+1)$ is true. However, further analysis reveals that $P(1)$ is actually false. What can you conclude about the truth of $P(n)$ for all $n > 1$?
A student is trying to prove that $n! > 2^n$ for all integers $n ≥ 3$ using mathematical induction. They correctly show the base case for $n=3$. However, in the inductive step, they struggle to manipulate the inequality to show that $(k+1)! > 2^{k+1}$ assuming $k! > 2^k$. What is a strategic step they could take?
A student is trying to prove that $n! > 2^n$ for all integers $n ≥ 3$ using mathematical induction. They correctly show the base case for $n=3$. However, in the inductive step, they struggle to manipulate the inequality to show that $(k+1)! > 2^{k+1}$ assuming $k! > 2^k$. What is a strategic step they could take?
Which statement best characterizes the relationship between mathematical induction and the well-ordering principle?
Which statement best characterizes the relationship between mathematical induction and the well-ordering principle?
Consider a scenario where you aim to prove a statement $P(n)$ for all integers $n ≥ 1$, but the base case $P(1)$ is demonstrably false. Despite this, you manage to show that if $P(k)$ is true, then $P(k+1)$ is also true. What is the most accurate conclusion you can draw?
Consider a scenario where you aim to prove a statement $P(n)$ for all integers $n ≥ 1$, but the base case $P(1)$ is demonstrably false. Despite this, you manage to show that if $P(k)$ is true, then $P(k+1)$ is also true. What is the most accurate conclusion you can draw?
When using mathematical induction to prove a statement $P(n)$ for all natural numbers $n$, which of the following is NOT a required component of a valid proof?
When using mathematical induction to prove a statement $P(n)$ for all natural numbers $n$, which of the following is NOT a required component of a valid proof?
In the context of strong induction, what is the crucial advantage of assuming $P(j)$ holds for all $j$ from a base case up to $k$, rather than just assuming $P(k)$?
In the context of strong induction, what is the crucial advantage of assuming $P(j)$ holds for all $j$ from a base case up to $k$, rather than just assuming $P(k)$?
Suppose you aim to prove that a specific algorithm has a time complexity of $O(n^2)$ using mathematical induction on the input size $n$. After establishing the base case, what would the inductive hypothesis typically state?
Suppose you aim to prove that a specific algorithm has a time complexity of $O(n^2)$ using mathematical induction on the input size $n$. After establishing the base case, what would the inductive hypothesis typically state?
While attempting to prove a statement of the form "For all $n \geq a$, $P(n)$," you find that the inductive step only works under the condition that $n$ is even. How should you adjust your proof?
While attempting to prove a statement of the form "For all $n \geq a$, $P(n)$," you find that the inductive step only works under the condition that $n$ is even. How should you adjust your proof?
Concerning mathematical induction, what is the most accurate interpretation of "inductive bias?"
Concerning mathematical induction, what is the most accurate interpretation of "inductive bias?"
You are attempting to prove a statement using mathematical induction, but you discover that your initial base case $P(1)$ is undefined. What is the most appropriate next step?
You are attempting to prove a statement using mathematical induction, but you discover that your initial base case $P(1)$ is undefined. What is the most appropriate next step?
In a proof by mathematical induction, a critical error is to treat the inductive hypothesis as something to be proven rather than assumed. How does this manifest itself?
In a proof by mathematical induction, a critical error is to treat the inductive hypothesis as something to be proven rather than assumed. How does this manifest itself?
What is a key difference in the structure of a proof using strong induction versus a proof using weak induction?
What is a key difference in the structure of a proof using strong induction versus a proof using weak induction?
Flashcards
Mathematical Induction
Mathematical Induction
A method of mathematical proof used to establish a given statement for all natural numbers greater than or equal to some starting value.
Basis Step
Basis Step
Prove the statement is true for the initial value (usually n = 1).
Inductive Step
Inductive Step
Assume the statement is true for an arbitrary natural number 'k'. Then, prove that the statement is also true for 'k+1', based on the assumption that it is true for 'k'.
Inductive Hypothesis
Inductive Hypothesis
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Divisibility Proofs
Divisibility Proofs
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Inequality Proofs
Inequality Proofs
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Base case failure
Base case failure
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Strong Induction
Strong Induction
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When to use Strong Induction
When to use Strong Induction
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Strong Induction - Inductive Step
Strong Induction - Inductive Step
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Base Case
Base Case
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State the Proposition P(n)
State the Proposition P(n)
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Conclusion (Induction)
Conclusion (Induction)
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Forgetting the Base Case
Forgetting the Base Case
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Misapplying the Inductive Hypothesis
Misapplying the Inductive Hypothesis
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Algebraic Errors
Algebraic Errors
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Proving P(k+1)
Proving P(k+1)
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Base Cases (Strong Induction)
Base Cases (Strong Induction)
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Study Notes
- Mathematical induction is useful for proving statements about sequences, series, and other mathematical structures defined recursively.
- It's a method for proving that a statement is true for all natural numbers, or a subset of them.
Principle of Mathematical Induction
- Base Case: Show the statement is true for the first natural number, usually n = 1.
- Inductive Hypothesis: Assume the statement is true for some arbitrary natural number k.
- Inductive Step: Prove that if the statement is true for k, then it also holds true for k + 1.
- Conclusion: If the base case and inductive step are proven, the statement is true for all natural numbers (or the specified subset).
Steps in Mathematical Induction
- Clearly state the proposition P(n) to be proven.
- Base Case: Verify that P(1) is true by substituting n = 1 into the proposition and showing that it holds.
- Inductive Hypothesis: Assume that P(k) is true for some arbitrary integer k ≥ 1; this means assuming the statement holds for n = k.
- Inductive Step: Prove that P(k + 1) is true, assuming that P(k) is true, which usually involves algebraic manipulation of P(k) to arrive at P(k + 1).
- Conclusion: Conclude that P(n) is true for all integers n ≥ 1 by the principle of mathematical induction.
Basis Step: Showing the Base Case
- The basis step involves verifying the statement for the smallest integer value in the domain, commonly n = 1.
- Substitute the initial value into the statement and assess that it holds true.
Inductive Hypothesis: Assuming True for n = k
- This assumption is called the inductive hypothesis and is the foundation for proving the statement for the next integer.
- Suppose the statement holds true for some arbitrary integer n = k, where k is greater than or equal to the initial value.
Inductive Step: Proving True for n = k + 1
- The inductive step aims to prove that if the statement holds true for n = k, then it must also hold true for n = k + 1.
- Start with the expression for n = k + 1 and manipulate it algebraically, using the inductive hypothesis to substitute the expression for n = k.
- Showcase that the expression for n = k + 1 satisfies the given statement, thus completing the inductive step.
Conclusion
- State clearly that, by the principle of mathematical induction, the statement is true for all integers n greater than or equal to the initial value.
Summation Formulae
- Mathematical induction can be used to prove summation formulae.
- Sum of first n natural numbers: 1 + 2 + 3 + ... + n = n(n+1)/2
- Sum of squares of first n natural numbers: 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6
- Sum of cubes of first n natural numbers: 1^3 + 2^3 + 3^3 + ... + n^3 = [n(n+1)/2]^2
Divisibility
- Mathematical induction can prove statements concerning divisibility.
- Example: Prove that n^3 - n is divisible by 3 for all positive integers n.
Sequences
- Mathematical induction is applicable to recurrence relations (sequences defined recursively).
- Example: Consider a sequence defined by u_1 = 1, u_(n+1) = 3u_n + 1. Prove that u_n = (5 * 3^(n-1) - 1) / 2.
Inequalities
- Mathematical induction can be used to prove inequalities.
- Example: Prove that 2^n > n for all positive integers n.
Common Errors in Mathematical Induction
- Failure to prove the base case.
- Incorrectly applying the inductive hypothesis.
- Making algebraic errors in the inductive step.
- Not explicitly stating the conclusion.
Example Proof: Sum of First n Natural Numbers
- Statement: 1 + 2 + 3 + ... + n = n(n+1)/2 for all positive integers n
- Basis Step: For n = 1, 1 = 1(1+1)/2 = 1, which is true
- Inductive Hypothesis: Assume 1 + 2 + 3 + ... + k = k(k+1)/2 for some positive integer k
- Inductive Step: Prove 1 + 2 + 3 + ... + (k+1) = (k+1)(k+2)/2.
- Starting with the left side: 1 + 2 + 3 + ... + k + (k+1)
- By the inductive hypothesis: k(k+1)/2 + (k+1)
- Factoring out(k+1): (k+1)(k/2 + 1) = (k+1)(k+2)/2
- Conclusion: By the principle of mathematical induction, the statement is true for all positive integers n.
Proof by Induction: Divisibility Example
- Statement: Prove that 7^n - 1 is divisible by 6 for all positive integers n.
- Basis Step: For n = 1, 7^1 - 1 = 6, which is divisible by 6.
- Inductive Hypothesis: Assume that 7^k - 1 is divisible by 6 for some positive integer k; this means that 7^k - 1 = 6m for some integer m, thus 7^k = 6m + 1.
- Inductive Step: Prove that 7^(k+1) - 1 is divisible by 6
- 7^(k+1) - 1 can be rewritten as 7 * 7^k - 1
- Substitute 7^k = 6m + 1: 7 * (6m + 1) - 1 = 42m + 7 - 1 = 42m + 6 = 6(7m + 1)
- Since 6(7m + 1) is a multiple of 6, it is divisible by 6
- Conclusion: By the principle of mathematical induction, 7^n - 1 is divisible by 6 for all positive integers n.
Proof by Induction: Inequality Example
- Statement: Prove that 2^n > n for all positive integers n
- Basis Step: For n = 1, 2^1 = 2 > 1, which is true
- Inductive Hypothesis: Assume that 2^k > k for some positive integer k
- Inductive Step: Prove that 2^(k+1) > k + 1
- 2^k > k by the inductive hypothesis
- Multiply both sides by 2: 2 * 2^k > 2 * k, simplifies to 2^(k+1) > 2k
- It is required to show that 2k > k + 1
- Since k is a positive integer, k >= 1
- k > 1 implies 2k > k + 1, so 2^(k+1) > k + 1
- Conclusion: By the principle of mathematical induction, 2^n > n for all positive integers n
Strong Induction
- Strong induction differs from weak induction in the inductive hypothesis.
- Instead of assuming the statement is true for n = k only, strong induction assumes the statement is true for all n ≤ k.
- This stronger assumption can be useful when the truth of the statement for n = k + 1 depends on the truth of the statement for multiple preceding values.
Example Where Strong Induction is Necessary
- Consider the Fibonacci sequence defined by F(0) = 0, F(1) = 1, and F(n) = F(n-1) + F(n-2) for n ≥ 2.
- To prove that F(n) < 2^n for all n ≥ 0
- Basis Step: For n = 0, F(0) = 0 < 2^0 = 1; for n = 1, F(1) = 1 < 2^1 = 2
- Inductive Hypothesis: Assume that F(i) < 2^i for all i ≤ k, where k ≥ 1
- Inductive Step: Prove that F(k+1) < 2^(k+1)
- By the definition of the Fibonacci sequence: F(k+1) = F(k) + F(k-1)
- By the inductive hypothesis: F(k) < 2^k and F(k-1) < 2^(k-1)
- Therefore, F(k+1) < 2^k + 2^(k-1)
- Since 2^(k-1) < 2^k, it follows that 2^k + 2^(k-1) < 2^k + 2^k = 2 * 2^k = 2^(k+1)
- Thus, F(k+1) < 2^(k+1)
- Conclusion: By the principle of strong mathematical induction, F(n) < 2^n for all n ≥ 0
Choosing Between Weak and Strong Induction
- Weak induction is typically sufficient for most problems.
- Strong induction is useful when the inductive step requires the truth of the statement for multiple preceding values, not just the immediately preceding value.
- If unsure, try weak induction first; consider strong induction if you get stuck.
Application of Mathematical Induction
- Used widely in computer science to prove the correctness and efficiency of algorithms.
- Crucial in mathematics for proving theorems in number theory, combinatorics, and analysis.
Example: Sum of the First n Natural Numbers
- Proposition: P(n): 1 + 2 + 3 +... + n = n(n + 1) / 2
- Base Case (n = 1): 1 = 1(1 + 1) / 2, which simplifies to 1 = 1. Thus, P(1) is true.
- Inductive Hypothesis: Assume P(k) is true for some k ≥ 1. That is, assume 1 + 2 + 3 +... + k = k(k + 1) / 2.
- Inductive Step: Prove P(k + 1) is true. We want to show that 1 + 2 + 3 +... + (k + 1) = (k + 1)(k + 2) / 2.
- Starting with the left side: 1 + 2 + 3 +... + (k + 1) = (1 + 2 + 3 +... + k) + (k + 1)
- By the inductive hypothesis, 1 + 2 + 3 +... + k = k(k + 1) / 2.
- Substituting, we get k(k + 1) / 2 + (k + 1) = [k(k + 1) + 2(k + 1)] / 2 = (k + 1)(k + 2) / 2.
- This is exactly P(k + 1), so P(k + 1) is true.
- Conclusion: By the principle of mathematical induction, P(n) is true for all n ≥ 1. Therefore, the sum of the first n natural numbers is n(n + 1) / 2.
Example: Divisibility
- Prove that 3^(2n) - 1 is divisible by 8 for all n ≥ 1.
- Proposition: P(n): 3^(2n) - 1 is divisible by 8.
- Base Case (n = 1): 3^(2*1) - 1 = 3^2 - 1 = 9 - 1 = 8, which is divisible by 8. Thus, P(1) is true.
- Inductive Hypothesis: Assume P(k) is true for some k ≥ 1. That is, assume 3^(2k) - 1 is divisible by 8. This means 3^(2k) - 1 = 8m for some integer m.
- Inductive Step: Prove P(k + 1) is true. We want to show that 3^(2(k + 1)) - 1 is divisible by 8.
- 3^(2(k + 1)) - 1 = 3^(2k + 2) - 1 = 3^(2k) * 3^2 - 1 = 9 * 3^(2k) - 1
- Using the inductive hypothesis, 3^(2k) = 8m + 1. Substituting, we get 9(8m + 1) - 1 = 72m + 9 - 1 = 72m + 8 = 8(9m + 1).
- Since 8(9m + 1) is divisible by 8, P(k + 1) is true.
- Conclusion: By the principle of mathematical induction, P(n) is true for all n ≥ 1. Therefore, 3^(2n) - 1 is divisible by 8 for all n ≥ 1.
Example: Inequality
- Prove that 2^n > n for all n ≥ 1.
- Proposition: P(n): 2^n > n
- Base Case (n = 1): 2^1 > 1, which simplifies to 2 > 1. Thus, P(1) is true.
- Inductive Hypothesis: Assume P(k) is true for some k ≥ 1. That is, assume 2^k > k.
- Inductive Step: Prove P(k + 1) is true. We want to show that 2^(k + 1) > k + 1.
- Starting with the left side: 2^(k + 1) = 2 * 2^k
- By the inductive hypothesis, 2^k > k. Therefore, 2 * 2^k > 2 * k.
- We want to show that 2k > k + 1.
- Since k ≥ 1, k > 1. Therefore, k + k > k + 1, which simplifies to 2k > k + 1.
- So, 2^(k + 1) > 2k > k + 1, which means 2^(k + 1) > k + 1. Thus, P(k + 1) is true.
- Conclusion: By the principle of mathematical induction, P(n) is true for all n ≥ 1. Therefore, 2^n > n for all n ≥ 1.
Common Mistakes
- Forgetting to prove the base case.
- Incorrectly applying the inductive hypothesis.
- Making algebraic errors in the inductive step.
- Not clearly stating the proposition P(n).
- Assuming the statement you are trying to prove; the inductive hypothesis is an assumption, not something you prove.
Strong Induction
- In strong induction, instead of assuming P(k) is true, assumption is P(1), P(2),..., P(k) are all true.
- This can be useful when the truth of P(k + 1) depends on multiple preceding cases, not just P(k).
- The basic structure remains similar: base case(s), inductive hypothesis (strong form), inductive step, and conclusion.
Example using Strong Induction: Fibonacci Numbers
- Let F(n) be the nth Fibonacci number, defined as F(1) = 1, F(2) = 1, and F(n) = F(n-1) + F(n-2) for n > 2.
- Proposition: F(n) < 2^n for all n ≥ 1.
- Base Cases:
- n = 1: F(1) = 1 < 2^1 = 2
- n = 2: F(2) = 1 < 2^2 = 4
- Inductive Hypothesis: Assume F(i) < 2^i for all i such that 1 ≤ i ≤ k, where k ≥ 2.
- Inductive Step: Show that F(k + 1) < 2^(k + 1).
- By definition, F(k + 1) = F(k) + F(k - 1).
- By the inductive hypothesis, F(k) < 2^k and F(k - 1) < 2^(k - 1).
- Therefore, F(k + 1) < 2^k + 2^(k - 1).
- Now, 2^k + 2^(k - 1) = 2^(k - 1) * (2 + 1) = 3 * 2^(k - 1).
- Since k ≥ 2, we know 3 < 4 = 2^2 = 2^(k+1-(k-1))
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