Advanced & Strong Induction Applications Quiz

1) To prove n! > 2^n for n ≥ 4, what is the base case?

1! > 2^1

0! > 2^0

4! > 2^4

K! > 2^k

The base case must be the first integer for which the statement is claimed to be true. Here, n ≥ 4, so we verify n=4: 4! = 24 and 2^4 = 16, and 24 > 16, so the base case holds.

Explanation

The base case must be the first integer for which the statement is claimed to be true. Here, n ≥ 4, so we verify n=4: 4! = 24 and 2^4 = 16, and 24 > 16, so the base case holds.

2) The inductive step is essentially a general version of what logical argument form?

Modus Ponens

Modus Tollens

Disjunctive Syllogism

Hypothetical Syllogism

The inductive step proves the implication P(k) → P(k+1). Then, by Modus Ponens, if P(k) is true, we can conclude P(k+1). In the chain of induction, we repeatedly apply Modus Ponens: from P(1) and P(1) → P(2), we get P(2), and so on.

Explanation

The inductive step proves the implication P(k) → P(k+1). Then, by Modus Ponens, if P(k) is true, we can conclude P(k+1). In the chain of induction, we repeatedly apply Modus Ponens: from P(1) and P(1) → P(2), we get P(2), and so on.

3) To prove that the number of subsets of a set with n elements is 2^n, what is P(k+1)?

A set with k elements has 2^k subsets.

A set with 1 element has 2^1 subsets.

A set with k+1 elements has 2^{k+1} subsets.

A set with n elements has 2^n subsets.

A set with k+1 elements has 2^{k+1} subsets

Explanation: P(k+1) is the statement for a set with k+1 elements. The goal is to show that such a set has 2^{k+1} subsets. This typically involves considering how adding a new element doubles the number of subsets, based on the inductive hypothesis for a set with k elements.

Explanation

A set with k+1 elements has 2^{k+1} subsets

Explanation: P(k+1) is the statement for a set with k+1 elements. The goal is to show that such a set has 2^{k+1} subsets. This typically involves considering how adding a new element doubles the number of subsets, based on the inductive hypothesis for a set with k elements.

4) If you can’t prove the inductive step, it means the statement is false.

True

False

Inability to prove the inductive step does not necessarily mean the statement is false. It may indicate that the proof approach is flawed or that induction is not the suitable method. The statement could still be true, and another proof technique might succeed.

Explanation

Inability to prove the inductive step does not necessarily mean the statement is false. It may indicate that the proof approach is flawed or that induction is not the suitable method. The statement could still be true, and another proof technique might succeed.

5) The two parts of an inductive proof are:

Assumption and Conclusion

Hypothesis and Antithesis

Base Case and Inductive Step

Premise and Proof

Every proof by induction must include both a base case, where the statement is verified for the initial value, and an inductive step, where we show that if the statement holds for an arbitrary case k, it also holds for k+1. These two parts are essential and non-negotiable.

Explanation

Every proof by induction must include both a base case, where the statement is verified for the initial value, and an inductive step, where we show that if the statement holds for an arbitrary case k, it also holds for k+1. These two parts are essential and non-negotiable.

6) To prove n^3 - n is divisible by 3, what is the base case for n ≥ 1?

1^3 - 1 is divisible by 3

0^3 - 0 is divisible by 3

K^3 - k is divisible by 3

2^3 - 2 is divisible by 3

For n ≥ 1, the base case is n=1: 1^3 - 1 = 0, and 0 is divisible by 3. This establishes the statement for the smallest value, and then the inductive step can extend it to all n ≥ 1.

Explanation

For n ≥ 1, the base case is n=1: 1^3 - 1 = 0, and 0 is divisible by 3. This establishes the statement for the smallest value, and then the inductive step can extend it to all n ≥ 1.

7) When is strong induction absolutely necessary over weak induction?

When the P(k+1) case depends on P(k-1)

For all inequality proofs

For all summation proofs

It is never absolutely necessary

Strong induction is necessary when proving P(k+1) requires the truth of one or more cases before k, such as P(k-1). In weak induction, we only assume P(k), which may be insufficient if the recurrence or property relies on earlier terms.

Explanation

Strong induction is necessary when proving P(k+1) requires the truth of one or more cases before k, such as P(k-1). In weak induction, we only assume P(k), which may be insufficient if the recurrence or property relies on earlier terms.

8) The logic of an inductive proof proceeds from:

The general to the specific

The specific to the general

An assumption to a contradiction

A statement to its contrapositive

Induction starts with a specific case (the base case) and then uses the inductive step to generalize to all cases. The base case provides a concrete example, and the inductive step shows that this truth propagates to all subsequent values, leading to a general conclusion.

Explanation

Induction starts with a specific case (the base case) and then uses the inductive step to generalize to all cases. The base case provides a concrete example, and the inductive step shows that this truth propagates to all subsequent values, leading to a general conclusion.

9) Structural Induction is a variant used to prove properties of:

Numbers

Recursively defined structures like trees

Geometric constructions

Chemical formulas

Structural induction is adapted for recursively defined objects, such as lists, trees, or expressions in computer science. It proves properties by showing they hold for base structures and are preserved under construction rules, similar to mathematical induction but for non-numeric domains.

Explanation

Structural induction is adapted for recursively defined objects, such as lists, trees, or expressions in computer science. It proves properties by showing they hold for base structures and are preserved under construction rules, similar to mathematical induction but for non-numeric domains.

10) The base case must always be n=1 or n=0.

True

False

The base case can be any integer, depending on the statement. For example, if a statement is claimed for n ≥ 5, the base case is n=5. The base case is simply the first integer in the domain for which the statement is to be proven.

Explanation

The base case can be any integer, depending on the statement. For example, if a statement is claimed for n ≥ 5, the base case is n=5. The base case is simply the first integer in the domain for which the statement is to be proven.

11) To prove that any postage amount ≥ 12 cents can be made with 4- and 5-cent stamps, what is a possible set of base cases?

P(1), P(2), P(3), P(4)

P(12)

P(12), P(13), P(14), P(15)

P(4) and P(5)

For strong induction, we often need multiple base cases to ensure the inductive step works. Here, to prove P(n) for n ≥ 12, we verify base cases 12, 13, 14, 15. Then, for n ≥ 16, we can use the inductive step that P(n) can be derived by adding a 4-cent stamp to P(n-4), which requires the truth of P(n-4), and since n-4 ≥ 12, the base cases cover the initial values.

Explanation

For strong induction, we often need multiple base cases to ensure the inductive step works. Here, to prove P(n) for n ≥ 12, we verify base cases 12, 13, 14, 15. Then, for n ≥ 16, we can use the inductive step that P(n) can be derived by adding a 4-cent stamp to P(n-4), which requires the truth of P(n-4), and since n-4 ≥ 12, the base cases cover the initial values.

12) The inductive step is sometimes called the:

Base step

Bridge step

Final step

Recursive step

The inductive step is often referred to as the recursive step because it defines the truth of P(k+1) in terms of P(k), similar to how recursive functions define values based on previous values. This term emphasizes the self-referential nature of induction.

Explanation

The inductive step is often referred to as the recursive step because it defines the truth of P(k+1) in terms of P(k), similar to how recursive functions define values based on previous values. This term emphasizes the self-referential nature of induction.

13) A proof by induction is a form of what kind of reasoning?

Deductive reasoning

Abductive reasoning

Inductive reasoning

Reductive reasoning

Despite its name, mathematical induction is a form of deductive reasoning because the conclusion necessarily follows from the premises (base case and inductive step). It is not inductive reasoning in the scientific sense, which involves making generalizations from observations.

Explanation

Despite its name, mathematical induction is a form of deductive reasoning because the conclusion necessarily follows from the premises (base case and inductive step). It is not inductive reasoning in the scientific sense, which involves making generalizations from observations.

14) A student proves P(1) and P(3) and P(k) → P(k+2). What have they proven?

P(n) for all n ≥ 1

P(n) for all odd n ≥ 1

P(n) for all even n ≥ 1

Nothing

By proving P(1) and the implication P(k) → P(k+2), they have shown that P(1) implies P(3), P(3) implies P(5), and so on. This covers all odd integers starting from 1. The even integers are not covered because the base case is odd and the step increases by 2.

Explanation

By proving P(1) and the implication P(k) → P(k+2), they have shown that P(1) implies P(3), P(3) implies P(5), and so on. This covers all odd integers starting from 1. The even integers are not covered because the base case is odd and the step increases by 2.

15) The conclusion of a proof by induction is that P(n) is true for:

At least one n

Infinitely many n

All n ≥ base case

Exactly k+1

After proving the base case for n=a and the inductive step P(k) → P(k+1) for all k ≥ a, we conclude that P(n) is true for every integer n ≥ a. This is the standard conclusion of a valid induction proof.

Explanation

After proving the base case for n=a and the inductive step P(k) → P(k+1) for all k ≥ a, we conclude that P(n) is true for every integer n ≥ a. This is the standard conclusion of a valid induction proof.