Math Problem Statement

A proof by mathematical induction is supposed to show that a given property is true for every integer greater than or equal to an initial value. In order for it to be valid, the property must be true for the initial value, and the argument in the inductive step must be correct for every integer greater than or equal to the initial value.

Consider the following statement.

For every integer

n ≥ 1,

3n − 2

is even.

The following is a proposed proof by mathematical induction for the statement.

Since the property is true for

n = 1,

the basis step is true. Suppose the property is true for an integer k, where

k ≥ 1.

That is, suppose that

3k − 2

is even. We must show that

3k + 1 − 2

is even. Observe that

3k + 1 − 2

=

3k · 3 − 2 = 3k(1 + 2) − 2

=

(3k − 2) + 3k · 2.

Now

3k − 2

is even by inductive hypothesis, and

3k · 2

is even by inspection. Hence the sum of the two quantities is even by (Theorem 4.1.1). It follows that

3k + 1 − 2

is even, which is what we needed to show.

Identify the error(s) in the proof. (Select all that apply.)

3k − 2 is odd by the inductive hypothesis.The property is not true for n = 1.(3k − 2) + 3k · 2 ≠ 3k(1 + 2) − 2The inductive hypothesis is assumed to be true.3k + 1 − 2 ≠ (3k − 2) + 3k · 2

Solution

Let's analyze the proposed proof and each option step by step.

1. "3k − 2 is odd by the inductive hypothesis."

  • False. The inductive hypothesis assumes that 3k23k - 2 is even, not odd. In mathematical induction, the hypothesis states that the property holds for kk, which means 3k23k - 2 should be even for k1k \geq 1.

2. "The property is not true for n=1n = 1."

  • False. We can verify this by substituting n=1n = 1 into 3n23n - 2: 3(1)2=32=1,3(1) - 2 = 3 - 2 = 1, which is odd, so the property does not hold for n=1n = 1. The error is that the base case is wrong.

3. "(3k − 2) + 3k \cdot 2 ≠ 3k(1 + 2) − 2."

  • True. The expression (3k2)+3k2(3k - 2) + 3k \cdot 2 does not simplify to 3k(1+2)23k(1 + 2) - 2. The manipulations shown in the proof are not valid algebraically, so there is a mistake in the algebraic steps here.

4. "The inductive hypothesis is assumed to be true."

  • False. This is not an error. In mathematical induction, we are supposed to assume the inductive hypothesis (i.e., assume the statement is true for kk) to prove it is true for k+1k + 1.

5. "3k + 1 − 2 ≠ (3k − 2) + 3k \cdot 2."

  • True. The expression 3k+123k + 1 - 2 simplifies to 3k13k - 1, while (3k2)+3k2(3k - 2) + 3k \cdot 2 simplifies to something different. This means the step where the proof compares these two expressions is incorrect.

Summary of Errors:

  • The property is not true for n=1n = 1.
  • The algebra in the step (3k2)+3k23k(1+2)2(3k - 2) + 3k \cdot 2 \neq 3k(1 + 2) - 2 is wrong.
  • The expression 3k+12(3k2)+3k23k + 1 - 2 \neq (3k - 2) + 3k \cdot 2.

Thus, the correct errors are:

  • "The property is not true for n=1n = 1."
  • "(3k − 2) + 3k \cdot 2 ≠ 3k(1 + 2) − 2."
  • "3k + 1 − 2 ≠ (3k − 2) + 3k \cdot 2."

Do you want more details or have any questions?
Here are 5 relative questions to expand on this information:

  1. What is the correct base case for a valid proof by induction for this problem?
  2. How do we correctly apply the inductive step for this statement?
  3. What role does algebraic manipulation play in proving statements using induction?
  4. Can you explain the concept of the inductive hypothesis in more detail?
  5. How would the proof change if the initial statement were true for n=1n = 1?

Tip: When proving statements using induction, always carefully check the base case first, as an incorrect base case will invalidate the whole proof.

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Math Problem Analysis

Mathematical Concepts

Mathematical Induction
Algebra
Even and Odd Numbers

Formulas

-

Theorems

Mathematical Induction

Suitable Grade Level

Grades 10-12