Contrapositive of Multiples of 6

Understanding the contrapositive is a fundamental logical tool in HSC Mathematics Extension 1, especially in the topic of Proof (ME-P1). In this problem, we examine a simple conditional statement about multiples of 6 and multiples of 2 to reinforce how to correctly form its contrapositive. By working through the logical structure, you'll learn to avoid common pitfalls like confusing the contrapositive with the inverse or the converse.

Problem Statement

Consider the statement: ``For all integers $n$, if $n$ is a multiple of 6, then $n$ is a multiple of 2.''

Which of the following is the contrapositive of this statement?

• There exists an integer $n$ such that $n$ is a multiple of 6 and not a multiple of 2.
• There exists an integer $n$ such that $n$ is a multiple of 2 and not a multiple of 6.
• For all integers $n$, if $n$ is not a multiple of 2, then $n$ is not a multiple of 6.
• For all integers $n$, if $n$ is not a multiple of 6, then $n$ is not a multiple of 2.

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Hints

Attempt the proof independently first. Focus on the key theorem, algebraic transformation, or contradiction setup that links the hypothesis to the target conclusion.

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Solutions

We'll break down the statement into its quantified conditional form to make the contrapositive transformation clear.

Logical Analysis of Contrapositive

Step 1: Identify the logical structure

The original statement has the form:

$$\forall n \in \mathbb{Z}, \quad P(n) \implies Q(n)$$

Where:

• $P(n)$: ``$n$ is a multiple of 6''
• $Q(n)$: ``$n$ is a multiple of 2''

Now we recall the definition of the contrapositive and apply it directly.

Step 2: Apply contrapositive definition

The contrapositive of $P \implies Q$ is $\neg Q \implies \neg P$.

Key: The universal quantifier (``for all'') remains unchanged.

Next, we need to negate each component carefully.

Step 3: Negate each part

• $\neg Q(n)$: ``$n$ is NOT a multiple of 2''
• $\neg P(n)$: ``$n$ is NOT a multiple of 6''

Combining the negated parts, we obtain the contrapositive.

Step 4: Construct the contrapositive

$$\forall n \in \mathbb{Z}, \quad \neg Q(n) \implies \neg P(n)$$

In words: ``For all integers $n$, if $n$ is not a multiple of 2, then $n$ is not a multiple of 6.''

This matches Option C.

To ensure you can spot common mistakes, let's analyse why the other options are wrong.

Analysis of incorrect options:

• Option A: Negation of original ($\exists n: P(n) \land \neg Q(n)$), not contrapositive
• Option B: Negation of converse
• Option D: Inverse ($\neg P \implies \neg Q$), not contrapositive

Answer: C}

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Takeaways

• Contrapositive Form: $P \implies Q$ has contrapositive $\neg Q \implies \neg P$ (swap and negate both parts)
• Logical Equivalence: A statement and its contrapositive are logically equivalent (same truth value)
• Quantifiers Unchanged: Universal quantifier (``for all'') stays when forming contrapositive
• Common Errors: Inverse ($\neg P \implies \neg Q$) and converse ($Q \implies P$) are NOT equivalent to original

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Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills: HSC Proofs, HSC Integrals, HSC Functions