In HSC Maths Extension 2, the ability to write a precise logical negation is not just a formality—it’s the key that unlocks proof by contradiction and contrapositive arguments. By working through this set of examples, you’ll internalise De Morgan’s laws and the negation of implications, building the rock‑solid foundation you need for every proof you’ll construct across the syllabus.

Mastering Negation & Logic: A Guide for HSC Maths Extension 2

Introduction

Welcome to the foundational world of formal logic! If you're tackling NSW HSC Mathematics Extension 2, you already know that rigorous proof is the backbone of the course. But before you can master Proof by Contradiction, Induction, or Contrapositive, you need to understand the absolute basics of logic—specifically, how to correctly construct and negate mathematical statements.

What is it, and why is it useful?

In logic, a proposition is a statement that is strictly either true or false. Negation is simply the mathematical equivalent of saying "Not." It flips the truth value of a proposition.

Why is this useful? Because higher-level mathematics rarely deals with straightforward calculations. Instead, we deal in implications ("If A, then B"). When we use Proof by Contradiction, our very first step is to assume the negation of the statement we want to prove. If you negate it incorrectly, your entire proof collapses before you even write your first equation!

Quick Review: AND, OR, and Negation

Let's define our basic logical operators for propositions $P$ and $Q$:

• AND (Conjunction, $\land$): True only if both $P$ and $Q$ are true.
• OR (Disjunction, $\lor$): True if at least one of $P$ or $Q$ is true.
• NOT (Negation, $\neg$): Flips true to false, and false to true.

De Morgan's Laws

Just like we expand brackets in arithmetic using the distributive law, we can "expand" a negation across an AND/OR statement. This is governed by De Morgan's Laws:

• $\neg(P \land Q) \equiv \neg P \lor \neg Q$
• $\neg(P \lor Q) \equiv \neg P \land \neg Q$

Notice the trick: Negate both individual statements, and flip the operator (AND becomes OR, OR becomes AND).

The Negation of an Implication

An implication statement takes the form "If $P$, then $Q$" (written as $P \rightarrow Q$). Logically, $P \rightarrow Q$ is equivalent to $\neg P \lor Q$.

If we apply De Morgan's Laws to negate this, we get a crucial result:

$$\neg(P \rightarrow Q) \equiv P \land \neg Q$$

In plain English: To disprove "If it rains, the ground is wet", you must find a scenario where it rains ($P$ is true) AND the ground is NOT wet ($\neg Q$ is true).

Solving the Problems (Practice)

Let's apply these rules to some practice examples similar to what you'll see in your Ext 2 textbook.

*Question 1: What is the negation of the statement "A is false OR B is true"?*

• Solution: We first interpret the given sentence as a logical combination: “$A$ is false” is one proposition, and “$B$ is true” is another, joined by OR.
  Let $P$ be "$A$ is false" and $Q$ be "$B$ is true".
• The original statement is $P \lor Q$. Because we are negating an OR statement, De Morgan’s Law tells us to change the OR to an AND and negate each part individually. Using De Morgan's Law, the negation is $\neg P \land \neg Q$.
• Finally, we translate the symbols back into English: $\neg P$ means “$A$ is true” and $\neg Q$ means “$B$ is false”. Therefore, the negated statement is: "$A$ is true AND $B$ is false."

*Question 2: What is the negation of "If $x$ is even, then $y$ is odd"?*

• Solution: Recognising this as an implication “$P \rightarrow Q$” is the crucial first step. The negation of an implication always breaks the “if… then” structure and forces the hypothesis to be true while the conclusion is false.
  This is an implication $P \rightarrow Q$. The negation is $P \land \neg Q$.
• Therefore, the negation is: "$x$ is even AND $y$ is not odd (i.e., $y$ is even)."
• (Note: It is NEVER "If $x$ is odd, then...")

Question 3: Write down the mathematical/logical negation of each statement:

1. All birds can fly.
2. $m > n$
3. John likes apples and bananas.
4. $0 < x \le 10$

• Solutions:

1. To negate a universal claim, we show that the property fails for at least one member of the group. "All" becomes "At least one". Negation: At least one bird cannot fly.
2. For an inequality, we must recall that the strict “greater than” flips to the non‑strict “less than or equal”. The opposite of strictly greater than is less than or equal to: $m \le n$
3. De Morgan’s Law instructs us to negate each simple statement inside the conjunction and switch AND to OR. Apply De Morgan's Law: John does not like apples OR he does not like bananas.
4. To handle a compound inequality, we first rewrite it as the intersection of two conditions, then apply De Morgan. First, expand it: $x > 0$ AND $x \le 10$. Negate both and flip AND to OR: $x \le 0$ OR $x > 10$.

Key Takeaways

• De Morgan's Laws are your best friend: Distribute the NOT, and flip the AND/OR signs.
• Implications are tricky: The negation of $P \rightarrow Q$ is $P \land \neg Q$. It completely drops the "If... then" structure!
• Watch your inequalities: The negation of $>$ is $\le$ (don't forget the 'or equals to' part!). Compound inequalities like $a < x < b$ must be split into AND statements before negating.
• Watch your quantifiers: "All" negates to "At least one" (or "Some"), and "None" negates to "At least one".

Conclusion

Negation isn't just about being contrary; it's a precise mathematical operation. By mastering De Morgan's laws and the negation of implications, you are laying down the bedrock required for the advanced proofs in the HSC Maths Extension 2 syllabus. Keep practicing these translations from English to logic and back again—they'll become second nature in no time!