Contradiction with a + b <= 5

Proof by contradiction is a cornerstone of logical reasoning in the HSC Extension 1 and Extension 2 courses. It works by assuming the opposite of what we want to prove and then chasing that assumption until it collides with a known fact. In this problem, you’ll learn how to negate an ‘or’ statement correctly, apply the integer property that turns strict inequalities into weak ones, and create a direct clash with the given hypothesis. Mastering these moves will make many later proofs much easier to unpick.

Problem Statement

Prove by contradiction that if $a, b$ are integers and $a + b \leq 5$, then $a \leq 2$ or $b \leq 2$.

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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

Proof by Contradiction

In a proof by contradiction we suppose the hypothesis is true but the conclusion is false, and then we look for a logical conflict. That conflict will force us to accept that the conclusion must actually be true.

Step 1: Assume the negation of the conclusion

We want to prove: If $a + b \leq 5$, then $a \leq 2$ or $b \leq 2$.

Assume the conclusion is false. The negation of ``$a \leq 2$ or $b \leq 2$'' is:

$$a > 2 \quad \text{AND} \quad b > 2$$

De Morgan’s law for logic tells us that the negation of an or is an and with both parts negated, so we get two strict inequalities.

Step 2: Apply the integer constraint

Since $a$ and $b$ are integers with $a > 2$ and $b > 2$, we must have:

$$a \geq 3 \quad \text{and} \quad b \geq 3$$

(The smallest integer greater than 2 is 3 — there is no integer strictly between 2 and 3, so the strict inequality forces the weak one.)

Step 3: Derive a contradiction

Now we combine the two lower bounds by addition:

$$\begin{aligned} a + b &\geq 3 + 3 \\ a + b &\geq 6 \end{aligned}$$

But this contradicts our hypothesis that $a + b \leq 5$.

We cannot have both $a + b \geq 6$ and $a + b \leq 5$ simultaneously. The two statements describe non‑overlapping sets of numbers, so the assumption that the conclusion is false has led us to an impossible situation.

Conclusion

Since assuming the negation of the conclusion leads to a contradiction, the conclusion must be true.

Therefore, if $a, b$ are integers with $a + b \leq 5$, then $a \leq 2$ or $b \leq 2$.

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Takeaways

• Negating "Or" Statements: The negation of $P$ or $Q$ is not $P$ AND not $Q$
• Integer Constraints: For integers, $a > 2$ implies $a \geq 3$ (no integers strictly between 2 and 3)
• Contradiction Structure: Derive statement that directly contradicts a hypothesis (here: $a+b \geq 6$ vs $a+b \leq 5$)
• Logical Form: Statement has form $(P \implies Q \lor R)$; negate conclusion to get $\neg Q \land \neg R$

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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 Last Resorts, HSC Sequences