Contradiction with a + b <= 5

Problem Statement

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

Hints

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

Solutions

Proof by Contradiction

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

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.)

Step 3: Derive a contradiction

Adding these inequalities:

\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.

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 $b \leq 2$ .

\hfill $\square$

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 $\neg Q \land \neg R$

Further Readings

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

HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
HSC Last Resorts: https://vumaths.com/booklets/hsc-last-resorts/
HSC Sequences: https://vumaths.com/booklets/hsc-sequences/

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