Comparing Radical Sums by Contradiction

Comparing the sum of two square roots to a single square root is a classic exercise in algebraic inequalities that appears in HSC Extension 1. In this problem we use proof by contradiction to determine whether $\sqrt{3}+\sqrt{5}$ exceeds $\sqrt{11}$. By working through the steps, you will strengthen your skill at manipulating radical inequalities and solidify the contradiction proof technique.

Problem Statement

Prove by contradiction that $\sqrt{3} + \sqrt{5} > \sqrt{11}$.

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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 start by assuming the opposite of what we want to prove. If that assumption forces a logical impossibility, then the original statement must be true.

Step 1: Assume the negation

Assume, for the sake of contradiction, that the statement is false. That is, assume:

$$\sqrt{3} + \sqrt{5} \leq \sqrt{11}$$

Step 2: Square both sides

Since both sides of the inequality are positive, squaring preserves the direction and removes the outer square roots.

$$\begin{aligned} (\sqrt{3} + \sqrt{5})^2 &\leq (\sqrt{11})^2 \\ 3 + 2\sqrt{3}\cdot\sqrt{5} + 5 &\leq 11 \\ 8 + 2\sqrt{15} &\leq 11 \\ 2\sqrt{15} &\leq 3 \end{aligned}$$

Step 3: Square again

We still have a square root term, so we square a second time to obtain a purely numerical inequality.

$$\begin{aligned} (2\sqrt{15})^2 &\leq 3^2 \\ 4 \cdot 15 &\leq 9 \\ 60 &\leq 9 \end{aligned}$$

Step 4: Establish contradiction

Now we have a simple numerical inequality to verify. The statement $60 \leq 9$ is clearly false.

This contradiction arose from our assumption that $\sqrt{3} + \sqrt{5} \leq \sqrt{11}$.

Therefore, our assumption must be false, and we conclude:

$$\sqrt{3} + \sqrt{5} > \sqrt{11}$$

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Takeaways

• Proof by Contradiction Structure: Assume the negation of what you want to prove, derive a logical impossibility, conclude original statement must be true
• Squaring Inequalities: When both sides are positive, squaring preserves the inequality direction
• Algebraic Manipulation: Expand $(\sqrt{3}+\sqrt{5})^2$ carefully: $(a+b)^2 = a^2 + 2ab + b^2$
• Clear Contradictions: A numerical impossibility like $60 \leq 9$ is an immediate and decisive contradiction

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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 Combinatorics, HSC Collections