Comparing Radical Sums by Contradiction

Problem Statement

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

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

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

Step 2: Square both sides

Since all terms are positive, we can square both sides without changing the inequality:

Step 3: Square again

Both sides are still positive, so square again:

Step 4: Establish contradiction

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:

\hfill $\square$

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

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 Combinatorics: https://vumaths.com/booklets/hsc-combinatorics/
• HSC Collections: https://vumaths.com/booklets/hsc-collections/

Connect with me

If you're eager for more HSC Maths insights, be sure to check out my YouTube - HSC Maths Extension 1+2. For deeper dives and regular tips, join my LinkedIn. Let's tackle these challenging math problems together! You can also catch my daily math content on Website - Vu's Maths Hub.