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Sum of Surds Irrationality - Proof by Contradiction

Authors
  • avatar
    Name
    Vu Hung
    Twitter

Problem Statement

Prove by contradiction that the real number x=2+3+5x = \sqrt{2} + \sqrt{3} + \sqrt{5} is irrational.


Hints

  • The Setup: Begin with the standard assumption for contradiction: let xQx \in \mathbb{Q}.
  • Strategic Isolation: Do not square the expression as it is. Squaring a trinomial creates a mess of cross-terms. Isolate one of the surds (e.g., move 5\sqrt{5} to the left) before squaring.
  • The Double Square: Squaring once will leave you with a 5\sqrt{5} term on the left and a 6\sqrt{6} term on the right. Isolate the 6\sqrt{6} term and square a second time to eliminate it.
  • The Contradiction Engine: Rearrange your final equation to make 5\sqrt{5} the subject. Rely on the closure property of rational numbers (addition, subtraction, multiplication, and non-zero division of rationals always yield a rational) to force the contradiction.

Solutions

Proof by Contradiction

Step 1: Assume the negation

Assume, for the purpose of contradiction, that xx is a rational number (xQx \in \mathbb{Q}).

Step 2: Isolate one surd

Rearrange the equation to isolate one surd: x5=2+3x - \sqrt{5} = \sqrt{2} + \sqrt{3}

Step 3: Square both sides

(x5)2=(2+3)2x22x5+5=2+26+3x22x5=26\begin{aligned} (x - \sqrt{5})^2 &= (\sqrt{2} + \sqrt{3})^2 \\ x^2 - 2x\sqrt{5} + 5 &= 2 + 2\sqrt{6} + 3 \\ x^2 - 2x\sqrt{5} &= 2\sqrt{6} \end{aligned}

Step 4: Square a second time

(x22x5)2=(26)2x44x35+20x2=24\begin{aligned} (x^2 - 2x\sqrt{5})^2 &= (2\sqrt{6})^2 \\ x^4 - 4x^3\sqrt{5} + 20x^2 &= 24 \end{aligned}

Step 5: Make 5\sqrt{5} the subject

4x35=x4+20x2245=x4+20x2244x3\begin{aligned} 4x^3\sqrt{5} &= x^4 + 20x^2 - 24 \\ \sqrt{5} &= \frac{x^4 + 20x^2 - 24}{4x^3} \end{aligned}

Since x>0x > 0, division by 4x34x^3 is valid.

Step 6: Derive the contradiction

Because the rational numbers are closed under addition, subtraction, multiplication, and non-zero division, the right-hand side evaluates to a rational number. However, the left-hand side, 5\sqrt{5}, is known to be irrational.

Conclusion

A rational number cannot equal an irrational number. This is a contradiction. Therefore, the initial assumption is false, and x=2+3+5x = \sqrt{2} + \sqrt{3} + \sqrt{5} must be irrational.


Takeaways

  • Algebraic Foresight: Recognising that squaring (a+b+c)2(a + b + c)^2 yields too many irrational cross-terms is the key separator for top-band students. Isolate one surd before squaring.
  • The Power of Closure: The entire proof hinges on knowing that manipulating rational xx through integer powers and coefficients guarantees a rational output on the right-hand side.
  • Structured Argumentation: The proof demands a clear logical flow from the initial assumption to an explicit statement of why the final line constitutes a mathematical contradiction.
  • Double Squaring Technique: This method extends the squaring strategy used in simpler surd proofs (such as comparing 3+5\sqrt{3} + \sqrt{5} with 11\sqrt{11}) to irrationality arguments involving multiple surds.
  • You are encouraged to prove that 2+3+6\sqrt{2} + \sqrt{3} + \sqrt{6} is irrational using a different method, which can be smarter than the technique used here. Note that 6=236 = 2 \cdot 3, which may be useful.