Irrationality of sum of square roots

Proving that the sum of square roots is irrational is a classic challenge that combines algebraic manipulation with logical structure. For HSC Extension 2 students, this problem reinforces the power of proof by contradiction and the careful technique of isolating surds before squaring. By working through this proof, you'll sharpen your ability to spot when squaring directly would create too many messy terms and learn how a double squaring strategy can be used to force an elegant contradiction.

Problem Statement

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

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Hints

• The Setup: Begin with the standard assumption for contradiction: let $x \in \mathbb{Q}$.
• 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 $\sqrt{5}$ to the left) before squaring.
• The Double Square: Squaring once will leave you with a $\sqrt{5}$ term on the left and a $\sqrt{6}$ term on the right. Isolate the $\sqrt{6}$ term and square a second time to eliminate it.
• The Contradiction Engine: Rearrange your final equation to make $\sqrt{5}$ 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.

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Solutions

Proof by Contradiction

Step 1: Assume the negation

We start by assuming the negation: assume, for the purpose of contradiction, that $x$ is a rational number ($x \in \mathbb{Q}$).

Step 2: Isolate one surd

The key insight is that squaring $( \sqrt{2} + \sqrt{3} + \sqrt{5} )^2$ directly would produce cross-terms like $\sqrt{6}$, $\sqrt{10}$, and $\sqrt{15}$, which are all irrational and would complicate the algebra. Instead, we isolate one surd, say $\sqrt{5}$:

$$x - \sqrt{5} = \sqrt{2} + \sqrt{3}$$

Step 3: Square both sides

Squaring both sides (using $(a-b)^2 = a^2 - 2ab + b^2$ on the left and $(c+d)^2 = c^2 + 2cd + d^2$ on the right) gives:

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

Notice that the integer terms cancel conveniently, leaving only $\sqrt{5}$ and $\sqrt{6}$ as the irrational parts.

Step 4: Square a second time

To remove $\sqrt{6}$, we square again, remembering that $(2\sqrt{6})^2 = 4 \times 6 = 24$:

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

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

We now treat $\sqrt{5}$ as the unknown and isolate it to see its relationship with $x$:

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

Since $x$ is the sum of positive square roots, $x > 0$, so $4x^3 \neq 0$ and the division is valid.

Step 6: Derive the contradiction

Now we examine the nature of this expression. Because the rational numbers are closed under addition, subtraction, multiplication, and non-zero division, if $x$ were rational, then $x^4$, $20x^2$, $24$, and $4x^3$ would all be rational, making the right-hand side a rational number. However, the left-hand side, $\sqrt{5}$, 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 $x = \sqrt{2} + \sqrt{3} + \sqrt{5}$ must be irrational.

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Takeaways

• Algebraic Foresight: Recognising that squaring $(a + b + c)^2$ 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 $x$ 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 $\sqrt{3} + \sqrt{5}$ with $\sqrt{11}$) to irrationality arguments involving multiple surds.
• You are encouraged to prove that $\sqrt{2} + \sqrt{3} + \sqrt{6}$ is irrational using a different method, which can be smarter thant the technique used here. Note that $6 = 2 \cdot 3$, which may be useful.

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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 Mechanics, HSC Trigonometry