Real Numbers | Exercise 1.2

Question 2

Prove that 3+253 + 2\sqrt{5} is irrational.

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Solution
Understand the Question
  • To prove that 3+253 + 2\sqrt{5} is irrational, we use the method of proof by contradiction.
  • We assume that 3+253 + 2\sqrt{5} is rational and can be expressed in the form ab\dfrac{a}{b}, where aa and bb are co-prime integers with b≠0b \neq 0.
  • Rearranging the expression isolates 5\sqrt{5} as a ratio of integers, showing that 5\sqrt{5} would have to be rational, which contradicts the known fact that 5\sqrt{5} is irrational.

Step 1 · Assume Rationality and Rearrange

Let us assume, to the contrary, that 3+253 + 2\sqrt{5} is rational.

Then, there exist co-prime integers aa and bb (b≠0b \neq 0) such that:

3+25=ab25=ab−325=a−3bb5=a−3b2b\begin{aligned} 3 + 2\sqrt{5} &= \dfrac{a}{b} \\[0.6em] 2\sqrt{5} &= \dfrac{a}{b} - 3 \\[0.6em] 2\sqrt{5} &= \dfrac{a - 3b}{b} \\[0.6em] \sqrt{5} &= \dfrac{a - 3b}{2b} \end{aligned}

Step 2 · Establish the Contradiction

Since aa and bb are integers, a−3ba - 3b and 2b2b are integers with 2b≠02b \neq 0.

Therefore, a−3b2b\dfrac{a - 3b}{2b} is a rational number, which implies that 5\sqrt{5} is rational.

This contradicts the fact that 5\sqrt{5} is irrational.

Hence, our assumption that 3+253 + 2\sqrt{5} is rational is false.

Answer

Hence, 3+253 + 2\sqrt{5} is irrational.

Common Mistakes
  • Omitting Co-prime / Integer Definitions: Forgetting to mention that aa and bb are integers with b≠0b \neq 0.
  • Re-proving 5\sqrt{5} from Scratch: Trying to prove that 5\sqrt{5} is irrational is unnecessary unless specifically asked; you can use it as an established fact.
  • Algebraic Transposition Errors: Incorrectly dividing or rearranging, such as writing 5=a−32b\sqrt{5} = \dfrac{a - 3}{2b} instead of a−3b2b\dfrac{a - 3b}{2b}.

More questions in Exercise 1.2

Q1

Prove that 5\sqrt{5} is irrational.

Q2

Prove that 3+253 + 2\sqrt{5} is irrational.

Q3

Prove that the following are irrationals :

(i) 12\dfrac{1}{\sqrt{2}}

(ii) 757\sqrt{5}

(iii) 6+26 + \sqrt{2}

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