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Real Numbers - Proofs of Irrationality

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Rational Numbers: A rational number is any number that can be expressed in the form pq\frac{p}{q}, where pp and qq are integers and q≠0q \neq 0. Visually, these numbers represent points on a number line that correspond to terminating decimals (like 0.50.5) or non-terminating repeating decimals (like 0.333...0.333...).

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Irrational Numbers: These are numbers that cannot be written in the form pq\frac{p}{q}. Their decimal expansions are non-terminating and non-repeating. On a number line, irrational numbers fill the gaps between rational numbers to form the set of real numbers.

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The Fundamental Theorem of Arithmetic: This theorem states that every composite number can be uniquely expressed as a product of prime numbers, regardless of the order. This uniqueness is critical when identifying common factors in proofs.

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Divisibility Theorem (Theorem 1.3): Let pp be a prime number. If pp divides a2a^2 (where aa is a positive integer), then pp also divides aa. This logical step is the cornerstone for proving that square roots of non-square integers are irrational.

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Co-prime Numbers: Two integers aa and bb are said to be co-prime if their Highest Common Factor (HCF) is 11. In irrationality proofs, we assume a rational fraction is in its simplest form, meaning the numerator and denominator share no common factors other than 11.

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Method of Contradiction: This is a proof technique where we begin by assuming the opposite of what we want to prove (e.g., assuming 2\sqrt{2} is rational). We then use logical steps to reach an impossible result or 'contradiction,' proving the initial assumption was false.

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Properties of Irrational Numbers: The sum, difference, product (with a non-zero rational), or quotient of a rational and an irrational number always results in an irrational number. Visually, adding a rational 'shift' to an irrational point on the number line results in another irrational point.

📐Formulae

x=pq,q≠0,HCF(p,q)=1x = \frac{p}{q}, q \neq 0, HCF(p, q) = 1

If p∣a2 then p∣a (where p is prime)\text{If } p | a^2 \text{ then } p | a \text{ (where } p \text{ is prime)}

Rational±Irrational=Irrational\text{Rational} \pm \text{Irrational} = \text{Irrational}

Rational (non-zero)×Irrational=Irrational\text{Rational (non-zero)} \times \text{Irrational} = \text{Irrational}

Rational÷Irrational=Irrational\text{Rational} \div \text{Irrational} = \text{Irrational}

💡Examples

Problem 1:

Prove that 2\sqrt{2} is irrational.

Solution:

  1. Assume 2\sqrt{2} is rational. Let 2=ab\sqrt{2} = \frac{a}{b} where aa and bb are co-prime integers and b≠0b \neq 0.
  2. Squaring both sides: 2=a2b22 = \frac{a^2}{b^2}, which implies a2=2b2a^2 = 2b^2.
  3. Therefore, 22 divides a2a^2, which means 22 must divide aa (by Theorem 1.3).
  4. Let a=2ca = 2c for some integer cc. Substituting this into the equation: (2c)2=2b2⇒4c2=2b2⇒2c2=b2(2c)^2 = 2b^2 \Rightarrow 4c^2 = 2b^2 \Rightarrow 2c^2 = b^2.
  5. This means 22 divides b2b^2, so 22 must also divide bb.
  6. Since 22 divides both aa and bb, they are not co-prime. This contradicts our assumption.
  7. Conclusion: 2\sqrt{2} is irrational.

Explanation:

This solution uses the method of contradiction by assuming the number is rational and showing that the numerator and denominator must share a common factor of 2, which violates the definition of a simplified fraction (co-prime).

Problem 2:

Prove that 7−237 - 2\sqrt{3} is irrational, given that 3\sqrt{3} is irrational.

Solution:

  1. Assume 7−237 - 2\sqrt{3} is rational. Let 7−23=r7 - 2\sqrt{3} = r, where rr is a rational number.
  2. Rearrange the equation to isolate the irrational term: 7−r=237 - r = 2\sqrt{3}.
  3. Divide by 22: 7−r2=3\frac{7 - r}{2} = \sqrt{3}.
  4. Since 77 and rr are rational, their difference (7−r)(7 - r) is rational. Dividing a rational number by 22 also results in a rational number.
  5. This implies 3\sqrt{3} is rational. However, it is given that 3\sqrt{3} is irrational.
  6. A rational number cannot equal an irrational number. This is a contradiction.
  7. Conclusion: 7−237 - 2\sqrt{3} is irrational.

Explanation:

This approach relies on the properties of rational numbers. By isolating the known irrational part on one side, we show that it would have to equal a rational expression, which is mathematically impossible.