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Real Numbers - Revisiting Irrational Numbers

Grade 10CBSE

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

🔑Concepts

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Definition: A number is called irrational if it cannot be written in the form pq\frac{p}{q}, where pp and qq are integers and q≠0q \neq 0.

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Theorem: Let pp be a prime number. If pp divides a2a^2, then pp divides aa, where aa is a positive integer. This is the fundamental property used in proofs of irrationality.

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Proof by Contradiction: To prove a number like 2\sqrt{2} is irrational, we assume it is rational (ab\frac{a}{b}), show that aa and bb share a common factor (violating the co-prime assumption), and conclude the original assumption was wrong.

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Arithmetic Properties: The sum or difference of a rational and an irrational number is always irrational. E.g., 3+23 + \sqrt{2} is irrational.

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Product Properties: The product or quotient of a non-zero rational number and an irrational number is irrational. E.g., 252\sqrt{5} or 37\frac{3}{\sqrt{7}} is irrational.

📐Formulae

p∣a2  ⟹  p∣a (where p is a prime number)p \mid a^2 \implies p \mid a \text{ (where } p \text{ is a prime number)}

If x∈Q and y∉Q, then x+y∉Q\text{If } x \in \mathbb{Q} \text{ and } y \notin \mathbb{Q}, \text{ then } x + y \notin \mathbb{Q}

If x∈Q,x≠0 and y∉Q, then x⋅y∉Q\text{If } x \in \mathbb{Q}, x \neq 0 \text{ and } y \notin \mathbb{Q}, \text{ then } x \cdot y \notin \mathbb{Q}

💡Examples

Problem 1:

Prove that 5\sqrt{5} is irrational.

Solution:

Assume to the contrary that 5\sqrt{5} is rational. Then there exist co-prime integers aa and bb (b≠0b \neq 0) such that 5=ab\sqrt{5} = \frac{a}{b}. Squaring both sides: 5=a2b2  ⟹  a2=5b25 = \frac{a^2}{b^2} \implies a^2 = 5b^2. This means 55 divides a2a^2, so 55 must divide aa. Let a=5ca = 5c for some integer cc. Substituting this: (5c)2=5b2  ⟹  25c2=5b2  ⟹  5c2=b2(5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies 5c^2 = b^2. This means 55 divides b2b^2, so 55 must divide bb. Since 55 divides both aa and bb, they are not co-prime. This contradicts our assumption. Thus, 5\sqrt{5} is irrational.

Explanation:

This is a proof by contradiction using the theorem that if a prime divides the square of an integer, it divides the integer itself.

Problem 2:

Show that 7−37 - \sqrt{3} is irrational, given that 3\sqrt{3} is irrational.

Solution:

Assume 7−37 - \sqrt{3} is rational. Let 7−3=r7 - \sqrt{3} = r, where rr is a rational number. Rearranging the equation: 3=7−r\sqrt{3} = 7 - r. Since 77 is rational and rr is rational, their difference 7−r7 - r must also be a rational number. This implies that 3\sqrt{3} is rational. However, this contradicts the given fact that 3\sqrt{3} is irrational. Therefore, our assumption is false, and 7−37 - \sqrt{3} is irrational.

Explanation:

This utilizes the property that the difference between two rational numbers is always rational.

Problem 3:

If a student performs the following calculation with rational approximations, find the difference: 1.73205081−1.414213560.31783725\begin{array}{r} 1.73205081 \\ - 1.41421356 \\ \hline 0.31783725 \end{array} If these were the exact values of 3\sqrt{3} and 2\sqrt{2}, would the result be rational or irrational?

Solution:

The result of the subtraction is 0.317837250.31783725. Since this result is a terminating decimal, it is a rational number. In general, if we subtract two irrational numbers, the result can be rational or irrational; however, the difference of two specific terminating decimals is always rational.

Explanation:

While 3−2\sqrt{3} - \sqrt{2} is irrational, any calculation using fixed decimal approximations results in a rational number because terminating decimals are rational.