Kaliyuga Ekalavya

NCERT Class X Chapter 1: Real Numbers Exercise 1.2 Question 3(i) Question: Prove that \( \frac{1}{\sqrt{2}} \) is irrational. Given: The number \( \frac{1}{\sqrt{2}} \). To Find: Prove that \( \frac{1}{\sqrt{2}} \) is irrational. Formula: We use the definitions of rational and irrational numbers. Definitions: Rational Number: A number that can be written as \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \). Irrational Number: A number that cannot be written as the ratio of two integers. Solution: Step 1: Assume, for contradiction, that \( \frac{1}{\sqrt{2}} \) is rational. Step 2: By definition, it can be written as \( \frac{p}{q} \), where \( p \) and \( q \) are integers, \( q \neq 0 \), and \( \gcd(p, q) = 1 \). $$ \frac{1}{\sqrt{2}} = \frac{p}{q} $$ Step 3: Cross-multiply to eliminate denominators. $$ 1 \times q = p \times \sqrt{2...

NCERT Class X Chapter 1: Real Numbers Question: Prove that \(6 + \sqrt{2}\) is irrational. Given: The number \(6 + \sqrt{2}\) To Find: Prove that \(6 + \sqrt{2}\) is irrational. Formula: If the sum of a rational number and an irrational number is rational, then the irrational number would also be rational, which is a contradiction. Solution: Step 1: Assume, for contradiction, that \(6 + \sqrt{2}\) is a rational number. Step 2: Let \(6 + \sqrt{2} = \frac{p}{q}\), where \(p\) and \(q\) are integers, \(q \ne 0\), and \(\gcd(p, q) = 1\). $$ 6 + \sqrt{2} = \frac{p}{q} $$ Step 3: Subtract 6 from both sides to isolate \(\sqrt{2}\). $$ 6 + \sqrt{2} - 6 = \frac{p}{q} - 6 $$ $$ \sqrt{2} = \frac{p}{q} - 6 $$ Step 4: Write 6 as \(\frac{6q}{q}\) to get a common denominator. $$ \sqrt{2} = \frac{p}{q} - \frac{6q}{q} $$ Step ...

NCERT Class X Chapter 1: Real Numbers Exercise 1.2 Question 3(ii) Question: Prove that \( 7\sqrt{5} \) is irrational. Given: The number \( 7\sqrt{5} \). To Find: Prove that \( 7\sqrt{5} \) is irrational. Formula: We will use the definition of rational and irrational numbers. Definitions: Rational Number: A number that can be written in the form \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \). Irrational Number: A number that cannot be written as the ratio of two integers. Solution: Step 1: Assume that \( 7\sqrt{5} \) is rational. Then it can be written as \( \frac{p}{q} \), where \( p \) and \( q \) are integers, \( q \neq 0 \), and \( \gcd(p, q) = 1 \). $$ 7\sqrt{5} = \frac{p}{q} $$ Step 2: Divide both sides by 7 to isolate \( \sqrt{5} \). $$ \sqrt{5} = \frac{p}{7q} $$ Step 3: Since \( p \) and \( q \) are integers and \( q \neq...

NCERT Class X Chapter 1: Real Numbers Question: Prove that \(3 + 2\sqrt{5}\) is irrational. Given: The number \(3 + 2\sqrt{5}\). To Find: To prove that \(3 + 2\sqrt{5}\) is irrational. Formula: We will use the method of contradiction to prove irrationality. Definitions: Rational Number: A number that can be written as \(\frac{a}{b}\), where \(a\) and \(b\) are integers and \(b \neq 0\). Irrational Number: A number that cannot be written as a fraction of two integers. Its decimal expansion is non-terminating and non-repeating. Solution: Step 1: Assume, to the contrary, that \(3 + 2\sqrt{5}\) is rational. Step 2: Let \(3 + 2\sqrt{5} = \frac{a}{b}\), where \(a\) and \(b\) are integers, \(b \neq 0\). Step 3: Subtract 3 from both sides to isolate the term with \(\sqrt{5}\). $$ 3 + 2\sqrt{5} - 3 = \frac{a}{b} - 3 $$ $$ 2\sqrt{5} = \frac{a}{b} - 3 $$ Step 4: Express the right side with a common denominator. $$ \frac{a}{b} - 3 = \frac{...

NCERT Class X Chapter 1: Real Numbers Question: Prove that $$\sqrt{5}$$ is irrational. Given: We need to prove that $$\sqrt{5}$$ is an irrational number. To Find: Show that $$\sqrt{5}$$ is irrational. Formula: Proof by Contradiction: Assume the opposite of what is to be proved and show that it leads to a contradiction. Prime Divisibility Property: If a prime number \( p \) divides \( a^2 \), then \( p \) divides \( a \). Coprime Integers: Two integers \( a \) and \( b \) are coprime if their greatest common divisor (GCD) is 1. Solution: Step 1: Assume, for contradiction, that $$\sqrt{5}$$ is rational. Then it can be written as $$\sqrt{5} = \frac{a}{b}$$, where \( a \) and \( b \) are coprime integers and \( b \ne 0 \). $$ \sqrt{5} = \frac{a}{b} $$ Step 2: Square both sides to remove the square root. $$ 5 = \frac{a^2}{b^2} $$ Step 3: Multiply both sides by \( b^2 \) to clear the...