Worksheet
Real Numbers encompass all rational and irrational numbers, forming a complete and continuous number line essential for various mathematical concepts.
Real Numbers - Practice Worksheet
Strengthen your foundation with key concepts and basic applications.
This worksheet covers essential long-answer questions to help you build confidence in Real Numbers from Mathematics for Class X (Mathematics).
Basic comprehension exercises
Strengthen your understanding with fundamental questions about the chapter.
Questions
Explain the Fundamental Theorem of Arithmetic and its significance.
Think about how prime numbers are the building blocks of all numbers and why their uniqueness in factorization matters.
Solution
The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur. This theorem is significant because it provides a foundational understanding of the structure of numbers, ensuring that any number can be broken down into its prime components in only one way. For example, the number 60 can be factorized into 2×2×3×5, and no other combination of primes will multiply to give 60. This uniqueness is crucial for various applications in mathematics, such as finding the HCF and LCM of numbers, simplifying fractions, and solving problems related to divisibility. The theorem also underpins the proof of the irrationality of certain numbers, like √2, by showing that assuming their rationality leads to a contradiction in the prime factorization. Understanding this theorem helps in grasping more complex concepts in number theory and algebra.
How does Euclid's division algorithm work, and how is it used to find the HCF of two numbers?
Remember, the key is to keep dividing the divisor by the remainder until the remainder is zero.
Solution
Euclid's division algorithm is a method to find the Highest Common Factor (HCF) of two positive integers. The algorithm is based on the principle that the HCF of two numbers also divides their difference. The steps involve dividing the larger number by the smaller number, then dividing the divisor by the remainder, and repeating this process until the remainder is zero. The last non-zero remainder is the HCF. For example, to find the HCF of 56 and 72: 72 = 56×1 + 16; 56 = 16×3 + 8; 16 = 8×2 + 0. The last non-zero remainder is 8, so the HCF is 8. This method is efficient and systematic, ensuring that the HCF is found without needing to factorize the numbers, which can be complex for large numbers. It's a fundamental tool in number theory and has applications in cryptography and computer science.
Prove that √2 is irrational using the Fundamental Theorem of Arithmetic.
Start by assuming √2 is rational and look for a contradiction in the prime factorization.
Solution
To prove that √2 is irrational, we assume the opposite, that √2 is rational, meaning it can be expressed as a fraction a/b where a and b are coprime integers. Squaring both sides gives 2 = a²/b², or 2b² = a². This implies that a² is even, and by the Fundamental Theorem of Arithmetic, a must also be even (since the square of an odd number is odd). Let a = 2k for some integer k. Substituting back, we get 2b² = (2k)² = 4k², simplifying to b² = 2k². This means b² is even, and so b must also be even. But if both a and b are even, they have a common factor of 2, contradicting our assumption that they are coprime. This contradiction arises from assuming √2 is rational, so √2 must be irrational. This proof highlights the power of the Fundamental Theorem of Arithmetic in establishing the nature of numbers.
Explain why the product of a non-zero rational number and an irrational number is irrational.
Assume the opposite and show that it leads to a contradiction with the definition of irrational numbers.
Solution
The product of a non-zero rational number and an irrational number is irrational because if it were rational, the irrational number could be expressed as the ratio of two rational numbers, which contradicts its definition. Let's consider a rational number p/q (where p and q are integers, q ≠ 0) and an irrational number r. Assume their product (p/q)×r is rational, say a/b (where a and b are integers, b ≠ 0). Then, r = (a/b)/(p/q) = (aq)/(bp), which is a ratio of integers, making r rational. This contradicts the assumption that r is irrational. Therefore, the product must be irrational. For example, 2 (rational) × √3 (irrational) = 2√3, which cannot be expressed as a simple fraction, confirming its irrationality.
Find the LCM and HCF of 12, 15, and 21 using the prime factorization method.
Break each number into its prime factors and use the highest powers for LCM and the lowest common powers for HCF.
Solution
To find the LCM and HCF of 12, 15, and 21 using prime factorization, first break each number down into its prime factors: 12 = 2² × 3; 15 = 3 × 5; 21 = 3 × 7. The HCF is found by taking the lowest power of each common prime factor. Here, the only common prime factor is 3, so HCF = 3. The LCM is found by taking the highest power of each prime factor present in the numbers: 2², 3, 5, and 7. Thus, LCM = 2² × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420. This method ensures that the LCM is the smallest number that all original numbers divide into without leaving a remainder, and the HCF is the largest number that divides all original numbers without leaving a remainder.
Why can't the number 6ⁿ end with the digit 0 for any natural number n?
Consider the prime factors needed for a number to end with 0 and check if 6ⁿ has those factors.
Solution
The number 6ⁿ cannot end with the digit 0 for any natural number n because a number ends with 0 only if it is divisible by 10, which requires the prime factors 2 and 5. The prime factorization of 6 is 2 × 3, so 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ. This shows that the only prime factors of 6ⁿ are 2 and 3, and it lacks the prime factor 5, which is necessary for divisibility by 10. Therefore, no power of 6 can include the prime factor 5, making it impossible for 6ⁿ to end with the digit 0. This conclusion is based on the Fundamental Theorem of Arithmetic, which guarantees the uniqueness of prime factorization.
Show that the sum of a rational number and an irrational number is irrational.
Assume the sum is rational and derive a contradiction.
Solution
To show that the sum of a rational number and an irrational number is irrational, assume the opposite: that the sum is rational. Let the rational number be p/q (where p and q are integers, q ≠ 0) and the irrational number be r. Assume their sum (p/q) + r is rational, say a/b (where a and b are integers, b ≠ 0). Then, r = (a/b) - (p/q) = (aq - bp)/(bq), which is a ratio of integers, implying r is rational. This contradicts the assumption that r is irrational. Therefore, the sum must be irrational. For example, 1/2 (rational) + √2 (irrational) = (1 + 2√2)/2, which cannot be expressed as a simple fraction, confirming its irrationality.
Explain the concept of terminating and non-terminating repeating decimal expansions of rational numbers.
Look at the denominator's prime factors to determine the nature of the decimal expansion.
Solution
A rational number has a terminating decimal expansion if its denominator, after simplifying, can be expressed as a product of powers of 2 and/or 5. For example, 1/2 = 0.5 (terminating) because the denominator is 2. A rational number has a non-terminating repeating decimal expansion if its denominator has prime factors other than 2 or 5. For example, 1/3 = 0.333... (non-terminating repeating) because the denominator is 3. This behavior is due to the way division works in base 10, where only denominators that divide 10 (i.e., 2 and 5) result in terminating decimals. The Fundamental Theorem of Arithmetic helps in determining the nature of the decimal expansion by analyzing the prime factors of the denominator.
Prove that 3 + 2√5 is irrational.
Assume the number is rational and show that it leads to √5 being rational.
Solution
To prove that 3 + 2√5 is irrational, assume the opposite: that it is rational, meaning it can be expressed as a/b where a and b are coprime integers. Then, 3 + 2√5 = a/b, which can be rearranged to 2√5 = (a/b) - 3 = (a - 3b)/b, and further to √5 = (a - 3b)/(2b). Since a and b are integers, (a - 3b)/(2b) is rational, implying √5 is rational. However, we know that √5 is irrational, leading to a contradiction. Therefore, our assumption that 3 + 2√5 is rational must be false, proving it is irrational. This proof relies on the fact that the sum or difference of a rational and an irrational number is irrational.
Find the HCF and LCM of 306 and 657, given that their HCF is 9.
Use the relationship HCF × LCM = Product of the two numbers to find the LCM.
Solution
Given the HCF of 306 and 657 is 9, we can find the LCM using the relationship between HCF and LCM of two numbers: HCF × LCM = Product of the numbers. So, 9 × LCM = 306 × 657. First, calculate 306 × 657 = 201,042. Then, divide by 9 to find LCM: LCM = 201,042 / 9 = 22,338. This method is efficient when the HCF is known, avoiding the need for prime factorization. It demonstrates how the Fundamental Theorem of Arithmetic underpins these calculations, ensuring that the product of HCF and LCM equals the product of the numbers themselves.
Real Numbers - Mastery Worksheet
Advance your understanding through integrative and tricky questions.
This worksheet challenges you with deeper, multi-concept long-answer questions from Real Numbers to prepare for higher-weightage questions in Class X Mathematics.
Intermediate analysis exercises
Deepen your understanding with analytical questions about themes and characters.
Questions
Explain the Fundamental Theorem of Arithmetic and its significance in the study of numbers.
Think about how prime factorization is used in various proofs and calculations.
Solution
The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur. This theorem is significant because it provides a foundational understanding of the structure of numbers, enabling the proof of irrationality of numbers like √2, √3, etc., and helps in finding the HCF and LCM of numbers efficiently.
Prove that √5 is irrational using the Fundamental Theorem of Arithmetic.
Use proof by contradiction and the theorem that if a prime divides a², it divides a.
Solution
Assume √5 is rational, so it can be written as a/b where a and b are coprime integers. Squaring both sides gives 5 = a²/b² ⇒ a² = 5b². This implies 5 divides a², and by the Fundamental Theorem of Arithmetic, 5 divides a. Let a = 5c. Substituting back gives 25c² = 5b² ⇒ 5c² = b², implying 5 divides b² and thus b. This contradicts a and b being coprime. Therefore, √5 is irrational.
Compare and contrast the methods to find HCF and LCM using prime factorization and Euclid's division algorithm.
Consider the computational efficiency and the nature of the results each method provides.
Solution
Prime factorization involves breaking down numbers into their prime factors to find HCF (product of smallest powers of common primes) and LCM (product of greatest powers of all primes). Euclid's division algorithm is a step-by-step division process to find HCF only, more efficient for large numbers. While prime factorization is straightforward, it can be cumbersome for large numbers, whereas Euclid's algorithm is more efficient but doesn't directly provide LCM.
Show that the product of three numbers is not necessarily equal to the product of their HCF and LCM.
Choose specific numbers to test the statement.
Solution
Consider the numbers 6, 72, and 120. Their HCF is 6 and LCM is 720. The product of the numbers is 6 × 72 × 120 = 51840, whereas the product of HCF and LCM is 6 × 720 = 4320. Clearly, 51840 ≠ 4320, demonstrating that the product of three numbers is not equal to the product of their HCF and LCM.
Explain why 7 × 11 × 13 + 13 is a composite number.
Factor out the common term to reveal the composite nature.
Solution
The expression can be factored as 13(7 × 11 + 1) = 13 × 78. Since it is the product of two integers greater than 1 (13 and 78), the number is composite.
Prove that 3 + 2√5 is irrational.
Assume the opposite and reach a contradiction.
Solution
Assume 3 + 2√5 is rational, so it can be written as a/b where a and b are coprime integers. Rearranging gives √5 = (a - 3b)/(2b). Since a and b are integers, (a - 3b)/(2b) is rational, implying √5 is rational, which contradicts the known irrationality of √5. Therefore, 3 + 2√5 is irrational.
Find the HCF and LCM of 12, 15, and 21 using prime factorization.
Identify the common and all prime factors with their highest powers.
Solution
Prime factorizations: 12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7. HCF is the product of the smallest powers of common primes: 3. LCM is the product of the greatest powers of all primes: 2² × 3 × 5 × 7 = 420.
Explain the concept of irrational numbers and give two examples not mentioned in the textbook.
Think beyond square roots and π.
Solution
Irrational numbers cannot be expressed as a fraction p/q where p and q are integers and q ≠ 0. They have non-terminating, non-repeating decimal expansions. Examples include Euler's number (e) ≈ 2.71828... and the golden ratio (φ) ≈ 1.61803...
How does the Fundamental Theorem of Arithmetic help in proving the irrationality of numbers?
Consider the role of unique factorization in contradiction proofs.
Solution
The theorem ensures unique prime factorization, allowing us to assume that a number can be expressed as a fraction in lowest terms (coprime numerator and denominator). When squaring leads to a contradiction (like both numerator and denominator being divisible by a prime), it proves the number cannot be rational, hence irrational.
Given that HCF (306, 657) = 9, find LCM (306, 657).
Use the formula connecting HCF, LCM, and the product of two numbers.
Solution
Using the relationship HCF × LCM = Product of the two numbers, we have LCM = (306 × 657) / 9 = 306 × 73 = 22338.
Real Numbers - Challenge Worksheet
Push your limits with complex, exam-level long-form questions.
The final worksheet presents challenging long-answer questions that test your depth of understanding and exam-readiness for Real Numbers in Class X.
Advanced critical thinking
Test your mastery with complex questions that require critical analysis and reflection.
Questions
Prove that the square root of any prime number is irrational using the Fundamental Theorem of Arithmetic.
Consider the uniqueness of prime factorization and how it applies to both sides of the equation.
Solution
Assume the contrary that √p is rational, express it as a/b in lowest terms, square both sides to get p = a²/b², leading to p*b² = a². By the Fundamental Theorem of Arithmetic, p must divide a, leading to a contradiction since a and b were supposed to be coprime.
Explain why the sum of a rational number and an irrational number is always irrational.
Think about the properties of rational and irrational numbers under addition.
Solution
Assume the sum is rational, then the irrational number could be expressed as the difference of two rational numbers, which is a contradiction since the difference of two rationals is rational.
Given two numbers, explain how to find their HCF and LCM using the Fundamental Theorem of Arithmetic.
Remember that the HCF is about commonality, and the LCM is about the union of all primes.
Solution
Factorize both numbers into their prime factors. The HCF is the product of the smallest power of each common prime, and the LCM is the product of the greatest power of each prime present in the numbers.
Why can't a number ending with the digit 0 be a perfect square unless it ends with an even number of zeros?
Consider the prime factorization of a number ending with zeros and the definition of a perfect square.
Solution
A number ending with 0 must have both 2 and 5 in its prime factorization. For it to be a perfect square, the exponents of 2 and 5 must be even, implying an even number of zeros.
Demonstrate that the product of two irrational numbers can be either rational or irrational, with examples.
Think about multiplying an irrational number by itself versus by another irrational number.
Solution
Example 1: √2 * √2 = 2 (rational). Example 2: √2 * √3 = √6 (irrational). The product depends on whether the irrational numbers are multiplicatively inverses or not.
Using Euclid's division lemma, show that the square of any positive integer is of the form 3m or 3m + 1.
Consider the possible remainders when an integer is divided by 3 and square each case.
Solution
Any integer can be expressed as 3k, 3k+1, or 3k+2. Squaring each form shows the result is either divisible by 3 or leaves a remainder of 1.
Explain the significance of the Fundamental Theorem of Arithmetic in proving the irrationality of numbers.
Think about how unique factorization leads to contradictions in proofs by contradiction.
Solution
It guarantees a unique prime factorization, allowing us to argue about the divisibility properties of numbers and contradictions when assuming rationality of irrational numbers.
How does the decimal expansion of a rational number relate to its denominator's prime factors?
Consider how the denominator's prime factors affect the divisibility in the base 10 system.
Solution
A rational number has a terminating decimal expansion if and only if the prime factors of the denominator, after simplifying, are only 2 and/or 5.
Prove that there are infinitely many prime numbers.
Use proof by contradiction and consider the properties of prime numbers.
Solution
Assume finitely many primes, multiply them all and add 1. The result is either a new prime or has a prime factor not in the original list, leading to a contradiction.
Discuss the application of the Fundamental Theorem of Arithmetic in cryptography.
Think about how prime factorization is used in public-key cryptography.
Solution
It underpins the security of RSA encryption, where the difficulty of factoring large numbers into primes ensures the encryption's strength.
Explore the world of Polynomials, understanding their types, degrees, and operations to solve algebraic expressions and equations effectively.
Explore the methods to solve a pair of linear equations in two variables, including graphical, substitution, elimination, and cross-multiplication techniques.
Explore the world of quadratic equations, learning to solve them using various methods like factoring, completing the square, and the quadratic formula.
A chapter that explores sequences where each term after the first is obtained by adding a constant difference, focusing on their properties, nth term, and sum formulas.
