Real Numbers: CBSE Class 10 Maths Chapter 1 Guide

Euclid's Division Lemma and Algorithm

Let's start with a concept that feels familiar but is very powerful when formalized. Euclid's Division Lemma states that for any two given positive integers, let's call them 'a' and 'b', there exist unique whole numbers 'q' and 'r' such that:

a = bq + r, where 0 ≤ r < b

This is nothing but a restatement of the long division you've always done! Here:

• a is the dividend (the number being divided).
• b is the divisor (the number we are dividing by).
• q is the quotient (the result of the division).
• r is the remainder (what's left over).

The crucial part is that the remainder 'r' is always greater than or equal to 0 and strictly less than the divisor 'b'.

This simple lemma is the key to Euclid's Division Algorithm, which is a technique or a series of well-defined steps to compute the Highest Common Factor (HCF) of two positive integers. The algorithm works by repeatedly applying the lemma until we get a remainder of zero. The last non-zero remainder's corresponding divisor is the HCF of the original two numbers.

How to Find HCF using Euclid's Division Algorithm

1. Step 1: Apply Euclid's Division Lemma — Given two positive integers, 'c' and 'd', with c > d, apply the lemma to find whole numbers 'q' and 'r' such that c = dq + r, where 0 ≤ r < d.
2. Step 2: Check the Remainder — If the remainder 'r' is zero, then the divisor 'd' is the HCF of c and d. The process stops.
3. Step 3: Continue the Process — If the remainder 'r' is not zero, apply the division lemma again, this time to the divisor 'd' and the remainder 'r'.
4. Step 4: Repeat until Remainder is Zero — Continue this process until the remainder becomes zero. The divisor at this final stage will be the required HCF.

The Fundamental Theorem of Arithmetic

This theorem is another cornerstone of number theory. It states that every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

For example, let's factorise 140:
140 = 2 × 70
140 = 2 × 2 × 35
140 = 2 × 2 × 5 × 7
140 = 2² × 5¹ × 7¹

No matter how you start factorising 140, you will always end up with two 2s, one 5, and one 7. The order might be different (e.g., 5 × 7 × 2²), but the prime factors are unique. This uniqueness is incredibly useful for finding HCF and LCM.

• HCF: The product of the smallest power of each common prime factor in the numbers.
• LCM: The product of the greatest power of each prime factor involved in the numbers.

Worked Example: Proving √3 is Irrational

• We will use the method of 'proof by contradiction'. Step 1: Assume the opposite. Let us assume that √3 is a rational number. This means it can be written in the form a/b, where 'a' and 'b' are integers, b ≠ 0, and 'a' and 'b' are co-prime (they have no common factors other than 1). √3 = a/b Step 2: Rearrange and square the equation. √3 b = a Squaring both sides, we get: (√3 b)² = a² => 3b² = a² Step 3: Analyse the result. Since 3b² = a², it means a² is divisible by 3. If a² is divisible by 3, then 'a' must also be divisible by 3. (This is a theorem: if a prime 'p' divides a², then 'p' divides 'a'). So, we can write 'a' as 3c for some integer 'c'. Step 4: Substitute and simplify. Substitute a = 3c back into the equation 3b² = a²: 3b² = (3c)² 3b² = 9c² b² = 3c² Step 5: Draw a conclusion and find the contradiction. This new equation, b² = 3c², means that b² is divisible by 3, and therefore 'b' is also divisible by 3. From Step 3, 'a' is divisible by 3. From Step 5, 'b' is divisible by 3. This means that 'a' and 'b' have a common factor of 3. This contradicts our initial assumption that 'a' and 'b' are co-prime. Final Conclusion: Our assumption was wrong. Therefore, √3 is an irrational number.

Exam Tips for Real Numbers

Students often make small but costly mistakes in this chapter. Here's what to watch out for:

• Euclid's Algorithm: Don't just write down the HCF. You must show all the steps of applying a = bq + r until the remainder is zero. Marks are awarded for the procedure.
• Proof of Irrationality: This is a very common board exam question. Memorize the structure of the proof, but understand the logic. Always end your proof by explicitly stating the contradiction: "This contradicts our assumption that 'a' and 'b' are co-prime. Hence, our assumption was incorrect..."
• HCF vs. LCM by Prime Factorisation: For HCF, you take the lowest power of common primes. For LCM, you take the highest power of all primes present in the factorisations.
• Terminating Decimals: A rational number p/q has a terminating decimal expansion if the prime factorization of the denominator 'q' is of the form 2ⁿ5ᵐ, where n and m are non-negative integers. Don't forget to simplify the fraction p/q to its lowest terms first!

Practice Questions with Solutions

• Q: Use Euclid's division algorithm to find the HCF of 867 and 255. A: Step 1: Since 867 > 255, we apply the division lemma to 867 and 255. 867 = 255 × 3 + 102 Step 2: The remainder is 102 ≠ 0. We apply the lemma to the divisor 255 and remainder 102. 255 = 102 × 2 + 51 Step 3: The remainder is 51 ≠ 0. We apply the lemma to the divisor 102 and remainder 51. 102 = 51 × 2 + 0 Step 4: The remainder is now 0. The process stops. The divisor at this stage is 51. Final answer: The HCF of 867 and 255 is 51.
• Q: Given that HCF(306, 657) = 9, find LCM(306, 657). A: Step 1: Recall the relationship between HCF and LCM for two numbers 'a' and 'b': HCF(a, b) × LCM(a, b) = a × b. Step 2: Substitute the given values into the formula. 9 × LCM(306, 657) = 306 × 657 Step 3: Solve for LCM. LCM(306, 657) = (306 × 657) / 9 LCM(306, 657) = 34 × 657 LCM(306, 657) = 22338 Final answer: The LCM of 306 and 657 is 22338.
• Q: Show that 5 - √3 is irrational. A: Step 1: Assume the contrary, that 5 - √3 is rational. Then, 5 - √3 = a/b, where a and b are co-prime integers and b ≠ 0. Step 2: Isolate the irrational part (√3). 5 - a/b = √3 (5b - a) / b = √3 Step 3: Analyze the equation. Since a and b are integers, (5b - a) and b are also integers. This means (5b - a) / b is a rational number. Therefore, √3 must be a rational number. Step 4: State the contradiction. This contradicts the fact that √3 is an irrational number. The contradiction has arisen because of our incorrect assumption. Final answer: Therefore, 5 - √3 is irrational.
• Q: Without actually performing the long division, state whether the rational number 13/3125 will have a terminating decimal expansion or a non-terminating repeating decimal expansion. A: Step 1: To determine the nature of the decimal expansion, we need to check the prime factorization of the denominator, q. Here, the number is 13/3125. The denominator is q = 3125. Step 2: Find the prime factors of the denominator. 3125 = 5 × 625 = 5 × 5 × 125 = 5 × 5 × 5 × 25 = 5 × 5 × 5 × 5 × 5 = 5⁵. Step 3: Check if the factorization is of the form 2ⁿ5ᵐ. The denominator 3125 can be written as 5⁵, which is the same as 2⁰ × 5⁵. This fits the form 2ⁿ5ᵐ where n=0 and m=5. Final answer: Since the denominator's prime factorization is in the form 2ⁿ5ᵐ, the rational number 13/3125 will have a terminating decimal expansion.

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