Welcome dear students! Today we are going to learn about Real Numbers from Class 10 Maths. In Class IX, you began your exploration of the world of real numbers and encountered irrational numbers. We continue our discussion on real numbers in this chapter. We begin with a very important property of positive integers in Section 1.2, namely the Fundamental Theorem of Arithmetic. The Fundamental Theorem of Arithmetic has to do with the multiplication of positive integers. You already know that every composite number can be expressed as a product of primes in a unique way. This important fact is the Fundamental Theorem of Arithmetic. While it is a result that is easy to state and understand, it has some very deep and significant applications in the field of mathematics. We use the Fundamental Theorem of Arithmetic to prove the irrationality of many of the numbers you studied in Class IX, such as √2, √3, and √5. So let us begin our exploration.
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In your earlier classes, you have seen that any natural number can be written as a product of its prime factors. For instance, 2 = 2, 4 = 2 × 2, 253 = 11 × 23, and so on. Now, let us try and look at natural numbers from the other direction. That is, can any natural number be obtained by multiplying prime numbers? Let us see. Take any collection of prime numbers, say 2, 3, 7, 11 and 23. If we multiply some or all of these numbers, allowing them to repeat as many times as we wish, we can produce a large collection of positive integers. In fact, there are infinitely many primes. So, if we combine all these primes in all possible ways, we will get an infinite collection of numbers, all the primes and all possible products of primes. The question is – can we produce all the composite numbers this way? Do you think that there may be a composite number which is not the product of powers of primes? Before we answer this, let us factorise positive integers, that is, do the opposite of what we have done so far. We are going to use the factor tree with which you are all familiar. Let us take some large number, say, 32760, and factorise it. In the diagram, 32760 is at the top, branching into 2 and 16380. 16380 branches into 2 and 8190. 8190 branches into 2 and 4095. 4095 branches into 3 and 1365. 1365 branches into 3 and 455. 455 branches into 5 and 91. Finally, 91 branches into 7 and 13. All the leaves of this tree are prime numbers.
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So we have factorised 32760 as 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 as a product of primes, i.e., 32760 = 2³ × 3² × 5 × 7 × 13 as a product of powers of primes. Let us try another number, say, 123456789. This can be written as 3² × 3803 × 3607. Of course, you have to check that 3803 and 3607 are primes! Try it out for several other natural numbers yourself. This leads us to a conjecture that every composite number can be written as the product of powers of primes. In fact, this statement is true, and is called the Fundamental Theorem of Arithmetic because of its basic crucial importance to the study of integers. Let us now formally state this theorem. Theorem 1.1 (Fundamental Theorem of Arithmetic): 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. The Fundamental Theorem of Arithmetic says that every composite number can be factorised as a product of primes. Actually it says more. It says that given any composite number it can be factorised as a product of prime numbers in a ‘unique’ way, except for the order in which the primes occur. That is, given any composite number there is one and only one way to write it as a product of primes, as long as we are not particular about the order in which the primes occur. So, for example, we regard 2 × 3 × 5 × 7 as the same as 3 × 5 × 7 × 2, or any other possible order in which these primes are written. This fact is also stated in the following form: The prime factorisation of a natural number is unique, except for the order of its factors.
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An equivalent version of Theorem 1.2 was probably first recorded as Proposition 14 of Book IX in Euclid’s Elements, before it came to be known as the Fundamental Theorem of Arithmetic. However, the first correct proof was given by Carl Friedrich Gauss in his Disquisitiones Arithmeticae. Carl Friedrich Gauss is often referred to as the ‘Prince of Mathematicians’ and is considered one of the three greatest mathematicians of all time, along with Archimedes and Newton. He has made fundamental contributions to both mathematics and science. In general, given a composite number x, we factorise it as x = p₁p₂ ... pₙ, where p₁, p₂,..., pₙ are primes and written in ascending order, i.e., p₁ ≤ p₂ ≤ . . . ≤ pₙ. If we combine the same primes, we will get powers of primes. For example, 32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 2³ × 3² × 5 × 7 × 13. Once we have decided that the order will be ascending, then the way the number is factorised, is unique. The Fundamental Theorem of Arithmetic has many applications, both within mathematics and in other fields. Let us look at some examples. Example 1: Consider the numbers 4ⁿ, where n is a natural number. Check whether there is any value of n for which 4ⁿ ends with the digit zero. Solution: If the number 4ⁿ, for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 4ⁿ would contain the prime 5. This is not possible because 4ⁿ = (2²)ⁿ; so the only prime in the factorisation of 4ⁿ is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4ⁿ. So, there is no natural number n for which 4ⁿ ends with the digit zero.
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You have already learnt how to find the HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic in earlier classes, without realising it! This method is also called the prime factorisation method. Let us recall this method through an example. Example 2: Find the LCM and HCF of 6 and 20 by the prime factorisation method. Solution: We have: 6 = 2¹ × 3¹ and 20 = 2 × 2 × 5 = 2² × 5¹. You can find HCF(6, 20) = 2 and LCM(6, 20) = 2 × 2 × 3 × 5 = 60, as done in your earlier classes. Note that HCF(6, 20) = 2¹ = Product of the smallest power of each common prime factor in the numbers. LCM(6, 20) = 2² × 3¹ × 5¹ = Product of the greatest power of each prime factor, involved in the numbers. From the example above, you might have noticed that HCF(6, 20) × LCM(6, 20) = 6 × 20. In fact, we can verify that for any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b. We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers. Example 3: Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM. Solution: The prime factorisation of 96 and 404 gives: 96 = 2⁵ × 3, 404 = 2² × 101. Therefore, the HCF of these two integers is 2² = 4. Also, LCM(96, 404) = (96 × 404) / HCF(96, 404) = (96 × 404) / 4 = 9696.
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Example 4: Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method. Solution: We have: 6 = 2 × 3, 72 = 2³ × 3², 120 = 2³ × 3 × 5. Here, 2¹ and 3¹ are the smallest powers of the common factors 2 and 3, respectively. So, HCF(6, 72, 120) = 2¹ × 3¹ = 2 × 3 = 6. 2³, 3² and 5¹ are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers. So, LCM(6, 72, 120) = 2³ × 3² × 5¹ = 360. Remark: Notice, 6 × 72 × 120 ≠ HCF(6, 72, 120) × LCM(6, 72, 120). So, the product of three numbers is not equal to the product of their HCF and LCM. EXERCISE 1.1. Question 1: Express each number as a product of its prime factors. For 140: 140 = 2² × 5 × 7. For 156: 156 = 2² × 3 × 13. For 3825: 3825 = 3² × 5² × 17. For 5005: 5005 = 5 × 7 × 11 × 13. For 7429: 7429 = 17 × 19 × 23. Question 2: Find the LCM and HCF of the following pairs and verify that LCM × HCF = product of the two numbers. For 26 and 91: 26 = 2 × 13, 91 = 7 × 13. HCF = 13. LCM = 2 × 7 × 13 = 182. Product = 26 × 91 = 2366. HCF × LCM = 13 × 182 = 2366. Verified. For 510 and 92: 510 = 2 × 3 × 5 × 17. 92 = 2² × 23. HCF = 2. LCM = 2² × 3 × 5 × 17 × 23 = 23460. Product = 510 × 92 = 46920. HCF × LCM = 2 × 23460 = 46920. Verified. For 336 and 54: 336 = 2⁴ × 3 × 7. 54 = 2 × 3³. HCF = 2 × 3 = 6. LCM = 2⁴ × 3³ × 7 = 3024. Product = 336 × 54 = 18144. HCF × LCM = 6 × 3024 = 18144. Verified.
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Question 3: Find the LCM and HCF by prime factorisation. For 12, 15 and 21: 12 = 2² × 3. 15 = 3 × 5. 21 = 3 × 7. HCF = 3. LCM = 2² × 3 × 5 × 7 = 420. For 17, 23 and 29: All are primes. HCF = 1. LCM = 17 × 23 × 29 = 11339. For 8, 9 and 25: 8 = 2³. 9 = 3². 25 = 5². HCF = 1. LCM = 2³ × 3² × 5² = 1800. Question 4: Given HCF(306, 657) = 9, find LCM(306, 657). Using HCF × LCM = product of numbers, we get 9 × LCM = 306 × 657. So LCM = (306 × 657) / 9 = 22338. Question 5: Check whether 6ⁿ can end with the digit 0 for any natural number n. If it ends in 0, it must be divisible by 5. But 6ⁿ = 2ⁿ × 3ⁿ. Its prime factors are only 2 and 3. By the Fundamental Theorem of Arithmetic, it cannot contain 5. So 6ⁿ can never end with 0. Question 6: Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. For the first: 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × 78. Since it has factors 13 and 78 other than 1 and itself, it is composite. For the second: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × 1009. It has factors 5 and 1009, so it is composite. Question 7: Sonia takes 18 minutes per round, Ravi takes 12 minutes per round. They start together. When will they meet again at the starting point? We need the LCM of 18 and 12. 18 = 2 × 3². 12 = 2² × 3. LCM = 2² × 3² = 36. They will meet again after 36 minutes.
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Now let us move to Section 1.3, Revisiting Irrational Numbers. In Class IX, you were introduced to irrational numbers and many of their properties. You studied about their existence and how the rationals and the irrationals together made up the real numbers. You even studied how to locate irrationals on the number line. However, we did not prove that they were irrationals. In this section, we will prove that √2, √3, √5 and, in general, √p is irrational, where p is a prime. One of the theorems, we use in our proof, is the Fundamental Theorem of Arithmetic. Recall, a number ‘s’ is called irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0. Some examples of irrational numbers, with which you are already familiar, are: √2, √3, √15, π, 0.10110111011110... Before we prove that √2 is irrational, we need the following theorem, whose proof is based on the Fundamental Theorem of Arithmetic. Theorem 1.2: Let p be a prime number. If p divides a², then p divides a, where a is a positive integer. *Note: This proof is not from the examination point of view.* Proof: Let the prime factorisation of a be as follows: a = p₁p₂ . . . pₙ, where p₁, p₂, . . ., pₙ are primes, not necessarily distinct. Therefore, a² = (p₁p₂ . . . pₙ)(p₁p₂ . . . pₙ) = p₁²p₂² . . . pₙ². Now, we are given that p divides a². Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a². However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a² are p₁, p₂, . . ., pₙ. So p is one of p₁, p₂, . . ., pₙ. Now, since a = p₁p₂ . . . pₙ, p divides a.
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We are now ready to give a proof that √2 is irrational. The proof is based on a technique called ‘proof by contradiction’. Theorem 1.3: √2 is irrational. Proof: Let us assume, to the contrary, that √2 is rational. So, we can find integers r and s (≠ 0) such that √2 = r/s. Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get √2 = a/b, where a and b are coprime. So, b√2 = a. Squaring on both sides and rearranging, we get 2b² = a². Therefore, 2 divides a². Now, by Theorem 1.2, it follows that 2 divides a. So, we can write a = 2c for some integer c. Substituting for a, we get 2b² = 4c², that is, b² = 2c². This means that 2 divides b², and so 2 divides b (again using Theorem 1.2 with p = 2). Therefore, a and b have at least 2 as a common factor. But this contradicts the fact that a and b have no common factors other than 1. This contradiction has arisen because of our incorrect assumption that √2 is rational. So, we conclude that √2 is irrational. Example 5: Prove that √3 is irrational. Solution: Let us assume, to the contrary, that √3 is rational. That is, we can find integers a and b (≠ 0) such that √3 = a/b. Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime. So, b√3 = a. Squaring on both sides, and rearranging, we get 3b² = a². Therefore, a² is divisible by 3, and by Theorem 1.2, it follows that a is also divisible by 3. So, we can write a = 3c for some integer c. Substituting for a, we get 3b² = 9c², that is, b² = 3c². This means that b² is divisible by 3, and so b is also divisible by 3 (using Theorem 1.2 with p = 3).
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Therefore, a and b have at least 3 as a common factor. But this contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that √3 is rational. So, we conclude that √3 is irrational. In Class IX, we mentioned that: the sum or difference of a rational and an irrational number is irrational and the product and quotient of a non-zero rational and irrational number is irrational. We prove some particular cases here. Example 6: Show that 5 – √3 is irrational. Solution: Let us assume, to the contrary, that 5 – √3 is rational. That is, we can find coprime a and b (b ≠ 0) such that 5 – √3 = a/b. Rearranging this equation, we get √3 = 5 – a/b = (5b – a)/b. Since a and b are integers, we get (5b – a)/b is rational, and so √3 is rational. But this contradicts the fact that √3 is irrational. This contradiction has arisen because of our incorrect assumption that 5 – √3 is rational. So, we conclude that 5 – √3 is irrational. Example 7: Show that 3√2 is irrational. Solution: Let us assume, to the contrary, that 3√2 is rational. That is, we can find coprime a and b (b ≠ 0) such that 3√2 = a/b. Rearranging, we get √2 = a/(3b). Since 3, a and b are integers, a/(3b) is rational, and so √2 is rational. But this contradicts the fact that √2 is irrational. So, we conclude that 3√2 is irrational.
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Now let us solve Exercise 1.2 completely. Question 1: Prove that √5 is irrational. Solution: Assume √5 is rational. Then √5 = a/b, where a and b are coprime integers and b ≠ 0. Squaring gives 5 = a²/b², so a² = 5b². This means 5 divides a². By Theorem 1.2, 5 divides a. Let a = 5c. Substituting gives 25c² = 5b², so b² = 5c². Thus 5 divides b², and by Theorem 1.2, 5 divides b. This means a and b share 5 as a common factor, contradicting that they are coprime. Hence, √5 is irrational. Question 2: Prove that 3 + 2√5 is irrational. Solution: Assume 3 + 2√5 is rational. Then 3 + 2√5 = a/b, where a and b are coprime and b ≠ 0. Rearranging gives 2√5 = a/b – 3 = (a – 3b)/b. So √5 = (a – 3b)/(2b). Since a, b, 3, and 2 are integers, (a – 3b)/(2b) is rational. This implies √5 is rational, which contradicts our proven result. Hence, 3 + 2√5 is irrational. Question 3: Prove that the following are irrationals. Part (i): 1/√2. Assume it is rational. Then 1/√2 = a/b, where a and b are coprime, b ≠ 0. Then √2 = b/a. Since b and a are integers, b/a is rational, implying √2 is rational. Contradiction. So 1/√2 is irrational. Part (ii): 7√5. Assume it is rational. Then 7√5 = a/b. Then √5 = a/(7b). The right side is rational, implying √5 is rational. Contradiction. So 7√5 is irrational. Part (iii): 6 + √2. Assume it is rational. Then 6 + √2 = a/b. Then √2 = a/b – 6 = (a – 6b)/b. The right side is rational, implying √2 is rational. Contradiction. So 6 + √2 is irrational.
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Let us now review the summary of this chapter. First, the Fundamental Theorem of Arithmetic: 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. Second, if p is a prime and p divides a², then p divides a, where a is a positive integer. Third, we have proved that √2, √3 are irrationals. A note to the reader: You have seen that HCF(p, q, r) × LCM(p, q, r) ≠ p × q × r, where p, q, r are positive integers. However, the following results hold good for three numbers p, q and r: LCM(p, q, r) = [p × q × r × HCF(p, q, r)] / [HCF(p, q) × HCF(q, r) × HCF(p, r)] and HCF(p, q, r) = [p × q × r × LCM(p, q, r)] / [LCM(p, q) × LCM(q, r) × LCM(p, r)].
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]