Wednesday, August 8, 2018

1 Real Numbers - Class 10 : Notes



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Real Numbers - Class 10 : Notes


Notes for real numbers chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.
(1) Euclid’s Division Lemma:Theorem: Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.
(2) Euclid’s division algorithm: To obtain the HCF of two positive integers, say c and d, with c > d, follow the steps below:Step 1: Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d.Step 2: If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.
For ExampleUse Euclid’s division algorithm to find the HCF of 135 and 225.Step 1: Here 225 > 135, on applying the division lemma to 225 and 135, we get 225 = 135 x 1 + 90Step 2: Since, remainder ≠ 0, we again apply division lemma to 135 and 90, we get 135 = 90 x 1 + 45Step 3: Again, applying division lemma to 90 and 45, we get 90 = 45 x 2 + 0The remainder has become zero. And since the divisor at this step is 45, the HCF of 135 and 225 is 45.
For ExampleShow that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.Let a be any positive odd integer. And we apply division algorithm with a and b = 4.As 0 ≤ r < 4, the possible remainders could be 0, 1, 2 and 3.So, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient.Now, since a is odd, so a cannot be 4q or 4q + 2 (as both are divisible by 2).Hence, any odd integer is of the form 4q + 1 or 4q + 3.
(3) The Fundamental Theorem of Arithmetic:Theorem: 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. In general, given a composite number x, we factorise it as x = p1 p2 ... pn , where p1, p2 ,..., pn are primes and written in ascending order, i.e., p1 ≤ p2 ≤ . . . ≤ p. If we combine the same primes, we will get powers of primes.
For ExampleThe prime factors of 32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 23 × 32 × 5 × 7 × 13.
For ExampleFind the HCF and LCM of 96 and 404 by prime factorisation method.The prime factorisation of 96 is 25 x 3. And that of 404 is 22 x 101.Hence, HCF of 96 and 404 will be 22 = 4.Now, LCM (96, 404) = (96 x 404)/(HCF (96, 404)) = (96 x 404)/4 = 9696.
For ExampleCheck whether 6n can end with the digit 0 for any natural number n.If a number ends with digit 0, then, it must be divisible by 10 or in other words, it will be divisible by 2 and 5 as 10 = 2 x 5.Now, prime factorisation of 6n = (2 x 3)n.Here, 5 is not in the prime factorisation of 6n. Hence, for any value of n, 6n will not be divisible by 5.Thus, 6n cannot end with the digit 0 for any natural number n.
(4) Revisiting Irrational Numbers:Irrational Number : are the numbers which cannot be written in p/q form, where p and q are integers and q≠ 0.Theorem 1: Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer.ProofSuppose the prime factorisation of a is as follows:(i) a = p1p2....pn, where p1,p2,....pn are primes.(ii) On squaring both the sides, we get,(iii) a2 = (p1p2....pn) ( p1p2....pn) = p12p22....pn2.(iv) It is given that p divides a2. Hence, we can say that p is one of the prime factors of a2 as per the Fundamental Theorem of Arithmetic.(v) However, as per the uniqueness part of the Fundamental Theorem of Arithmetic, we can deduce that the only prime factors of aare p1p2....pn. Thus, p is one of p1p2....pn.Since, a = p1p2....pn,p divides a.  
Theorem 2: √2 is irrational.ProofWe shall start by assuming √2 as rational. In other words, we need to find integers x and y such that √2 = x/y.(i) Let x and y have a common factor other than 1, and so we can divide by that common factor and assume that x and y are coprime. So, y√2 = x.(ii) Squaring both side, we get, 2y2 = x2.(iii) Thus, 2 divides x2. and by theorem we can say that 2 divides x.(iv) Hence, x = 2z for some integer z.(v) Substituting x, we get, 2x2 = 4z2e. y2 = 4z2; which means yis divisible by 2, and so y will also be divisible by 2.(vi) Now, from theorem, x and y will have 2 as a common factor. But, it is opposite to fact that x and y are co-prime.(vii) Hence, we can conclude √2 is irrational.
For ExampleProve that √3 is irrational.We shall start by assuming √3 as rational. In other words, we need to find integers x and y such that √3 = x/y.Let x and y have a common factor other than 1, and so we can divide by that common factor and assume that x and y are coprime. So, y√3 = x.Squaring both side, we get, 3y2 = x2.Thus, x2 is divisible by 3, and by theorem we can say that x is also divisible by 3.Hence, x = 3z for some integer z.Substituting a, we get, 3x2 = 9z2 i.e. y2 = 3z2; which means yis divisible by 3, and so y will also be divisible by 3.Now, from theorem, x and y will have 3 as a common factor. But, it is opposite to fact that x and y are co-prime.Hence, we can conclude √3 is irrational.
For ExampleProve that 6 + √2 is irrational.Let us assume 6 + √2 to be rational.Therefore, we must find two integers a, b (b ≠ 0) such that6 + √2 = a/b i.e. √2 = a/b – 6.Since, a and b are integers, a/b – 6 is also rational and hence √2 must be rational.Now, this contradicts the fact that √2 is irrational.Hence, 6 + √2 is irrational.
(5) Revisiting Rational Numbers and Their Decimal Expansions:Theorem 1: Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form, p/q where p and q are co-prime, and the prime factorisation of q is of the form 2n 5m, where n, m are non-negative integers.For Example13/125 = 13/53 = (13 x 23)/(2x 53) = 104/103 = 0.104
Theorem 2: Let x = p/q be a rational number, such that the prime factorisation of q is of the form 25m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.
Theorem 3: Let x = p/q be a rational number, such that the prime factorisation of q is not of the form 2n 5m, where n, m are non-negative integers. Then, x has a decimal expansion which is non-terminating repeating (recurring).For ExampleWithout actually performing the long division, state whether 6/15 will have a terminating decimal expansion or a non-terminating repeating decimal expansion.The prime factorisation of 6/15 can be written as6/15 = (2 x 3)/(3 x 5) = 2/5Here, the denominator is of the form 5n.Hence, decimal expansion of 6/15 is terminating.
For ExampleWrite down the decimal expansions of 17/8.The decimal expansion of 17/8 is
For ExampleThe following real number has decimal expansions as given below. Decide whether it is rational or not. If it is rational, and of the form, p/q what can you say about the prime factors of q?Here, as the decimal expansion is non-terminating recurring, the given number is a rational number of the form p/q.Moreover, q is not of the form 2n 5m, hence, prime factors of q will also have factors other than 2 or 5

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