Maths class 10th ppt.pptx

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1. POWER POINT PR E SENTA T ION FOR CLASS X MATHEMATICS RANJITH B S CLASS 10 th B ROLL NO 26

2. CHAPTER 1 REAL NUMBERS

3. TOPICS The fundamental theorem of Arithmetic. Revisiting irrational numbers. Revisiting rational numbers and their decimal representation.

4. HIGHEST COMMON FACTOR(HCF) HCF of (two positive integers a and b)is the largest positive integer that divides both a and b

5. THE FUNDAMENTAL THEOREM OF ARITHMETIC Every composite number can be expressed (factorised) as a product of primes, and this factorization is unique.

6. The method of finding the HCF and LCM of two positive numbers by the prime factorization method. Example: Find HCF and LCM of 108 and 150 108 =2² X 3³ and 150 =2 X 3 X 5² HCF(108,150) =2 X3 = Product of SMALLEST power of each common prime factor in the numbers. LCM(108,150)= 2² X 3³ X5² = Product of GREATEST power of each common prime factor in the numbers. Notice that HCF(108,150) X LCM(108,150)= 108 X150

7. For any two positive integers a and b, HCF (a,b) X LCM(a,b) = a X b This result can be used to find the LCM of two numbers .

8. R EVISITING IRRATIONAL NUMBERS In this section, we will prove that numbers of the form √ p are irrational where p is a prime. Example: Prove √ 2 is irrational. Proof: Assume √ 2 is rational. Then √ 2 =a/b ,where a and b are co-prime and b ≠ 0. Squaring both sides, we get 2b² = a² , i.e. 2 divides a²,implies 2 divides a. Let a=2c.Then , substituting for a, we get 2b²=4c² i.e.b² = 2c² This means that 2 divides b²,and so divides b.

9. Therefore, a and b have at least 2 as a common factor. This contradicts the fact a and b have no common factors other than 1. So, we conclude that √ 2 is irrational. Similarly, we can prove that √ 3 , √ 5 etc are irrational. Example: Show that 3 − √ 5 is irrational. Proof: Assume 3 − √ 5 is rational. Then 3 − √ 5=a/b, where and b are co-prime,b ≠ 0. Rearranging the equation, we get √ 5=3 − (a/b) = (3b − a)/b Since a and b are integers (3b − a)/b is rational, and so , √ 5 is rational. This contradicts the fact that √ 5 is irrational. Therefore, our assumption is wrong.

10. REVISITING RATIONAL NUMBERS AND THEIIR DECIMAL REPRESENTATION . 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 in the form 2 n 5 m where n and m are non-negative integers. Example : 0.107 = 107/1000= 107/(2³ x 5³ ) Example: 7.28 = 728/100 = 728 / 10²

11. THEOREM 2 Let x = p/q be a rational number such that prime factorisation of q is of the form 2 n 5 m wh ere n and m are non-negative integers. Then x has a decimal representation that terminates. Example: 3/8 = 3/2³ = 0.375 Example: 13/250=13/ 2 x 5³ = 0.052

12. THEOREM 3 Let x=p/q ,where p and q are co- primes be a rational number such that prime factorisation of q is not of the form 2 n 5 m ,where n and m are non-negative integers. Then ,x has a decimal expansion which is non- terminating repeating .

13. THANK YOU