Chapetr 1 real number class 10 th
- 1. GLORY ENGLISH MEDIUM SCHOOL SUBJECT: MATHEMATICS CLASS: 10TH TOPIC:REAL NUMBER(CHAPTER 1) PRESENTED BY POOJA VAISH
- 11. THE FUNDAMENTAL THEOREM OF ARITHMATIC The fundamental theorem of arithmetic -states, "Every composite number can be factorized as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur". This is also called unique factorization theorem. So we have factorised 1092 as 2*3*2*7*13 as a product of prime. Qu: Express each no as a product of its prime factor. (i) 140 By Taking the LCM of 140, we will get the product of its prime factor. Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 22×5×7 (ii) 156 By Taking the LCM of 156, we will get the product of its prime factor.
- 12. HCF(A,B)=PRODUCT OF THE SMALLEST POWER OF EACH COMMON PRIME FACTORS IN THE NUMBERS LCM(A,B)=PRODUCT OF THE GREATEST POWER OF EACH PRIME FACTOR INVOLVED IN NUMBERS FOR ANY TWO POSITIVE INTEGAR ,HCF(A,B)*LCM(A,B) =A*B Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two number. 510 and 92 510 and 92 Expressing 510 and 92 as product of its prime factors, we get, 510 = 2 × 3 × 17 × 5 × 1 92 = 2 × 2 × 23 × 1 Therefore, LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460 And HCF (510, 92) = 2 Verification Now, product of 510 and 92 = 510 × 92 = 46920 And Product of LCM and HCF = 23460 × 2 = 46920 Hence, LCM × HCF = product of the 510 and 92.
- 13. • FIND LCM & HCF BY APPLYING THE PRIME FACTORISATION METHOD 12,15,21 Writing the product of prime factors for all the three numbers, we get, 12=2×2×3 15=5×3 21=7×3 Therefore, HCF(12,15,21) = 3 LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420 • REVISITING IRRATIONAL NUMBER RATIONAL NUMBER +IRRATIONAL NUMBER=REAL NUMBER A number is called Irrational if it cannot be written in the form of p/q , where p and q are integers and q ≠0 √2,√3,√15,Π,0.11011101111…….,
- 14. Theorem 1.3-Let p be a prime number ,if p divides 𝒂𝟐 ,then p divides a,where a is a positive integars. Solution Let the prime factorisation of a be as below a=𝒑𝟏𝒑𝟐𝒑𝟑𝒑𝟒 … … . . 𝒑𝒏 ,where 𝒑𝟏𝒑𝟐𝒑𝟑𝒑𝟒 … … . . 𝒑𝒏 are primes ,not necessarily distinct. therefore 𝒂𝟐 = (𝒑𝟏𝒑𝟐….𝒑𝒏 )(𝒑𝟏𝒑𝟐𝒑𝟑 … . 𝒑𝒏)= 𝒑𝟏 𝟐 𝒑𝟐 𝟐 ……….𝒑𝒏 𝟐 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 How ever using the uniqueness part of the fundamental theorem of arithmetic we realise that the only prime factors of 𝒂𝟐 = 𝒑𝟏𝒑𝟐….𝒑𝒏 so is one of the 𝒑𝟏,𝒑𝟐,𝒑𝟑, 𝒑𝟒 … … . . 𝒑𝒏 Now ,since a= 𝒑𝟏,𝒑𝟐,𝒑𝟑, 𝒑𝟒 … … . . 𝒑𝒏, p divides a Hence proved
- 15. Prove that √5 is irrational. Solutions: Let us assume, that √5 is rational number. i.e. √5 = x/y (where, x and y are co-primes) y√5= x Squaring both the sides, we get, (y√5)2 = x2 ⇒5y2 = x2……………………………….. (1) Thus, x2 is divisible by 5, so x is also divisible by 5. Let us say, x = 5k, for some value of k and substituting the value of x in equation (1), we get, 5y2 = (5k)2 ⇒y2 = 5k2 is divisible by 5 it means y is divisible by 5. Therefore, x and y are co-primes. Since, our assumption about is rational is incorrect. Hence, √5 is irrational number.
- 19. *35/50 35/50 = 7/10, Factorising the denominator, we get, 10 = 2×5 Since, the denominator is in the form of 2m × 5n thus, 35/50 has a terminating decimal expansion. *77/210 77/210 = (7× 11)/ (30 × 7) = 11/30 Factorising the denominator, we get, 30 = 2 × 3 × 5 As you can see, the denominator is not in the form of 2m × 5n .Hence, 77/210 has a non- terminating decimal expansion. *Is it rational or irrational i)0.120120012000120000. . . Since, it has non-terminating and non- repeating decimal expansion, it is an irrational number. ii) 43.123456789 Since it has non-terminating but repeating decimal expansion, it is a rational number in the form of p/q and q has factors other than 2 and 5.
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