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CBSE Class X Chapter 1 Real Numbers

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Published in: Mathematics
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Short notes on the concepts and Solution to the exercises

Pushkar K / Pune

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  1. CBSE Mathematics Chapter 1 Real Numbers Rational numbers: Numbers which can be written in the form of (p/q) where q is not equal to zero and the division process is terminating or repeating. In case of irrational numbers, division process is non-terminating. Euclid's division Lemma (EDL): EDL says that any number can be represented in the form a=bq+r...r is the remainder of the division a/b EDL is mostly used to calculate highest common factor (HCF) of big numbers. Prove that every even integer can be represented in the form of 2q and every odd integer can be represented in the form 2q+1 Solution: let "a" be any integer and let b=2 So, using EDL, a=2q+r...(1) Now, since r is the remainder, the value of r has to be between 0 and 2, i.e. Hence, r can be O or 1 Substituting the value of r in equation (1) a=2q or a=2q+1 Now, when any number is multiplied by 2, it becomes an even number. So 2q is an even number. At the same time, when 1 is added to any even number it becomes an odd number, so 2q+1 is an even number. Hence proved that, every even integer can be represented in the form of 2q and every odd integer can be represented in the form 2q+1. Exercise 1.1 1. (i) HCF of 135 and 225 Using EDL 1+45 Since r=O, the division process is complete. HCF is 45 (ii) HCF and 867 and 255 I-ICF is 51
  2. 2. 3. 4. CBSE Mathematics Chapter 1 Real Numbers Show that any positive odd integer can be expressed in the form of 6q+1 or 6q+3 or 6q+5 Solution: Let a be any positive integer and let b=6 Hence, using EDL, a=6q+r.... (1) Now, since r is the remainder of the division (a/6), r has to be between O and 6, i.e. Substituting in equation (1) a=6q or a=6q+1 or a=6q+2 or a=6q+3 or a=6q+4 or a=6q+5 Now, a=6q=2*3*q We know that when any number is multiplied by 2, it becomes an even number. So, 6q is an even number when 1 is added to any even number, it becomes an odd number Hence, a=6q+1 is odd number When an even number is added to odd number, it remains an odd number So, a=6q+3=6q+1+2 is an odd number Similarly, a=6q+5=6q+3+2 is an odd number. Maximum number of columns can be found out by calculating HCF of number of contingent members and number of army band members Calculating HCF of 616 and 32 Using EDL Hence, HCF=8 Hence, maximum number of columns is 8. Using EDL show that square of any positive integer is either of the form 3m or 3m+1 Solution: let a be any positive integer and let b=3 Using EDL, Now, 0
  3. 5. CBSE Mathematics Chapter 1 Real Numbers Case 3 a=3q+2 aA2=9qA2+12q+4 =9qA2+12q+3+1 Let (3qA2+4q+1) be m Hence, aA2=3m+1 Use EDL to show that the cube of any positive integer is of the form 9m or 9m+1 or 9m+8 Solution: Let a be any positive integer and b=3 Hence, a=3q+r so, Hence, or r=l or r=2; i.e. a=3q or a=3q+1 or a=3q+2 Case 1 a=3q Taking cube on both sides aA3=27qA3 Let Hence, aA3=9m Case 2 a=3q+1 Taking cube on both sides aA3=27qA3+27qA2+9q+1 Let (3qA3+3qA2+q) +m Hence, aA3=9m+1 Case3 a=3q+2 Taking cube on both sides Let Hence, aA3=9m+8
  4. CBSE Mathematics Chapter 1 Real Numbers Fundamental theorem of arithmetic Any composite number can be expressed as multiplication of primes and this factorization is unique apart from the order in which prime factors occur. HCF of (a,b)*LCM of (a,b) = a* b Exercise 1.2 1. 2. Express as product of prime (iii) (iv) (v) 1001 Verify HCF*LCM=product of the two numbers 26 & 91 HCF=13 Product of the numbers=26*91=2366 510 & 92 HCF=2 Product of the two
  5. (iii) CBSE Mathematics Chapter 1 Real Numbers 336 & 54 Product of the two 3. 4. 5. Prime factorization method (iii) 12, 15 HCF=3 17, 23 & 29 As all of the numbers are prime, HCF=I 8, 9 & 25 HCF=I HCF=9 Numbers given 306 and 657 HCF*LCM=Product of two numbers 306*657 LCM-22338 Check whether 6An can end with zero For any number to end with zero, it should get factorized by 2 & 5 Since, 5 is not one of the factors of 6An, it will not end with a zero.
  6. 6. 7. CBSE Mathematics Chapter 1 Real Numbers Since, the number has factors other than 1 and itself, it is a composite number Since, the number has factors than 1 and itself, it is a composite number To calculate the minutes after they meet again, let's calculate LCM of 18* 12 LCM=36 Hence, both of them will meet after 36 minutes. Solved Example Show that if 'p' is a prime number and p divides aA2 then p divides a as well Solution: According to fundamental theorem of arithmetic, pn.....pl, p2, p3, pn are the prime factors Hence, . pn) * (pl*p2*p3* .*pn) Now, if p divides aA2, it must be one of the numbers PI, p2, p3...pn But, a = pl*p2*p3* . *pn Hence, p divides a as well. Prove that 1/2 is irrational Solution: We know that any rational number can be represented as (a/b) where a and b are co-prime numbers. Co-prime numbers are the numbers which do not have any factor in common other than 1. Hence, 112 = (a/b) bi12 = a Squaring on both sides, 2bA2 = Hence, 2 divides aA2 and a Let a=2c...c is some positive integer Hence, 2bA2=4cA2 bA2 = 2cA2 Hence, 2 divides bA2, so 2 divides b as well So, 2 divides a as well as b, but this fails the initial assumption that a and b are co-prime Hence 1/2 is not rational.
  7. CBSE Mathematics Chapter 1 Real Numbers Exercise 1.3 1. 2. 3. Prove that 1/5 is irrational Let, 115 = (a/b) where a and b are co-prime Squaring on both sides, 5bA2 = Hence, 5 divides aA2 and a Let a=5c...c is some positive integer Hence, 5bA2=25cA2 bA2 = 5cA2 Hence, 5 divides bA2, so 5 divides b as well So, 5 divides a as well as b, but this fails the initial assumption that a and b are co-prime Hence 115 is not rational. Prove that 3+21/5 is irrational Let 3+21/5 is rational Hence, 3+21/5 = (a/b) where and b are co-prime 21/5 = (a/b)-3 Since, and b are co-prime, (a/b) is rational So ((a/b)-3)/2 is also rational However, ((a/b)-3)/2 = 1/5 where we know that 115 is irrational So, our assumption 3+21/5 is rational is false. Prove that the following are irrational 1/1/2 Let 1/1/2 = (a/b) where a and b are co-prime Hence, ax/2=b Squaring on both sides Hence 2 divides bA2 So, 2 divides b as well. Let b=2c where c is some integer Substituting in equation 1 2aA2=4cA2 Hence, 2 divides a as well. However, this contradicts our assumption that a and b are co-prime. 7115 Let 7115 be rational Let where a and b are co-prime Hence, 7bV5=a Now, a/ 7b is rational, since (a/b) is rational and 7 is a rational number Hence, 1/5 must also be rational according to equation (1) However, 115 is irrational
  8. CBSE Mathematics Chapter 1 Real Numbers (iii) This contradicts our assumption that 7115 is rational. Hence, 7V5 is irrational. 6+v2 Let 6+1/2 is rational Hence, where a and b are co-prime 6b+bV2=a Hence, bV2=a-6b Now, since a and b are rational, (a-6b)/b should also be rational So, from equation 1, 1/2 is rational. But, 112 is irrational This is because our assumption than 6+1/2 is rational. Hence, 6+1/2 is irrational. In case of rational numbers, the decimal expansion can be terminating or non-terminating repeating. In case of terminating expansion, the prime factorization will always be in the form of 2Am5An. It can be observed that 2 and 5 are the factors of 10. Hence, if any rational number can be expressed in the form of (p/q) and prime factorization of q is in the form of 2Am5An, then the division p/q is terminating. In the other case, where, q cannot be expressed in the form 2Am5An, then the division is non- terminating repeating. Exercise 1.4 1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion. (iii) 13/3125 -5 5555 Since, it can be represented in the form 2Am5An; division is terminating. 17/8 8=2A3 Since, it can be represented in the form 2Am5An; division is terminating. 64/455 Since, it cannot be represented in the form 2Am5An; division is non-terminating repeating.
  9. (iv) (v) (vi) (vii) (viii) (ix) (x) CBSE Mathematics Chapter 1 Real Numbers 15/1600 Since, it can be represented in the form 2Am5An; division is terminating. 29/343 343=7A3 Since, it cannot be represented in the form 2Am5An; division is non-terminating repeating. Since, it can be represented in the form 2Am5An; division is terminating. Since, it cannot be represented in the form 2Am5An; division is non-terminating repeating. 6/15 Since, it cannot be represented in the form 2Am5An; division is non-terminating repeating. 35/50 Since, it can be represented in the form 2Am5An; division is terminating. 77/210 2. Write down the decimal expansion of those rational numbers which have terminating decimal expansion 13/3125 -o 130 -o 1300 13000 -12500 5000 -3125
  10. (ii) (iii) (iv) (v) (vi) CBSE Mathematics Chapter 1 Real Numbers 18750 -18750 17/8 -16 10 -8 20 -2 40 -40 15/1600 =0.009375 -0.115 6/15=0.4 33/50=0.7 3. The following real numbers have decimal expansion as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p/q, what can you say about prime factorization of q. (iii) 43.123456789 Since, the division is terminating repeating, given number is rational and can be expressed in the form p/q where q can be expressed in the form 2Am5An, i.e. the prime factors include 2 or 5 or both. 0.120120012000120000 Since the division is non-terminating non-repeating, the number is irrational. 43.123456789 Note that the line above the numbers after the decimal point represents repetition. The numbers with the line above will be repeated in the same order. So the above number can be written as 43.123456789123456789123456789
  11. CBSE Mathematics Chapter 1 Real Numbers Since, the division is non-terminating repeating, the number is rational and can be expressed in the form p/q where q there is no limitation on the prime factors of q, i.e. factors can be 2 or 5 or both or any other prime numbers.

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