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Unit-1(Examples)

Example-1 :-  Use Euclid’s algorithm to find the HCF of 4052 and 12576.

Solution :-
Step 1 : Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get 12576 = 4052 × 3 + 420
Step 2 : Since the remainder 420 ≠ 0, we apply the division lemma to 4052 and 420, to get 4052 = 420 × 9 + 272 
Step 3 : We consider the new divisor 420 and the new remainder 272, and apply the division lemma to get 420 = 272 × 1 + 148
         We consider the new divisor 272 and the new remainder 148, and apply the division lemma to get 272 = 148 × 1 + 124 
         We consider the new divisor 148 and the new remainder 124, and apply the division lemma to get 148 = 124 × 1 + 24 
         We consider the new divisor 124 and the new remainder 24, and apply the division lemma to get 124 = 24 × 5 + 4 
         We consider the new divisor 24 and the new remainder 4, and apply the division lemma to get 24 = 4 × 6 + 0 
         The remainder has now become zero, so our procedure stops. 
         Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4. 

Example-2 :-  Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer.

Solution :-
 Let a be any positive integer and b = 2. Then, by Euclid’s algorithm, a = 2q + r,
 for some integer q ≥ 0, and r = 0 or r = 1, because 0 ≤ r < 2. So, a = 2q or 2q + 1.
 If a is of the form 2q, then a is an even integer.
 Also, a positive integer can be either even or odd.
 Therefore, any positive odd integer is of the form 2q + 1.

Example-3 :-  Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.

Solution :-
 Let a, where a is a positive odd integer.
 We apply the division algorithm with a and b = 4. Since 0 ≤ r < 4, the possible remainders are 0, 1, 2 and 3. 
 That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient. 
 However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2). 
 Therefore, any odd integer is of the form 4q + 1 or 4q + 3
    

Example-4 :-  A sweetseller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the number of that can be placed in each stack for this purpose?

Solution :-
 Let us use Euclid’s algorithm to find their HCF.
 We have : 420 = 130 × 3 + 30 
           130 = 30 × 4 + 10 
           30 = 10 × 3 + 0 
 So,the HCF of 420 and 130 is 10.
 Therefore, the sweetseller can make stacks of 10 for both kinds of barfi.

Example-5 :- 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 = (2)2n; 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. 

Example-6 :-  Find the LCM and HCF of 6 and 20 by the prime factorisation method.

Solution :-
 We have :  6 = 2 × 3 and 20 = 2 × 2 × 5 = 2² × 5.
            HCF(6, 20) = 2  =  Product of the smallest power of each common prime factor in the numbers
            LCM (6, 20) = 2² × 3 × 5 =  Product of the greatest power of each prime factor, involved in the numbers.
    

Example-7 :-  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 x 404)/HCF(96,404) = (96 x 404)/4 = 9696

Example-8 :- 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, 2 and 3 are the smallest powers of the common factors 2 and 3, respectively.
 So, HCF (6, 72, 120) = 2 × 3 = 2 × 3 = 6 
 23 32 and 5 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 × 5 = 360

Example-9 :-  Prove that √3 is irrational.

Solution :-
 Let us assume, to the contrary, that  √3 is rational.
 i.e., 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,  √3 = b/a. Squaring on both sides, and rearranging, we get 3b2 = a2. 
 Therefore, a2 is divisible by 3, and 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, i.e., b2 = 3c2. 
 This means that b2 is divisible by 3, and so b is also divisible by 3. 
 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. 
    

Example-10 :-  Prove that √2 is irrational.

Solution :-
 Let us assume, to the contrary, that  √2 is rational.
 i.e., we can find integers a and b (≠ 0) such that  √2 = 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,  √2 = b/a. Squaring on both sides, and rearranging, we get 2b2 = a2. 
 Therefore, a2 is divisible by 2, and it follows that a is also divisible by 2.
 So, we can write a = 2c for some integer c. Substituting for a, we get 2b2 = 4c2, i.e., b2 = 2c2. 
 This means that b2 is divisible by 2, and so b is also divisible by 2. 
 Therefore, a and b have at least 2 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  √2 is rational. So, we conclude that  √2 is irrational. 
    

Example-11 :-  Show that 5 – √3 is irrational.

Solution :-
 Let us assume, to the contrary, that 5 – √3 is rational.
 i.e., we can find coprime a and b (b ≠ 0) such that 5 - √3 = a/b.
 Therefore, 5 - (a/b) = √3⋅ 
 Rearranging this equation, we get √3 = 5 – (a/b) = (5b-a)/b⋅ 
 Since a and b are integers, we get 5 – (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-12 :-  Show that 3√2 is irrational

Solution :-
 Let us assume, to the contrary, that 3√2 is rational. 
 i.e., 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.
    

Example-13 :-  Change following decimal expansion into p/q form :
(i) 0.375,  (ii) 0.104,  (iii) 0.0875,  (iv) 23.3408

Solution :-
 (i)   0.375 = 375/1000 = 375/103 = (3 x 53)/(23 x 53) = 3/23 = 3/8
 (ii)  0.104 = 104/1000 = 104/103 = (13 x 23)/(23 x 53) = 13/53 = 13/125
 (iii) 0.0875 = 875/10000 = 875/104 = 7/(24 x 5) = 7/80
 (iv)  23.3408 = 233408/10000 = 233408/104 = (22 x 7 x 521)/54 = 14588/625
    

Example-14 :-  Change following p/q into decimal expansion form :
(i) 3/8,  (ii) 13/125,  (iii) 7/80,  (iv) 14588/625

Solution :-
 (i)   3/8 = 3/23 = (3 x 53)/(23 x 53) = 375/103 = 0.375
 (ii)  13/125 = 13/53 = (13 x 23)/(23 x 53) = 104/103 = 0.104
 (iii) 7/80 = 7/(24 x 5) = (7 x 53)/(24 x 54) = 875/104 = 0.0875
 (iv)  14588/625 = (22 x 7 x 521)/54 = (26 x 7 x 521)/(24 x 54) = 233408/104  = 23.3408
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