Q1. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
View AnswerAns:
Let x be any positive integer and y = 3.
By Euclid’s division algorithm;
x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3)
Therefore,
x = 3q, 3q + 1 and 3q + 2
As per the given question, if we take the square on both the sides, we get;
x^{2} = (3q)^{2} = 9q^{2} = 3.3q^{2}
Let 3q^{2} = m
Therefore,
x^{2} = 3m ………………….(1)
x^{2} = (3q + 1)^{2}
= (3q)^{2} + 1^{2} + 2 × 3q × 1
= 9q^{2} + 1 + 6q
= 3(3q^{2} + 2q) + 1
Substituting 3q2+2q = m we get,
x^{2} = 3m + 1 ……………………………. (2)
x^{2} = (3q + 2)^{2}
= (3q)^{2} + 2^{2} + 2 × 3q × 2
= 9q^{2} + 4 + 12q
= 3(3q^{2} + 4q + 1) + 1
Again, substituting 3q^{2} + 4q + 1 = m, we get,
x^{2} = 3m + 1…………………………… (3)
Hence, from eq. 1, 2 and 3, we conclude that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
Q2. Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Ans:
(i) 140
Using the division of a number by prime numbers method, we can get the product of prime factors of 140.
Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 2^{2} × 5 × 7
(ii) 156
Using the division of a number by prime numbers method, we can get the product of prime factors of 156.
Hence, 156 = 2 × 2 × 13 × 3 = 2^{2} × 13 × 3
(iii) 3825
Using the division of a number by prime numbers method, we can get the product of prime factors of 3825.
Hence, 3825 = 3 × 3 × 5 × 5 × 17 = 3^{2} × 5^{2} × 17
(iv) 5005
Using the division of a number by prime numbers method, we can get the product of prime factors of 5005.
Hence, 5005 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13
(v) 7429
Using the division of a number by prime numbers method, we can get the product of prime factors of 7429.
Hence, 7429 = 17 × 19 × 23 = 17 × 19 × 23
Q3. Given that HCF (306, 657) = 9, find LCM (306, 657).
View AnswerAns: As we know that,
HCF × LCM = Product of the two given numbers
So,
9 × LCM = 306 × 657
LCM = (306 × 657)/9 = 22338
Therefore, LCM(306,657) = 22338
Q4. Prove that 3 + 2√5 is irrational.
View AnswerAns: Let 3 + 2√5 be a rational number.
Then the coprimes x and y of the given rational number where (y ≠ 0) is such that:
3 + 2√5 = x/y
Rearranging, we get,
2√5 = (x/y) – 3
√5 = 1/2[(x/y) – 3]
Since x and y are integers, thus, 1/2[(x/y) – 3] is a rational number.
Therefore, √5 is also a rational number. But this confronts the fact that √5 is irrational.
Thus, our assumption that 3 + 2√5 is a rational number is wrong.
Hence, 3 + 2√5 is irrational.
Q.5: Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a nonterminating repeating decimal expansion:
(i) 13/3125
(ii) 17/8
(iii) 64/455
(iv) 15/1600
Ans:
Note: If the denominator has only factors of 2 and 5 or in the form of 2m × 5n then it has a terminating decimal expansion.
If the denominator has factors other than 2 and 5 then it has a nonterminating repeating decimal expansion.
(i) 13/3125
Factoring the denominator, we get,
3125 = 5 × 5 × 5 × 5 × 5 = 55
Or
= 20 × 55
Since the denominator is of the form 2m × 5n then, 13/3125 has a terminating decimal expansion.
(ii) 17/8
Factoring the denominator, we get,
8 = 2× 2 × 2 = 23
Or
= = 23 × 50
Since the denominator is of the form 2m × 5n then, 17/8 has a terminating decimal expansion.
(iii) 64/455
Factoring the denominator, we get,
455 = 5 × 7 × 13
Since the denominator is not in the form of 2m × 5n, therefore 64/455 has a nonterminating repeating decimal expansion.
(iv) 15/1600
Factoring the denominator, we get,
1600 = 26 × 52
Since the denominator is in the form of 2m × 5n, 15/1600 has a terminating decimal expansion.
Q.6: The following real numbers have decimal expansions 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 the prime factors of q?
(i) 43.123456789
(ii) 0.120120012000120000. . .
Ans:
(i) 43.123456789
Since it has a terminating decimal expansion, it is a rational number in the form of p/q and q has factors of 2 and 5 only.
(ii) 0.120120012000120000. . .
Since it has a nonterminating and nonrepeating decimal expansion, it is an irrational number.
Q.7: Check whether 6n can end with the digit 0 for any natural number n.
View AnswerAns: If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with a unit place as 0 or 5 is divisible by 5.
Prime factorization of 6^{n} = (2 × 3)^{n}
Therefore, the prime factorization of 6n doesn’t contain the prime number 5.
Hence, it is clear that for any natural number n, 6n is not divisible by 5 and thus it proves that 6n cannot end with the digit 0 for any natural number n.
Q.8: What is the HCF of the smallest prime number and the smallest composite number?
View AnswerAns:The smallest prime number = 2
The smallest composite number = 4
Prime factorisation of 2 = 2
Prime factorisation of 4 = 2 × 2
HCF(2, 4) = 2
Therefore, the HCF of the smallest prime number and the smallest composite number is 2.
Q.9: Using Euclid’s Algorithm, find the HCF of 2048 and 960.
View AnswerAns:
2048 > 960
Using Euclid’s division algorithm,
2048 = 960 × 2 + 128
960 = 128 × 7 + 64
128 = 64 × 2 + 0
Therefore, the HCF of 2048 and 960 is 64.
Q.10: Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers.
View AnswerAns: Prime factorisation of 404 = 2 × 2 × 101
Prime factorisation of 96 = 2 × 2 × 2 × 2 × 2 × 3 = 2^{5} × 3
HCF = 2 × 2 = 4
LCM = 2^{5} × 3 × 101 = 9696
HCF × LCM = 4 × 9696 = 38784
Product of the given two numbers = 404 × 96 = 38784
Hence, verified that LCM × HCF = Product of the given two numbers.
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1. What is a real number in class 10? 
2. What is the difference between a rational number and an irrational number? 
3. How do you classify a number as a real number? 
4. Give an example of a rational number and an irrational number. 
5. How are real numbers used in everyday life? 

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