Chapter 1 : Real Numbers |
Exercise 1.1 Complete solution Exercise 1.2 Complete solution |
Algebraic Identities (i) (ii) (iii) (iv)
(v)
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Important notes(ii) Fundamental Theorem of Arithmetic : Every composite number can be expressed ( factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. (iii) The prime factorisation of a natural number is unique, except for the order of its factors. (iv) HCF = Product of the smallest power of each common prime factor in the numbers. (vi) If two positive integers and ,then . i.e ., HCF of two numbers × LCM of two numbers = One number × Other number (vii) The product of three numbers is not equal to the product of their HCF and LCM. (viii) A number is called rational if it can be written in the form , where p and q are integers . Example : 0 , – 5 , , …… etc . (ix) A number is called irrational if it cannot be written in the form , where p and q are integers . Example : ……… etc. (x) Let be a prime number. If divides , then divides , where is a positive integer. (xi) The sum or difference of a rational and an irrational number is irrational . |
Solution : (i) We have,
(ii) We have,
(iii) We have,
(iv) We have ,
(v) We have ,
Solution : (i) 26 and 91
We have , and
HCF (26 , 91) 13
and LCM (26 , 91)
Now ,
Verified .
(ii) 510 and 92
We have ,
and
HCF(510 , 92)
and LCM(510 , 92)
Now ,
Verified .
(iii) 336 and 54
We have,
and
HCF(336 , 54)
and LCM(336 , 54)
Now ,
Verified .
Solution : (i) 12 , 15 and 21
We have ,
HCF (12 , 15 , 21) 3
and LCM (12 , 15 , 21)
(ii) 17 , 23 and 29
We have ,
HCF (17 , 23 , 29)
and LCM (17 , 23 , 29)
(iii) 8 , 9 and 25
We have ,
HCF(8 , 9 , 25)
and LCM (8 , 9 , 25)
Solution : We have ,
Solution : We have ,
The prime factors of does not contain in factor , where are positive integers .Therefore, does not end with the digit 0 .
Solution: We have ,
is a composite number.
and
is a composite number .
Solution : The required number of minutes is LCM (18 , 12) . We find the LCM by prime factorization method ,
and
Therefore, the LCM (18 , 12)
Hence, Sonia and Ravi will meet again at the starting point after 36 minutes .
Solution : let, us assume to the contrary that is rational .There exists co-prime integers and () such that
Therefore , is divisible by 5 . So, is also divisible by 5 .
Let , for some integer c .
From and , we get
So, is divisible by 5 . So, is also divisible by 5 .
Therefore, and have at least as a common factor . But this contradicts the fact that and are co-prime. This contradiction has arisen because of our incorrect assumption that is rational . So, we conclude that is irrational .
Solution : let us assume , to the contrary that is rational .
We can find co-prime and ( ) such that
Since, 2 , and are integers , is rational and so, is rational . But this contradicts the fact that is irrational . So , we conclude that is irrational .
Solution : (i) Let us assume , to the contrary , that is rational . We can find co-prime and ( ) such that
Since 2 , and are integers, is rational and so, is rational . But this contradicts the fact that is irrational . So , is an irrational .
(ii) Let us assume , to the contrary , that is rational . We can find co-prime and ( ) such that
Since 7 , and are integers , is rational and so, is rational . But this contradicts the fact that is irrational .So , is an irrational .
(iii) Let us assume , to the contrary , that is rational . We can find co-prime and ( ) such that
Since and are integers , is rational and so, is rational . But this contradicts the fact that is irrational . So , is an irrational .