NCERT Solutions For Class 10 Maths Real Numbers
Real Numbers CBSE Class 10 Maths Chapter 1 Solutions
NCERT Solutions For Class 10 Maths Real Numbers Exercise 1.1
NCERT Solutions For Class 10 Maths Real Numbers Exercise 1.2
NCERT Solutions For Class 10 Maths Real Numbers Exercise 1.3
NCERT Solutions For Class 10 Maths Real Numbers Exercise 1.4
Rational numbers and irrational numbers taken together form the set of real numbers. The set of real numbers is denoted by R. Thus every real number is either a rational number or an irrational number. In either case, it has a nonâ€“terminating decimal representation. In case of rational numbers, the decimal representation is repeating (including repeating zeroes) and if the decimal representation is nonâ€“repeating, it is an irrational number. For every real number, there corresponds a unique point on the number line â€˜l’ or we may say that every point on the line â€˜l’ corresponds to a real number (rational or irrational).
From the above discussion we may conclude that:
To every real number there corresponds a unique point on the number line and conversely, to every point on the number line there corresponds a real number. Thus we see that there is oneâ€“toâ€“one correspondence between the real numbers and points on the number line â€˜l’, that is why the number line is called the â€˜real number line’.
The students will be able to ;
prove Euclid’s Division Lemma
state fundamental theorem of arithmetic
find HCF and LCM using prime factorisation
establish the given number as an irrational number
conclude the decimal expansion of a rational number is either terminating or non-terminating repeating.
We have studied the following points:
1. Euclid’s Division Lemma : Given positive integers a and b, there exist whole numbers q and r satisfying a = bq + r where 0 = r = b.
2. Euclid’s Division Algorithm: According to this, which is based on Euclid’s division lemma, the HCF of any two positive integers a and b with a > b is obtained as follows:
Step 1 Apply the division lemma to find q and r where a = bq + r, O = r < b.
Step 2 If r = 0, the HCF is b . If r ? 0 apply Euclid Lemma to b and r
Step 3 Continue the process till the remainder is zero. The divisor at this stage will be HCF (a, b). Also HCF (a, b) = HCF (b, r)
3. The 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.