NCERT Revision Notes for Chapter 1 Real Numbers Class 10 Maths

Answer

Algorithm: An algorithm is a series of well-defined steps which gives a procedure for solving a type of problem.

Lemma: A lemma is a proven statement used for proving another statement.

Euclid’s Division Lemma
For any two given positive integers ‘a’ and ‘b’ there exists unique whole numbers ‘q’ and ‘r’ such that:

a = bq + r, where 0 ≥ r < b
Here,
a = Dividend,
b = Divisor
q = Quotient,
r = Remainder
Means, Dividend = (Divisor × Quotient) + Remainder

Example:
When 33 ÷ 7 then, 4 = quotient and 5 = remainder such that:
33 = 7×4 + 5

Answer

Euclid’s Division Algorithm is a technique to compute the HCF of two positive integers ‘a’ and ‘b’ where, a > b. It includes following steps:

Step-I: Applying Euclid’s Lemma to a and b to find whole numbers ‘q’ and ‘r’ such that:
a = bq + r, 0 ≥ r < b
Step-II: If r = 0 then ‘b’ is the HCF of ‘a’ and ‘b’. If r ≠ 0 then apply the Euclid’s division lemma to ‘b’ and ‘r’.
Step-III: Continue the process till the remainder is zero, i.e., repeat the step-II again and again until r = 0. Then, the divisor at this stage will be the required H.C.F.

Note: 
I. We state Euclid’s Divison Algorithm for positive integers only but it can be extended for all integers except zero i.e., b ≠ 0.
II. When ‘a’ and ‘b’ are two positive integers such that a = bq + r, 0 ≤ r < b then HCF (a, b) = HCF (b, r).

Example:
HCF of 135 and 225:
Applying the Euclid’s lemma to 225 and 135, (where 225 > 135), we get
225 = (135 × 1) + 90
since 90 ≠ 0, therefore, applying the Euclid’s lemma to 135 and 90, we have:
135 = (90 × 1) + 45
But 45 ≠ 0
∴ Applying Euclid’s Lemma to 90 and 45, we get
90 = (45 × 2) + 0
Here, r = 0, so our process stops. Since, the divisor at the last step is 45,
∴ HCF of 225 and 135 is 45.

Answer

Composite Numbers: The number which have more than two factors.
For example: Factors of 4 are 1, 2 and 4.

Statement: Every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
This is also known as prime factorisation method.

For example:

60 is a composite numbers which can be expressed in the form of its prime factors as

60= 2² ×3¹×5¹

Fundamental Theorem of Arithmetic is used to Find HCF and LCM of positive integers.

If a and b are two positive integers, then

HCF (a, b) = product of the smallest power of each common prime factor in the numbers a and b.

LCM (a, b) = product of the greatest power of each prime factor involved in the numbers a and b. 

Answer

HCF ×LCM = product of two natural numbers.

For example:

HCF of 306 and 657 = 9

LCM = 22338

LCM × HCF = 9 × 22338 = 201042

Product of the numbers = 306 × 657 = 201042
We have:
LCM × HCF = Product of the numbers

Note: This relationship is not true for more than two natural numbers.

For example:

HCF of 24, 60 and 112 = 4

LCM of 24, 60 and 112 = 1680

Notice that: HCF × LCM ≠ 24 × 60 × 112.

Answer

If x is rational number whose simplest form is p/q, (p and q are co-prime integers q≠0 ) will have terminating decimal expansion when

Or

 q= 2 ^m ×5 ⁿ , Where m and n are non-negative integers.

Otherwise, the rational number will have non-terminating reccuring decimal expansion.

Answer

Let the prime factorization of a is

a=(p₁)×(p₂)×(p₃)…….(p_n)

Since every numbers prime factorization is unique, this is only prime factorization a is having.

a²={(p₁)×(p₂)×(p₃)…….(pn)}× {(p₁)×(p₂)×(p₃)…….(p_n)}

So p is a prime number factor of a² which is among p_1, p₂, ......p_n which are also prime factor of a

Answer

Let us assume that it is a rational number
Then, √2=p/q
Where p and q are co-primes.
Or, q√2=p
Squaring both sides, we get
2q²=p²
Hence, 2 divide p².
From theorem, we know that,
2 will divide p also. p = 2c, where c is any constant
2q²=4c²
or q²=2c²
So q is divisible by 2 also
So 2 divide both p and q, which contradicts our assumption.
So √2 is an irrational number.