# CBSE Class 10th Maths

## Ch-1 Real Numbers

Euclid’s Division Lemma

It is basically the restatement of the usual division system. The formal statement for this is-

For each pair of given positive integers a and b, there exist unique whole numbers q and r which satisfies the relation

a = bq + r, 0 ≤ r < b, where q and r can also be Zero.

where ‘a’ is a dividend, ‘b' is divisor, ‘q’ is quotient and ‘r’ is remainder.

∴ Dividend = (Divisor x Quotient) + Remainder

### Natural Numbers

Non-negative counting numbers excluding zero are known as natural numbers.

i.e. 5, 6, 7, 8,  ……….

### Whole numbers

All non-negative counting numbers including zero are known as whole numbers.

i.e. 0, 1, 2, 3, 4, 5, …………….

### Integers

All negative and non-negative numbers including zero altogether known as integers.

i.e. ………. – 3, – 2, – 1, 0, 1, 2, 3, 4, …………..

### Algorithm

An algorithm gives us some definite steps to solve a particular type of problem in a well-defined manner.

### Lemma

A lemma is a statement which is already proved and is used for proving other statements.

## Euclid’s Division Algorithm

This concept is based on Euclid’s division lemma. This is the technique to calculate the HCF (Highest common factor) of given two positive integers m and n,

To calculate the HCF of two positive integers’ m and n with m > n, the following steps are followed:

Step 1: Apply Euclid’s division lemma to find q and r where m = nq + r, 0 ≤ r < n.

Step 2: If the remainder i.e. r = 0, then the HCF will be ‘n’ but if r ≠ 0 then we have to apply Euclid’s division lemma to n and r.

Step 3: Continue with this process until we get the remainder as zero. Now the divisor at this stage will be HCF(m, n). Also, HCF (m, n) = HCF (n, r), where HCF (m, n) means HCF of m and n.

## The Fundamental Theorem of Arithmetic

We can factorize each composite number as a product of some prime numbers and of course, this prime factorization of a natural number is unique as the order of the prime factors doesn’t matter.

• HCF of given numbers is the highest common factor among all which is also known as GCD i.e. greatest common divisor.
• LCM of given numbers is their least common multiple.
• If we have two positive integers  ‘m’ and ‘n’ then the property of their  HCF and LCM will be:

HCF (m, n) × LCM (m, n) = m × n.

### Rational Numbers

The number ‘s’  is known as a rational number if we can write it in the form of m/n where  ‘m' and ‘n’ are integers and n ≠ 0, 2/3, 3/5 etc.

Rational numbers can be written in decimal form also which could be either terminating or non-terminating. E.g. 5/2 = 2.5 (terminating) and(non-terminating).

### Irrational Numbers

The number ‘s’ is called irrational if it cannot be written in the form of m/n, where m and n are integers and n≠0 or in the simplest form, the numbers which are not rational are called irrational numbers. Example - √2, √3 etc.

• If p is a prime number and p divides a2 , then p is one of the prime factors of a2 which divides a, where a is a positive integer.
• If p is a positive number and not a perfect square, then √n is definitely an irrational number.
• If p is a prime number, then √p is also an irrational number.

### Rational Number and their Decimal Expansions

• Let y be a real number whose decimal expansion terminates into a rational number which we can express in the form of a/b, where a and b are coprime, and the prime factorization of the denominator b has the powers of 2 or 5 or both like 2n5m, where n, m are non-negative integers.
• Let y be a rational number in the form of y = a/b, so that the prime factorization of the denominator b is of the form 2n5m, where n, m are non-negative integers then y has a terminating decimal expansion.
• Let y = a/b be a rational number, if the prime factorization of the denominator b is not in the form of 2n2m, where n, m are non-negative integers then y has a non-terminating repeating decimal expansion.
• The decimal expansion of every rational number is either terminating or a non-terminating repeating.
• The decimal form of irrational numbers is non-terminating and non-repeating.

## Some Important Quetions for class 10 Maths

Question 1.
Apply Euclid’s division algorithm to find HCF of numbers 4052 and 420.
Solution

4052 = 420 × 9 + 272
420 = 272 × 1 + 148
272 = 148 × 1 + 124
148 = 124 × 1 + 24
124 = 24 × 5 + 4
24 = 4 × 6 + 0

H.C.F = 4

Question 2.
Show that 2√2 is an irrational number.
Solution:

Question 3.

Prove that the product of two consecutive positive integers is divisible by 2.

Solution:

Let n and n + 1 are two consecutive positive integer
We know that n is of the form n = 2q and n + 1 = 2q + 1
n (n + 1) = 2q (2q + 1) = 2 (2q2 + q)
Which is divisible by 2
If n = 2q + 1, then
n (n + 1) = (2q + 1) (2q + 2)
= (2q + 1) x 2(q + 1)
= 2(2q + 1)(q + 1)
Which is also divisible by 2
Hence the product of two consecutive positive integers is divisible by 2

Question 4.

Prove that the square of any positive, integer is of the form 3m or, 3m + 1 but not of the form 3m + 2.

Solution:

Let a be any positive integer
Let it be in the form of 3m or 3m + 1
Let a = 3q, then

Question 5.

Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers ?

Solution:

We know that a composite number is that number which can be factorize. It has more factors other than itself and one
Now, 7 x 11 x 13 + 13 = 13 (7 x 11 + 1) = 13 x 78
Which is composite number
Similarly,
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5(7 x 6 x 4 x 3 x 2 x 1 + 1)
= 5 x 1009
Which is a composite number