**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 a
^{2}, then p is one of the prime factors of a^{2}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 2
^{n}5^{m}, 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 2
^{n}5^{m}, 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 2
^{n}2^{m}, 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 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 (2q^{2} + 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

NCERT SOLUTIONS CH - 1 MCQ SOLUTIONS CH - 1