1.

This result was perhaps known for a long time, but was first recorded in Book VII of Euclid's Elements. Euclid's division algorithm is based on this lemma.

An

A

**Euclid’s division lemma**states that “*Given positive integers a and b, there exist whole numbers q and r satisfying a = bq + r, 0**≤ r < b.*”This result was perhaps known for a long time, but was first recorded in Book VII of Euclid's Elements. Euclid's division algorithm is based on this lemma.

An

**algorithm**is a series of well defined steps which gives a procedure for solving a type of problem.A

**lemma**is a proven statement used for proving another statement.##
2. **Euclid’s division algorithm:** Euclid’s division algorithm is based
on Euclid’s division lemma. According to this algorithm, 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, 0*

*≤ r < b.*

**Step 2:**If r = 0, the HCF is b and if r ≠ 0, apply Euclid’s lemma to b and r.

**Step 3:**Continue the process till the remainder is zero. At this stage, the divisor will be HCF of a and b. Also, HCF of a and b equals HCF of b and r.

**Example:**Use Euclid's division algorithm to find the HCF of 210 and 350.

**Solution:**

**Step 1:**Since 350 > 210, we apply the division lemma to 350 and 210, to get

350 = 210 × 1 + 140

**Step 2:**Since the remainder 140 ≠ 0, we apply the division lemma to 210 and 140, to get

**Step 3:**Since the remainder 70 ≠ 0, we apply the division lemma to 140 and 70, to get

The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 70, the HCF of 350 and 210 is 70.

3.

**The Fundamental Theorem of Arithmetic:**Every composite number can be expressed as a product of prime numbers, and this factorization is unique, apart from the order in which the prime factors occur.
4. If p is a prime number and p divides a

^{2}, then p divides a, where a is a positive integer.
5. We can prove that √2, √3 and √5 are irrational
numbers.

Let us assume that √2 is a rational number.

We can find integers a and b (b ≠ 0) such that √2 = a/b.

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprimes.

So, b√2 = a

Squaring on both sides, we get 2b

Therefore, a

Then, a is divisible by 2.

So, we can write a = 2c for some integer c.

Substituting for a in above, we get 2b

This means that 2 divides b

Therefore, a and b have at least 2 as a common factor.

But this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that √2 is a rational number.

So, we conclude that √2 is an irrational number.

Similarly, we can prove that √3, √5, etc. are all irrational numbers.

**Example:**Prove that √2 is an irrational number.**Solution:**Let us assume that √2 is a rational number.

We can find integers a and b (b ≠ 0) such that √2 = a/b.

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprimes.

So, b√2 = a

Squaring on both sides, we get 2b

^{2}= a^{2}Therefore, a

^{2}is divisible by 2.Then, a is divisible by 2.

So, we can write a = 2c for some integer c.

Substituting for a in above, we get 2b

^{2}= 4c^{2}implies b^{2}= 2c^{2}This means that 2 divides b

^{2}and also 2 divides b.Therefore, a and b have at least 2 as a common factor.

But this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that √2 is a rational number.

So, we conclude that √2 is an irrational number.

Similarly, we can prove that √3, √5, etc. are all irrational numbers.

**Note: 1.**The sum and difference of a rational and an irrational number is irrational.**2.**The product and quotient of a non-zero rational and irrational number is irrational.
6. If x be a rational number whose decimal
expansion terminates, then we can express x in the form of p/q, where p and q
are co-prime, and the prime factorization of q is of the form

*2*, where n, m are non-negative integers.^{n}5^{m}
7. If x = p/q be a rational number such that the
prime factorization of q is of the form

*2*, where n, m are non-negative integers, then x has a decimal expansion which is a terminating decimal.^{n}5^{m}
For example,
3/20 is a terminating decimal because 20 = 2

^{2}× 5^{1}, which is of the form*2*, where n, m are non-negative integers.^{n}5^{m}
8. If x = p/q be a rational number such that the
prime factorization of q is not of the form

*2*, where n, m are non-negative integers, then x has a decimal expansion which is a non-terminating repeating (recurring) decimal.^{n}5^{m}
For example,
24/45 is a non-terminating decimal because 45 = 3

^{2}× 5^{1}, which is not of the form*2*, where n, m are non-negative integers.^{n}5^{m}