Euclid’s Division Lemma

What is a dividend? Let us understand it with the help of a simple example.

Can you divide 14 by 6?

After division, we get 2 as the quotient and 2 as the remainder.

Thus, we can also write 14 as 6 × 2 + 2.

A dividend can thus be written as:

Dividend = Divisor × Quotient + Remainder

Let us try it out with some other sets of dividends and divisors.Can you think of any other number which, when multiplied with 6, gives 14 as the dividend and 2 as the remainder?

(1) Divide 100 by 20: 100 = 20 × 5 + 0

(2) Divide 117 by 15: 117 = 15 × 7 + 12

(3) Divide 67 by 17: 67 = 17 × 3 + 16

Thus, if we have a dividend and a divisor, then there will be a unique pair of a quotient and a remainder that will fit into the above equation.

This brings us to Euclid’s division lemma.

Let us look at some more examples.This lemma is very useful for finding the H.C.F. of large numbers where breaking them into factors is difficult. This method is known as Euclid’s Division Algorithm.

Example 1: 

Find the H.C.F. of 4032 and 262 using Euclid’s division algorithm. 

Solution:

Step 1:

First, apply Euclid’s division lemma on 4032 and 262.

4032 = 262 × 15 + 102

Step 2:

As the remainder is non-zero, we apply Euclid’s division lemma on 262 and 102.

262 = 102 × 2 + 58

Step 3:

Apply Euclid’s division lemma on 102 and 58.

102 = 58 × 1 + 44

Step 4:

Apply Euclid’s division lemma on 58 and 44.

58 = 44 × 1 + 14

Step 5:

Apply Euclid’s division lemma on 44 and 14.

44 = 14 × 3 + 2

Step 6:

Apply Euclid’s division lemma on 14 and 2.

14 = 2 × 7 + 0

In the problem given above, to obtain 0 as the remainder, the divisor has to be taken as 2. Hence, 2 is the H.C.F. of 4032 and 262.

Note that Euclid’s division algorithm can be applied to polynomials also.

Example 2:

A rectangular garden of dimensions 190 m × 60 m is to be divided in square blocks to plant different flowers in each block. Into how many blocks can this garden be divided so that no land is wasted? 

Solution:

If we do not want to waste any land, we need to find the largest number that completely divides both 190 and 60 and gives the remainder 0, i.e., the H.C.F. of (190, 60).

To find the H.C.F., let us apply Euclid’s algorithm.

190 = 60 × 3 + 10

60 = 10 × 6 + 0

Therefore, the H.C.F. of 190 and 60 is 10.

 Therefore, there will be  = 19 square blocks along the length of the garden and = 6 blocks along its breadth.

Hence, the total number of blocks in the garden will be 19 × 6 = 114.

Example 3:

Find the H.C.F. of 336 and 90 using Euclid’s division algorithm.

Solution:

As 336 > 90, we apply the division lemma to 336 and 90.

336 = 90 × 3 + 66

Applying Euclid’s division lemma to 90 and 66:

90 = 66 × 1 + 24

Applying Euclid’s division lemma to 66 and 24:

66 = 24 × 2 + 18

Applying Euclid’s division lemma to 24 and 18:

24 = 18 × 1 + 6

Applying Euclid’s division lemma to 18 and 6:

18 = 6 × 3 + 0

As the remainder is zero, we need not apply Euclid’s division lemma anymore. The divisor (6) is the required H.C.F.

Example 4:

Find the H.C.F. of 45, 81, and 117 using Euclid’s division algorithm.

Solution:

Let us begin by choosing any two out of the three given numbers, say 45 and 81.

As 81 > 45, we apply Euclid’s division lemma to 81 and 45.

81 = 45 × 1 + 36

Applying Euclid’s division lemma to 45 and 36:

45 = 36 × 1 + 9

Applying Euclid’s division lemma to 36 and 9:

36 = 9 × 4 + 0

As the remainder is zero, the H.C.F. of 45 and 81 is 9.

Now, we again need to apply Euclid’s division algorithm on the H.C.F. of the two numbers and the remaining number.

Since the H.C.F. of 45 and 81 is 9 and the third number is 117, we apply Euclid’s division lemma to 117 and 9.

117 = 9 × 13 + 0

As the remainder is zero, the H.C.F. of 9 and 117 is 9.

Here, the second H.C.F. (the H.C.F. of the H.C.F. of the first two numbers and the third number) is the H.C.F. of the three numbers.

Thus, we can say that the H.C.F. of the three numbers is 9.

Example 5:

In an inter-school essay writing competition, the numbers of participants from schools A, B, and C are 20, 16, and 28 respectively. If the participants in each room are from the same school, then find the minimum number of rooms required such that each room has the same number of participants.

Solution:

If we need to find the minimum number of rooms, then we need to keep the maximum number of participants in each room i.e., we need to find the largest number that completely divides 20, 16, and 28.

Thus, we start by choosing any two out of the given three numbers, say 20 and 16.

Applying Euclid’s division lemma to 20 and 16:

20 = 16 × 1 + 4

Applying Euclid’s division lemma to 16 and 4:

16 = 4 × 4 + 0

Hence, 4 is the H.C.F. of 20 and 16.

Applying Euclid’s division algorithm to 28 and 4:

28 = 4 × 7 + 0

Hence, 4 is the H.C.F. of 28 and 4.

∴H.C.F. of 20, 16, 28 = 4

Therefore, each room would have 4 participants.

The number of rooms in which the participants from schools A, B, and C can be accommodated is 

= 5, = 4, and = 7 respectively.

Therefore, a total of 5 + 4 + 7 =16 rooms are required.

The Use Of Euclid's Division

 Lemma To Prove Mathematical

 Relationships

Whenever we divide 32 by 5, we will get 6 as the quotient and 2 as the remainder. Therefore, we can also write 32 as 6 x 5 + 2.

We can thus write this rule about the dividend as

This expression is unique, i.e., whenever we divide an integer (dividend) by another integer (divisor), we always get a fixed quotient and a fixed remainder.

The above statement can be proven by taking an example.

Whenever we divide 54 by 11, we will get 4 as the quotient and 10 as the remainder. We will never get a quotient other than 4 and a remainder other than 10 when we divide 54 by 11.

This brings us to Euclid’s division lemma.

This lemma has several applications, one of which is to prove mathematical relationships among numbers.

Let us discuss this concept with the help of a few examples.

Example 1:

Prove that every positive integer is of the form 3p, 3p + 1, or 3p + 2, where p is any integer.

Solution:

Let a be any positive integer and let b = 3.

Applying Euclid’s algorithm to a and 3:

a = 3p + r; for some integer p and 0 r < 3

Therefore, a can be 3p, 3p + 1, or 3p + 2.

As a is a positive integer, we can say that any positive integer is of the form 3p, 3p + 1, or 3p + 2.

Example 2:

Prove that every positive even integer is of the form 2m and every positive odd integer is of the form 2m + 1, where m is any integer.

Solution:

Let a be any positive integer and let b = 2.

According to Euclid’s division lemma, there exist two unique integers m and r such that

a = bm + r = 2m + r, where 0  r < 2.

Thus, r = 0 or 1

If r = 0, i.e., if a = 2m, then the expression is divisible by 2. Thus, it is an even number.

If r = 1, i.e., if a = 2m + 1, then the expression is not divisible by 2. Thus, it is an odd number.

Thus, every positive even integer is of the form 2m and every positive odd integer is of the form 2m + 1.

Example 3:

Prove that the expression y(y + 1) always represents an even number, where y is any positive integer.

Solution:

Let y be an integer. Thus, it may be either odd or even.

When y is an odd number, i.e., when y is of the form 2p + 1, where p is an integer:

y(y + 1) = (2p + 1)(2p + 1 + 1) = (2p + 1)(2p + 2) = 2(2p + 1)(p + 1)

The above expression is a multiple of 2. Thus, the expression y(y + 1) represents an even number.

When y is an even number, i.e., when y is of the form 2q, where q is an integer:

y(y + 1) = (2q)(2q + 1) = 2q(2q + 1)

The above expression is a multiple of 2. Thus, the expression y(y + 1) represents an even number.

Thus, for a positive integer y, the expression y(y + 1) always represents an even number.

Example 4: 

Show that

(a) The sum, the difference, and the product of two even numbers is always 

even.

(b) The sum and the difference of two odd numbers is always even, whereas 

their product is always odd.

Solution:

(a) Let there be two even numbers, x and y, such that x = 2m and y = 2n, where m and n are two positive integers.

Now, x + y = 2m + 2n = 2(m + n) = 2p, where p = m + n is an integer.

Thus, x + y is always even.

Similarly, x − y = 2m − 2n = 2(m − n) = 2q, where q = m − n is an integer.

Thus, x − y is always even.

Likewise, xy = 2m × 2n = 4mn = 2(2mn) = 2t, where t = 2mn is an integer.

Thus, xy is always even.

Hence, the sum, the difference, and the product of two even numbers is always even.

(b) Let there be two odd numbers x and y such that x = 2m + 1 and y = 2n + 1, where m and n are two positive integers.

Now, x + y = (2m + 1) + (2n + 1) = 2(m + n + 1) = 2p, where p = m + n + 1 is an integer.

Thus, x + y is even.

Similarly, x − y = (2m + 1) − (2n + 1) = 2(m − n) = 2q, where q = m − n is an integer.

Thus, x − y is even.

Likewise, xy = (2m + 1)(2n + 1) = 2(2mn + m + n) + 1 = 2t + 1, where t = 2mn + m + n is an integer.

Thus, xy is always odd.

Hence, the sum and the difference of two odd numbers is always even, whereas their product is always odd.

Prime Factorisation of Numbers Using Fundamental Theorem of

 Arithmetic

We can see that whichever way we factorise the number 36, it will be broken down as the product of the same prime numbers, which is unique. The only difference is that the ordering of the prime numbers will be different for different ways of factorising the number. In fact, this is true for all numbers. We can check this by taking the example of a larger number, say 21560, which can be uniquely broken down into its prime factors as

2³ × 5 × 7² × 11

Hence, we can say that any composite number can be written in the form of the product of prime numbers, which is unique, except the order in which they occur. By this, we mean that 2 × 3 × 7 × 11 is the same as 

7 × 11 × 2 × 3.

This is the fundamental theorem of arithmetic. It can be formally stated as:

Thus, this theorem can be used to write the prime factorisation of any number. Let us try to build on this concept with the help of some examples.

Example 1:

Write the prime factorization of 31250. What are its prime factors?

Solution:

31250 = 2 × 15625

= 2 × 5 × 3125

= 2 × 5 × 5 × 625

= 2 × 5 × 5 × 5 × 125

= 2 × 5 × 5 × 5 × 5 × 25

= 2 × 5 × 5 × 5 × 5 × 5 × 5

= 2 × 5⁶

Hence, 2 × 5⁶ is the prime factorisation of 31250. Its prime factors are 2 and 5.

Example 2:

If it is given that 13125 = 2^a ×3^b× 5^c× 7^d, then find the value of a +2b +7c + 11d.

Solution:

2^a × 3^b × 5^c × 7^d = 13125

= 3 × 4375

= 3 × 5 × 875

= 3 × 5 × 5 ×175

= 3 × 5 × 5 × 5 × 35

= 3 × 5 × 5 × 5 × 5 × 7

= 3¹ × 5⁴ × 7¹

∴ 2^a × 3^b × 5^c × 7^d = 2⁰ × 3¹ × 5⁴ × 7¹

Comparing exponents of the bases (integers): a = 0, b = 1, c = 4, and d =1

Hence, a +2b +7c +11d = 0 + 2 ×1 + 7 × 4 + 11 × 1

= 0 + 2 + 28 + 11

= 41

Example 3:

Show that the expressions given below are composite numbers.

(a) 3 × 5 × 7 × 23 + 2 × 7 × 11 × 13

(b) 29 × 35 + 14

(c) 3⁴ + 6³

Solution:

(a) 3 × 5 × 7 × 23 + 2 × 7 × 11 × 13 = 7( 3 × 5 × 23 + 2 × 11 × 13)

= 7(345 + 286)

= 7 × 631

Since both 7 and 631 are prime numbers, we have expressed the given expression as the product of two prime numbers. We know that according to the fundamental theorem of arithmetic, every composite number can be uniquely written as the product of its prime factors. Thus, the given expression represents a composite number.

(b) 29 × 35 + 14 = 29 × 5 × 7 + 2 × 7

= 7(29 × 5 + 2)

= 7 × 147

= 7 × 3 × 7 × 7

= 3 × 7³

Since both 3 and 7 are prime numbers, we have expressed the given expression as the product of its prime factors. We know that according to the fundamental theorem of arithmetic, every composite number can be uniquely written as the product of its prime factors. Thus, the given expression represents a composite number.

(c) 3⁴ + 6³ = 3⁴ + (2 × 3) ³

= 3⁴ + 2³ × 3³

= 3³(3 + 2³)

= 3³ × 11

Since both 3 and 11 are prime numbers, we have expressed the given expression as the product of its prime factors. It is known that according to the fundamental theorem of arithmetic, every composite number can be uniquely written as the product of its prime factors. Thus, the given expression represents a composite number.

Application Of The Fundamental Theorem Of Arithmetic To Find

 The HCF And LCM Of Numbers

All composite numbers can be written as the product of two or more prime numbers. For example, 20 can be written as 2² × 5; 54 can be written as 2 × 3³, and so on.

Note that if we do not consider the way in which the prime factors are written, then we can prime factorise every number in only one way. This applies to other numbers as well.

This leads to the fundamental theorem of arithmetic, which states that:

Even though we did not notice it before, whenever we prime factorise a number, we use the fundamental theorem of arithmetic to do so.

For example, the prime factorisation of 980 is represented as

980 = 2 × 490

= 2 × 2 × 245

= 2 × 2 × 5 × 49

= 2 × 2 × 5 × 7 × 7

= 2² × 5¹ × 7²

Hence, 2² × 5¹ × 7² is the prime factorisation of 980; and 2, 5, and 7 are its prime factors.

By applying the fundamental theorem of arithmetic to the prime factorized numbers, we can also find their HCF and LCM.

This is known as the prime factorisation method, which states that:

Example 1:

Find the LCM and the HCF of 432 and 676 using the prime factorization method.

Solution:

We can write these numbers as

432 = 2⁴ × 3³

676 = 2² × 13²

To calculate the HCF

We observe that the only common prime factor is 2 and the smallest power of this prime factor is also 2.

Thus, HCF (432, 676) = 2² = 4

To calculate the LCM

We observe that the prime factors of 432 and 676 are 2, 3, and 13. The greatest powers of these factors are 4, 3, and 2 respectively.

LCM is the product of the greatest power of each prime factor.

Thus, LCM (432, 676) = 2⁴ × 3³ × 13² = 73008

Example 2:

Find the HCF and the LCM of 28, 42, and 64 using the prime factorization method.

Solution:

We can write these numbers as

28 = 2² × 7¹

42 = 2 × 3¹ × 7¹

64 = 2⁶

HCF is the product of the smallest power of each common prime factor.

Here, the only common prime factor is 2 and its power is 1.

Thus, HCF (28, 42, 64) = 2¹ = 2

LCM is the product of the greatest power of each prime factor.

Thus, LCM (28, 42, 64) = 2⁶ × 3¹ × 7¹ = 1344

Example 3:

Find the HCF and the LCM of 1080 and 900 using the prime factorization and show that HCF × LCM = Product of two numbers.

Solution:

1080 = 2³ × 3³ × 5

900 = 2² × 3² × 5²

Hence, HCF (1080, 900) = 2² × 3² × 5 = 180

LCM (1080, 900) = 2³ × 3³ × 5² = 5400

HCF × LCM = 180 × 5400 = 972000

Product of numbers = 1080 × 900 = 972000

Hence, HCF × LCM = Product of two numbers

Example 4:

The HCF of 273 and another number is 7, while their LCM is 3003. Find the other number.

Solution:

Let the first number (a) be 273 and the second number be b.

It is given that HCF (a, b) = 7 and LCM (a, b) = 3003.

We know that HCF × LCM = Product of two numbers.

⇒ HCF (a, b) × LCM (a, b) = a × b

⇒ 7 × 3003 = 273 × b

⇒ b = 77

Hence, the other number is 77.

Example 5:

Anurag takes 6 minutes to complete one round of jogging around the circular track of a park, while Twinkle takes 8 minutes to do the same. If both of them start jogging at the same time from the same point, then how much time will they take before they meet at the point from which they started?

Solution:

Since Anurag and Twinkle take 6 minutes and 8 minutes respectively to complete one round of the circular track, the time after which they will meet at the starting point will be the lowest multiple of 6 and 8, i.e., their LCM.

6 = 2 × 3

8 = 2³

∴ LCM (6, 8) = 2³ × 3 = 24

Thus, they will meet at the starting point after 24 minutes.

Example 6:

There are 120 students in a class. When the students were arranged according to their roll numbers, it was observed that every second student got distinction in Mathematics, every third student got distinction in Science, and every fifth student got distinction in English. How many students got distinction in all three subjects?

Solution:

As every second, third, and fifth student got distinction in Math, Science, and English respectively, the roll numbers of the students who got distinction in all three subjects will be equal to the multiples of the LCM of 2, 3, and 5.

LCM (2, 3, 5) = 2 × 3 × 5 = 30

Thus, every 30^th student got distinction in all three subjects.

Thus, a total of  students got distinction in all three subjects.

Properties Of Prime Numbers

Irrational Numbers

We know that a number which cannot be written in the form of , where p and q are integers and q ≠ 0, is known as an irrational number.

For example: all numbers of the form , where p is a prime number such as   etc., are irrational numbers.

How can we prove that these are irrational numbers?

We can prove this by making use of a theorem which can be stated as follows.

“If p divides a2, then p divides a (where p is a prime number and a is a positive integer)”.

Proof of theorem: Let a=p1×p2×...×pn$a=p_1\times p_2\times ...\times p_n$                     ...(1)
where p1,p2,...pn$p_1,p_{2,...}{p}_n$ are the prime factors of a. 
Squaring (1) we get
a2=(p1×p2×.....×pn)2$a^2={\left(p_1\times p_2\times .....\times p_n\right)}^2$
But p divides a² 
p is a factor of p12×p22×.....×pn2${p_1}^2\times {p_2}^2\times .....\times {p_n}^2$
By the fundamental theorem of arithmetic, the primes in the factorisation of p12×p22×.....×pn2${p_1}^2\times {p_2}^2\times .....\times {p_n}^2$ are unique. 
So, p is one out of the primes p1,p2,....pn$p_1,p_{2,....}{p}_n$.
If p = p_k where k has value from 1 to n, p_k divides p1×p2×.....pn$p_1\times p_2\times .....p_n$
So, pk$p_k$ divides a.
Thus, p divides a as pk$p_k$ = p. 

So go through the given video to understand the application of the above stated property.

Similarly, we can prove that square roots of other prime numbers ,  ,etc. are irrational numbers.

Besides these irrational numbers, there are some other irrational numbers like etc.

We can also prove why these numbers are irrational. Before this, let us first see what happens to irrational numbers, when we apply certain mathematical operations on them.

Addition or subtraction of two irrational numbers gives a rational or an irrational number. Addition or subtraction of a rational and an irrational number gives an irrational number. Multiplication of a non-zero rational number and an irrational number gives an irrational number. Multiplication of two irrational numbers gives a rational or an irrational number.

We will now prove that is irrational.

We know that is irrational (as proved before).

Now, the multiplication of a rational and an irrational number gives an irrational number.

Therefore, is an irrational number.

Let us now try to understand the concept further through some more examples.

Example 1:

Prove that is irrational.

Solution:

Let us assume that  is not irrational, i.e. is a rational number.

Then we can write , where a and b are integers and b ≠ 0.

Let a and b have a common factor other than 1.

After dividing by the common factor, we obtain

, where c and d are co-prime numbers.

As c, d and 2 are integers,  and are rational numbers.

Thus, is rational.

is rational as the difference of two rational numbers is again a rational number.

This is a contradiction as is irrational.

Therefore, our assumption that is rational is wrong.

Hence, is irrational.

Example 2:

Prove thatis irrational.

Solution:

Let us assume  is rational. Then, we can write

,where a and b are co-prime and b ≠ 0.

⇒=

Now, as a and b are integers, is rational or is a rational number.

This means thatis rational.

This is a contradiction as is irrational.

Therefore, our assumption that   is rational is wrong.

Hence,  is an irrational number.

Decimal Expansions of Rational Numbers

The Need for Converting Rational Numbers into Decimals 

A carpenter wishes to make a point on the edge of a wooden plank at 95 mm from any end. He has a centimeter tape, but how can he use that to mark the required point? 

Simple! He should convert 95 mm into its corresponding centimeter value, i.e., 9.5 cm and then measure and mark the required length on the wooden plank. 

This is just one of the many situations in life when we face the need to convert numbers into decimals. In this lesson, we will learn to convert rational numbers into decimals, observe the types of decimal numbers, and solve a few examples based on this concept. 

Two rational numbers a/b and c/d are equal if and only if ad = bc.

Take, for example, the rational numbersand. and  Let us see if they are equal or not.

Here, a = 2, b = 4, c = 3 and d = 6

Now, we have:

ad = 2 × 6 = 12

bc = 4 × 3 = 12

Since ad = bc, we obtain=