Mathematics REAL NUMBERS - Introduction , Euclid’s Division Lemma

### Introduction

\color{green} ✍️ In Class IX, you began your exploration of the world of real numbers and encountered irrational numbers.

\color{green} ✍️ We continue our discussion on real numbers in this chapter. We begin with two very important properties of positive integers namely the color(blue)("Euclid’s division algorithm") and the color(green)("Fundamental Theorem of Arithmetic.")

color(red)(ul★"Euclid’s division algorithm"), as the name suggests, has to do with divisibility of integers.

Stated simply, it says any positive integer a can be divided by another positive integer b in such a way that it leaves a remainder r that is smaller than b.

Many of you probably recognise this as the usual long division process. Although this result is quite easy to state and understand, it has many applications related to the divisibility properties of integers.

color(red)(ul★"The Fundamental Theorem of Arithmetic,") on the other hand, has to do something with multiplication of positive integers.

\color{green} ✍️You already know that every composite number can be expressed as a product of primes in a unique way—this important fact is the Fundamental Theorem of Arithmetic.

Again, while it is a result that is easy to state and understand, it has some very deep and significant applications in the field of mathematics.

\color{green} ✍️ We use the Fundamental Theorem of Arithmetic for two main applications. First, we use it to prove the irrationality of many of the numbers you studied in Class IX, such as sqrt2 , sqrt3 and sqrt5 .

Second, we apply this theorem to explore when exactly the decimal expansion of a rational number, say p/q (q ne 0) , is terminating and when it is non - terminating repeating.

We do so by looking at the prime factorisation of the denominator q of p/q . You will see that the prime factorisation of q will completely reveal the nature of the decimal expansion of p/q .

### Euclid’s Division Lemma

\color{green} ✍️ A trader was moving along a road selling eggs. An idler who didn’t have much work to do, started to get the trader into a wordy duel. This grew into a fight, he pulled the basket with eggs and dashed it on the floor. The eggs broke. The trader requested the Panchayat to ask the idler to pay for the broken eggs. The Panchayat asked the trader how many eggs were broken. He gave the following response:

" "color(green)" If counted in pairs, one will remain;"

" "color(green)" If counted in threes, two will remain;"

" "color(green)" If counted in fours, three will remain;"

" "color(green)" If counted in fives, four will remain;"

" "color(green)" If counted in sixes, five will remain;"

" "color(green)" If counted in sevens, nothing will remain;"

" "color(green)" My basket cannot accomodate more than 150 eggs."

So, how many eggs were there? Let us try and solve the puzzle. Let the number of eggs be a. Then working backwards, we see that a is less than or equal to 150:

If counted in sevens, nothing will remain, which translates to a = 7p + 0, for some natural number p. If counted in sixes, a = 6q+5, for some natural number q.

If counted in fives, four will remain. It translates to a = 5w + 4, for some natural number w.

If counted in fours, three will remain. It translates to a = 4s + 3, for some natural number s.

If counted in threes, two will remain. It translates to a = 3t + 2, for some natural number t.

If counted in pairs, one will remain. It translates to a = 2u + 1, for some natural number u.

That is, in each case, we have a and a positive integer b (in our example, b takes values 7, 6, 5, 4, 3 and 2, respectively) which divides a and leaves a remainder r (in our case, r is 0, 5, 4, 3, 2 and 1, respectively), that is smaller than b.

The moment we write down such equations we are using Euclid’s division lemma,

Getting back to our puzzle, do you have any idea how you will solve it? Yes! You must look for the multiples of 7 which satisfy all the conditions. By trial and error

(using the concept of LCM), you will find he had 119 eggs.

\color{green} ✍️ In order to get a feel for what Euclid’s division lemma is, consider the following pairs of integers:

17, \ \ \ 6; \ \ \ 5, \ \ \ 12; \ \ \ 20, \ \ \ 4

Like we did in the example, we can write the following relations for each such pair:

17 = 6 × 2 + 5 (6 goes into 17 twice and leaves a remainder 5)

5 = 12 × 0 + 5 (This relation holds since 12 is larger than 5)

20 = 4 × 5 + 0 (Here 4 goes into 20 five-times and leaves no remainder)

That is, for each pair of positive integers a and b, we have found whole numbers q and r, satisfying the relation.

\color{green} ✍️ a = bq + r, 0 ≤ r < b

Note that q or r can also be zero.

Why don’t you now try finding integers q and r for the following pairs of positive integers a and b?

(i) 10, 3;" " (ii) 4, 19; " " (iii) 81, 3

Did you notice that q and r are unique? These are the only integers satisfying the conditions a = bq + r, where 0 ≤ r < b.

You may have also realised that this is nothing but a restatement of the long division process you have been doing all these years, and
that the integers q and r are called the quotient and remainder, respectively.

\color{green} ✍️ color(blue)(ul"A formal statement of this result is as follows :")

color(red)(★ ul"Theorem ( Euclid’s Division Lemma) :") Given positive integers a and b, there exist unique integers 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.

Euclid’s division algorithm is a technique to compute the Highest Common Factor (HCF) of two given positive integers.

Recall that the HCF of two positive integers a and b is the largest positive integer d that divides both a and b.

color(red)(=>"To obtain the HCF of two positive integers"), say c and d, with c > d, follow the steps below:

color(red)("Step 1 :") Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and such that color(blue)(c = dq + r,\ \ 0 ≤ r < d).

color(red)("Step 2 :") If color(blue)(r = 0),\ \ d is the HCF of c and d.

" " If color(blue)(r ≠ 0,) apply the division lemma to d and r.

color(red)("Step 3 :") Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

This algorithm works because color(blue)(HCF (c, d) = HCF (d, r)) where the symbol HCF (c, d) denotes the HCF of c and d, etc.

color(red)(=>"Let us see how the algorithm works, through an example").

Suppose we need to find the HCF of the integers 455 and 42. We start with the larger integer, that is, 455.

Then we use Euclid’s lemma to get

 " " 455 =42 xx 10+35

Now consider the divisor 42 and the remainder 35, and apply the division lemma to get

" " 42 = 35 × 1 + 7

Now consider the divisor 35 and the remainder 7, and apply the division lemma to get

" " 35 = 7 × 5 + 0

Notice that the remainder has become zero, and we cannot proceed any further.

We claim that the HCF of 455 and 42 is the divisor at this stage, i.e., 7.

You can easily verify this by listing all the factors of 455 and 42. Why does this method work? It works because of the following result.

So, let us state Euclid’s division algorithm clearly.

Q 3179656516

Use Euclid’s algorithm to find the HCF of 4052 and 12576.
Class 10 Chapter 1 Example 1
Solution:

Step 1 : Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get
12576 = 4052 × 3 + 420
Step 2 : Since the remainder 420 ≠ 0, we apply the division lemma to 4052 and 420, to
get
4052 = 420 × 9 + 272
Step 3 : We consider the new divisor 420 and the new remainder 272, and apply the
division lemma to get
420 = 272 × 1 + 148

We consider the new divisor 272 and the new remainder 148, and apply the division
lemma to get
272 = 148 × 1 + 124
We consider the new divisor 148 and the new remainder 124, and apply the division
lemma to get
148 = 124 × 1 + 24
We consider the new divisor 124 and the new remainder 24, and apply the division
lemma to get
124 = 24 × 5 + 4
We consider the new divisor 24 and the new remainder 4, and apply the division
lemma to get
24 = 4 × 6 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this
stage is 4, the HCF of 12576 and 4052 is 4.
Notice that 4 = HCF (24, 4) = HCF (124, 24) = HCF (148, 124) = HCF (272, 148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576, 4052).
Euclid’s division algorithm is not only useful for calculating the HCF of very
large numbers, but also because it is one of the earliest examples of an algorithm that a computer had been programmed to carry out.

text(Remarks) :
1. Euclid’s division lemma and algorithm are so closely interlinked that people often
call former as the division algorithm also.
2. Although Euclid’s Division Algorithm is stated for only positive integers, it can be
extended for all integers except zero, i.e., b ≠ 0. However, we shall not discuss this
aspect here.
Q 3109656518

Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer.
Class 10 Chapter 1 Example 2
Solution:

Let a be any positive integer and b = 2. Then, by Euclid’s algorithm,
a = 2q + r, for some integer q ≥ 0, and r = 0 or r = 1, because 0 ≤ r < 2. So,
a = 2q or 2q + 1.
If a is of the form 2q, then a is an even integer. Also, a positive integer can be
either even or odd. Therefore, any positive odd integer is of the form 2q + 1.
Q 3119756610

Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.
Class 10 Chapter 1 Example 3
Solution:

Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b = 4.
Since 0 ≤ r < 4, the possible remainders are 0, 1, 2 and 3.
That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient.
However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).
Therefore, any odd integer is of the form 4q + 1 or 4q + 3.
Q 3139756612

A sweetseller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the number of that can be placed in each stack for this
purpose?
Class 10 Chapter 1 Example 4
Solution:

This can be done by trial and error. But to do it systematically, we find HCF (420, 130). Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least. The area of the tray that is used up will be the least.
Now, let us use Euclid’s algorithm to find their HCF. We have :

420 = 130 × 3 + 30
130 = 30 × 4 + 10
30 = 10 × 3 + 0

So, the HCF of 420 and 130 is 10.
Therefore, the sweetseller can make stacks of 10 for both kinds of barfi.