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R EAL NUMBERS

1

REAL NUMBERS

1

1
...
We continue our discussion on real numbers in this chapter
...
2 and 1
...

Euclid’s division algorithm, as the name suggests, has to do with divisibility of
integers
...
Many of you
probably recognise this as the usual long division process
...
We touch upon a few of them, and use it mainly to compute the HCF of
two positive integers
...
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
...

We use the Fundamental Theorem of Arithmetic for two main applications
...
Second, we apply this theorem to explore when exactly the decimal
p
expansion of a rational number, say q ( q  0) , is terminating and when it is nonterminating repeating
...
You will see that the prime factorisation of q will completely reveal the nature
q
p
of the decimal expansion of
...


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MATHEMATICS

1
...

A trader was moving along a road selling eggs
...
This grew into a
fight, he pulled the basket with eggs and dashed it on the floor
...

The trader requested the Panchayat to ask the idler to pay for the broken eggs
...
He gave the
following response:
If counted in pairs, one will remain;
If counted in threes, two will remain;
If counted in fours, three will remain;
If counted in fives, four will remain;
If counted in sixes, five will remain;
If counted in sevens, nothing will remain;
My basket cannot accomodate more than 150 eggs
...
Let the number
of eggs be a
...
If counted in sixes, a = 6 q+ 5, for some natural number q
...
It translates to a = 5 w + 4, for some natural
number w
...
It translates to a = 4s + 3, for some natural
number s
...
It translates to a = 3t + 2, for some natural
number t
...
It translates to a = 2 u + 1, for some natural
number u
...
The
* This is modified form of a puzzle given in ‘Numeracy Counts!’ by A
...


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R EAL NUMBERS

3

moment we write down such equations we are using Euclid’s division lemma,
which is given in Theorem 1
...

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
...

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:
a = bq + r, 0 ≤ r < b
Note that q or r can also be zero
...
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
...
1 (Euclid’s Division Lemma) : Given positive integers a and b,
there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b
...
Euclid’s division algorithm is based on this lemma
...

The word algorithm comes from the name
of the 9th century Persian mathematician
al-Khwarizmi
...

A lemma is a proven statement used for Muhammad ibn Musa al-Khwarizmi
proving another statement
...
E
...
Recall that the HCF of two positive integers a
and b is the largest positive integer d that divides both a and b
...
Suppose we need
to find the HCF of the integers 455 and 42
...
Then we use Euclid’s lemma to get
455 = 42 × 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
...
e
...
You can easily
verify this by listing all the factors of 455 and 42
...

So, let us state Euclid’s division algorithm clearly
...
So, we find whole numbers, q and
r such that c = dq + r, 0 ≤ r < d
...
If r ≠ 0, apply the division lemma to d and r
...
The divisor at this stage will
be the required HCF
...

Example 1 : Use Euclid’s algorithm to find the HCF of 4052 and 12576
...
Since the divisor at this
stage is 4, the HCF of 12576 and 4052 is 4
...

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
...
Euclid’s division lemma and algorithm are so closely interlinked that people often
call former as the division algorithm also
...
Although Euclid’s Division Algorithm is stated for only positive integers, it can be
extended for all integers except zero, i
...
, b ≠ 0
...


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Euclid’s division lemma/algorithm has several applications related to finding
properties of numbers
...

Solution : Let a be any positive integer and b = 2
...
So,
a = 2q or 2q + 1
...
Also, a positive integer can be
either even or odd
...

Example 3 : Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where
q is some integer
...
We apply the
division algorithm with a and b = 4
...

That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient
...

Therefore, any odd integer is of the form 4q + 1 or 4q + 3
...
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
...
But to do it systematically, we find
HCF (420, 130)
...
The area of the tray that is used
up will be the least
...
We have :
420 = 130 × 3 + 30
130 = 30 × 4 + 10
30 = 10 × 3 + 0
So, the HCF of 420 and 130 is 10
...


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R EAL NUMBERS

7

EXERCISE 1
...
Use Euclid’s division algorithm to find the HCF of :
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
2
...

3
...
The two groups are to march in the same number of columns
...
Use Euclid’s division lemma to show that the square of any positive integer is either of
the form 3m or 3m + 1 for some integer m
...
Now square
each of these and show that they can be rewritten in the form 3m or 3m + 1
...
Use Euclid’s division lemma to show that the cube of any positive integer is of the form
9m, 9m + 1 or 9m + 8
...
3 The Fundamental Theorem of Arithmetic
In your earlier classes, you have seen that any natural number can be written as a
product of its prime factors
...

Now, let us try and look at natural numbers from the other direction
...

Take any collection of prime numbers, say 2, 3, 7, 11 and 23
...

Let us list a few :
7 × 11 × 23 = 1771

3 × 7 × 11 × 23 = 5313

2 × 3 × 7 × 11 × 23 = 10626

2 3 × 3 × 73 = 8232

2 2 × 3 × 7 × 11 × 23 = 21252
and so on
...

What is your guess about the size of this collection? Does it contain only a finite
number of integers, or infinitely many? Infact, there are infinitely many primes
...
The question is – can we
produce all the composite numbers this way? What do you think? Do you think that
there may be a composite number which is not the product of powers of primes?
Before we answer this, let us factorise positive integers, that is, do the opposite of
what we have done so far
...
Let us take
some large number, say, 32760, and factorise it as shown :
32760

2

16380

2

8190

4095

2

1365

3

3

455

5

91

7

13

So we have factorised 32760 as 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 as a product of
primes, i
...
, 32760 = 23 × 32 × 5 × 7 × 13 as a product of powers of primes
...
This can be written as 32 × 3803 × 3607
...
) This leads us to a conjecture that every composite number can be
written as the product of powers of primes
...
Let us now formally state this theorem
...
2 (Fundamental Theorem of Arithmetic) : Every composite number
can be expressed ( factorised ) as a product of primes, and this factorisation is
unique, apart from the order in which the prime factors occur
...
2 was probably first
recorded as Proposition 14 of Book IX in Euclid’s
Elements, before it came to be known as the Fundamental
Theorem of Arithmetic
...

Carl Friedrich Gauss is often referred to as the ‘Prince of
Mathematicians’ and is considered one of the three
greatest mathematicians of all time, along with Archimedes
and Newton
...

(1777 – 1855)
The Fundamental Theorem of Arithmetic says that every composite number
can be factorised as a product of primes
...
It says that given
any composite number it can be factorised as a product of prime numbers in a
‘unique’ way, except for the order in which the primes occur
...
So, for
example, we regard 2 × 3 × 5 × 7 as the same as 3 × 5 × 7 × 2, or any other
possible order in which these primes are written
...

In general, given a composite number x, we factorise it as x = p 1 p 2
...
, p n are primes and written in ascending order, i
...
, p 1 ≤ p 2
≤
...
If we combine the same primes, we will get powers of primes
...

The Fundamental Theorem of Arithmetic has many applications, both within
mathematics and in other fields
...

Example 5 : Consider the numbers 4n , where n is a natural number
...

Solution : If the number 4n, for any n, were to end with the digit zero, then it would be
divisible by 5
...
This is

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MATHEMATICS

not possible because 4n = (2)2n; so the only prime in the factorisation of 4n is 2
...
So, there is no natural number n for which 4 n
ends with the digit zero
...
Let us recall this method
through an example
...

Solution : We have :

6 = 21 × 31 and 20 = 2 × 2 × 5 = 22 × 51
...

Note that HCF(6, 20) = 21 = Product of the smallest power of each common
prime factor in the numbers
...

From the example above, you might have noticed that HCF(6, 20) × LCM(6, 20)
= 6 × 20
...
We can use this result to find the LCM of two
positive integers, if we have already found the HCF of the two positive integers
...
Hence,
find their LCM
...

Also,

LCM (96, 404) =

96  404
96  404

 9696
HCF(96, 404)
4

Example 8 : Find the HCF and LCM of 6, 72 and 120, using the prime factorisation
method
...


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R EAL NUMBERS

So,

11

HCF (6, 72, 120) = 21 × 3 1 = 2 × 3 = 6

2 3, 32 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively
involved in the three numbers
...
So, the
product of three numbers is not equal to the product of their HCF and LCM
...
2
1
...
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF =
product of the two numbers
...
Find the LCM and HCF of the following integers by applying the prime factorisation
method
...
Given that HCF (306, 657) = 9, find LCM (306, 657)
...
Check whether 6n can end with the digit 0 for any natural number n
...
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers
...
There is a circular path around a sports field
...
Suppose they both start at the
same point and at the same time, and go in the same direction
...
4 Revisiting Irrational Numbers
In Class IX, you were introduced to irrational numbers and many of their properties
...
You even studied how to locate irrationals on the number
line
...
In this section, we will
prove that 2 , 3 , 5 and, in general, p is irrational, where p is a prime
...

p
Recall, a number ‘s’ is called irrational if it cannot be written in the form ,
q
where p and q are integers and q ≠ 0
...
10110111011110
...

3
Before we prove that 2 is irrational, we need the following theorem, whose
proof is based on the Fundamental Theorem of Arithmetic
...
3 : Let p be a prime number
...

*Proof : Let the prime factorisation of a be as follows :
a = p 1p 2
...
, p n are primes, not necessarily distinct
...
pn)( p 1 p 2
...
p2
...
Therefore, from the Fundamental Theorem of
Arithmetic, it follows that p is one of the prime factors of a2
...
, p n
...
, p n
...
pn , p divides a
...

The proof is based on a technique called ‘proof by contradiction’
...

Theorem 1
...


Proof : Let us assume, to the contrary, that

2 is rational
...

s
Suppose r and s have a common factor other than 1
...

b

So, b 2 = a
...
Therefore, 2 divides a 2
...
3, it follows that 2 divides a
...

* Not from the examination point of view
...

This means that 2 divides b 2, and so 2 divides b (again using Theorem 1
...

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

This contradiction has arisen because of our incorrect assumption that
So, we conclude that



2 is irrational
...


3 is irrational
...


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 coprime
...

Therefore, a2 is divisible by 3, and by Theorem 1
...

So, we can write a = 3c for some integer c
...

This means that b 2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1
...

Therefore, a and b have at least 3 as a common factor
...

This contradiction has arisen because of our incorrect assumption that
So, we conclude that

3 is rational
...


In Class IX, we mentioned that :


the sum or difference of a rational and an irrational number is irrational and



the product and quotient of a non-zero rational and irrational number is
irrational
...


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Example 10 : Show that 5 – 3 is irrational
...


a
That is, we can find coprime a and b (b ≠ 0) such that 5  3  
b
a
Therefore, 5   3 
b
a 5b  a
Rearranging this equation, we get 3  5 – 

b
b
a
Since a and b are integers, we get 5 – is rational, and so 3 is rational
...

This contradiction has arisen because of our incorrect assumption that 5 –
rational
...

Example 11 : Show that 3 2 is irrational
...

a
That is, we can find coprime a and b (b ≠ 0) such that 3 2  
b
a
Rearranging, we get 2  
3b
a
Since 3, a and b are integers,
is rational, and so 2 is rational
...


So, we conclude that 3 2 is irrational
...
3
1
...


2
...

3
...
5 Revisiting Rational Numbers and Their Decimal Expansions
In Class IX, you studied that rational numbers have either a terminating decimal
expansion or a non-terminating repeating decimal expansion
...
We do so by considering several examples
...
375
Now

(ii) 0
...
0875

375 375

1000 10 3
875
875
(iii) 0
...
375 

(iv) 23
...


104 104

1000 10 3
233408 233408
(iv) 23
...
104 

As one would expect, they can all be expressed as rational numbers whose
denominators are powers of 10
...
375 

375 3  53
3
 3
 3
3
3
10
2 5
2

(iii) 0
...
104 

104 13  23 13


10 3 2 3  53 53

(iv) 23
...
We should expect the denominator to look like
this, since powers of 10 can only have powers of 2 and 5 as factors
...
Also the only prime factors of 10
are 2 and 5
...

Let us write our result formally:

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Theorem 1
...

p
Then x can be expressed in the form , where p and q are coprime, and the
q
prime factorisation of q is of the form 2n5 m, where n, m are non-negative integers
...
5
...
You will surely agree
a
that any rational number of the form , where b is a power of 10, will have a terminating
b
decimal expansion
...
Let us go back to our examples above and work backwards
...
375
3
8 2
2 5
10

(ii)

13 13 13  23 104



 0
...
0875
80 2  5 2  54 10 4

(iv)

14588 2 2  7  521 2 6  7  521 233408



 23
...
Therefore, the decimal expansion of such a rational number
terminates
...

p
be a rational number, such that the prime factorisation
q
of q is of the form 2n5 m , where n, m are non-negative integers
...
6 : Let x =

decimal expansion which terminates
...

Once again, let us look at an example to see what is going on
...
Here, remainders are 3, 2, 6, 4, 5, 1, 3,
7
2, 6, 4, 5, 1,
...


0
...
e
...
Therefore, from Theorems 1
...
6, we
1
know that will not have a terminating decimal expansion
...
So, we
will have a block of digits, namely, 142857, repeating in the
1
quotient of
...
5 and 1
...
For such numbers we have :
p
be a rational number, such that the prime factorisation
q
of q is not of the form 2n5 m, where n, m are non-negative integers
...

Theorem 1
...


EXERCISE 1
...
Without actually performing the long division, state whether the following rational
numbers will have a terminating decimal expansion or a non-terminating repeating decimal
expansion:
(i)

13
3125

(ii)

(v)

29
343

(vi)

(ix)

35
50

(x)

17
8
23
3 2

25

(iii)
(vii)

64
455

(iv)

129
2 7

25 7

5

(viii)

15
1600
6
15

77
210

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MATHEMATICS
2
...

3
...
In each case,
decide whether they are rational or not
...
123456789

(ii) 0
...


(iii) 43
...
6 Summary
In this chapter, you have studied the following points:
1
...

2
...
According to this,
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
...
If r ≠ 0, apply Euclid’s lemma to b and r
...
The divisor at this stage will be
HCF (a, b)
...

3
...

4
...

5
...

6
...
Then we can express x
p
in the form , where p and q are coprime, and the prime factorisation of q is of the form
q
2n5m, where n, m are non-negative integers
...
Let x = p be a rational number, such that the prime factorisation of q is of the form 2n5m,
q
where n, m are non-negative integers
...

p
be a rational number, such that the prime factorisation of q is not of the form
q
2n 5m , where n, m are non-negative integers
...


8
...
However, the following results hold good for three numbers
p, q and r :
LCM (p, q, r) =

p q r HCF(p , q, r )
HCF( p , q)  HCF( q ,r )  HCF( p, r )

HCF (p, q, r) =

p q r  LCM(p, q, r )
LCM( p, q)  LCM(q, r )  LCM( p , r )

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