1

Real Nunnbers

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1.1 FUNDAMENTAL THEOREM OF ARITHMETIC

Ruall th.at a natunal nunlr in valled a opww umber it it han atlealone more lactor other
than ladthe numleriteli Somevonposite umben ae 4, 6, 15, 42, 0,
In carier classes, we have lvant thal every cooite nalural nunber can be expressed
nunber.
Tor evample:

Wecan also use wtor tee o tactorise o) into prie taclos.

60 |

2 30

|2 I5

Real Numbers 1
 5040
x 3x 5 x7
5040 =2 x 2 x 2 x 2x3 2
2520
(i) z!
x 32x 5 x
= 24
2
1260
2 630
315
3 105
5 35
7

32760
2 16380
13 2 8190
2 x 2 x 3x3 x5x7x
x
(iüi) 32760 = 2 131 4095
32 x 51x 7l x
= 23 x 3 1365
5 455
7 91
13

composite natural number can be evn

examples, we see that a
From the aboveproduct of primenumbers. primes is unique except for the order
(factorised) as a product of
factorisation into natural. number there is one and only one Wa
Moreover, such a composite
i.e. given any regard2 x 2 x3 x 5the same as 2 x 3 x 5x 2, or am
occur
whichthe primesproduct of primes. Here, we are written. This leads to:
to writeit as a prime factors
in which these
other possible order
Arithmetic
Fundamental Theorem of
epressed as a product of primes, and thË
O integer n can be which the primefactorsoccur i.e.
positive
Every composite apart fromthe order in are pos1ti
factorisation is unique,
distinct primes and a, a2 , a7
PP;..p, where p, p, , , are
n= unique. by using th
integers, and such a factorisation is HCF and LCM of two positive integers
already learnt how tofind This method is also called prin
we nave Arithmetic in earlier classes.
Fundamental Theorem of
factorisation method.
number
Ifa and bare two positive integers, then prime factor in the
HCF (a, b) =product of the smallest pOwer of each common
a and b.
involved in the number
LCM (a, b) = product of the factor
a and b. greatest power of each prime
O
Relationship between HCF and LCM of two natural numbers
HCF × LCM = product of
This two natural
For example,
is not numbers.
relationship true for more than two
HCF natural numbers.
(find it) of 24, 60
and 112 is 4 (find it) and LCM of 24, 60 and
112 is 168
Note that 4 x 1680
24 × 60 x
 Itlustrative Examples
prie factorisation method.
EXAMPLE 1. Find the HCF and LCM of 84 and 144 by
are:
SOLUTION. Prime factorisation of 84 and 144
84 = 2 x 2 x3 x 7= 22 x 3! x 7' and
144 =2 x 2 x 2 x 2 x 3 x 3 = 24 x 32
HCF (84, 144) = 22 x 3l = 4x 3 = 12 and
LCM (84, 144) = 24 x 32 x 7= 16x 9 x 7= 1008.
prime factorisation method.
EXAMPLE 2. Find the HCF and LCM of 6, 72 and 120 by using (NCERT)

SOLUTION. Prime factorisation of given numbers are:

6= 2x3 - 2l x 31.
72 = 2 x 2 x 2 x3 x 3 = 23 x 32.
120 =2x 2 x 2 x 3 x5= 23x3l x 51
common prime factors 2 and 3 in the given
Here, 2 and 3' are the smallest powers of the
numbers.
HCF (6, 72, 120) = 2 x 3 = 2x3 =6
of the prime factors 2, 3 and 5 involved in
We note that 2°, 3, 5 are the greatest powers
the given numbers.
LCM (6, 72, 120) =23 x 32 x 51 =8x 9x5=360.
method. Hence, find their LCM.
EXAMPLE 3. Find HCF of 96 and 404 by prime factorisation
SOLUTION. Prime factorisation of 96 and 404 are:
96 =2 x 2 x 2 x2x2 x3= 25 x 31 and
404 = 2 x 2 x 101 = 22 x (101)1
HCF (96,404) = 22 = 4.
numbers,
We know that HCF x LCM= product of twonatural
HCF (96, 404) × LCM (96, 404) = 96 x 404
4 x LCM (96,404) = 96 x 404
96 x 404 = 96 x 101 = 9696.
LCM (96, 404) = 4
There are 48 students in section A and
EXAMPLE 4. In a school there are tw0 sections A and B of class X.
required for the school library that books can
60 students in section B. Determine the least number of books (CBSE 2017)
be distributed equally among the students of section Aor section B. section A or
SOLUTION. Since the books are to be distributed equally among the students of As the least
well as 60.
section B, therefore,the number of books must be multiple of 48 as
numbers of books is required, it is the LCM of 48 and 60.
48 = 24 x 31, 60 = 22 x 31 x 51,
LCM of 48and 60= 24 x 3l x 5 = 240.
Hence, the required number of books = 240.
EXAMPLE 5. Dudhnath has two vessels containing 720 1nL and 405 mL of milk. Milk from these
containers is poured into glasses of equal capacity to their brim. Find the minimum number of glasses
(CBSE 2014)
that can be filled.
SOLUTION. Two vessels contain 720 mL and 405 mL of milk respectively. Since we need the
minimum number of glasses of equal capacity, so the capacity of each glass should be
maximum. Therefore, we have to find the HCF of 720 mL and 405 mL.
 Number of glasses filled fronm Ist glass 45 m
vessel 720
number of glasses filled from 2nd 45
=l6 and
Total number of glasses filled vessel=0405
= l6 + 9
45

= 25.
EXAMPLE 6. If to pOsitive
y=p'q. what cn you say about integers nd y are
heir LCM and HCE?
Is in expressible terms of
p and q LCM
Solution. Givenx=p and y =p'g a primes as X=p
where
LCM (, y) =pq' and HCF (r, ) =p. are primes. mltiple of HCP? Explain.
We note that LCM =p°q=pq' x p'q =pg² x HCF
LCM is amultiple of HCE.
Hence, the LCM of twOpositive integers is always a multiple of their HCE
EXAMPLE 7. Explain why tle umber 7x 11x 13 +13 is composite.
SOLUTION. 7 x 11 x 13 + 13= 13(7 x11 + 1) = 13 x 78
the given number is divisible by 13
than 1and the number itself
thegiven number has 13 as a facto other
Therefore, the given number is composite.
factorisation of thedenominator of rationalnumber expressed as 6.l
EXAMPLE8. Find the prine (CBSE 201:
the simplest form.
Solution. Let x= 6.12 =6.121212 ...
Multiplying both sides of (i) by 100, we get
..

100x = 612.1212
Subtracting (i) from (ii), we get
99x = 606
form.
, which is in simplest
606 202
99 33
Its denominator = 33.
Prime factorisation of denominator = 3 x 11. Funda-
using
numbers by
3 Practical difficulty in finding HCF of two natural
mental Theorem of Arithmetic notoriously difficult.
numbersin
Factorisation of a natural nunber into its prime factorsis factorisingthese
Forexample, consider 8633 and 9167. Findinggtheir HICF by tactors dl
product of primes is a difficut job. all the
that
IHowever, if you know that 8633 =89 x 97 and 9167 = 89 x 103and
Primes then
IHCF(8633, 9167) = 89
Remarks
. LCM of two (or more) natural numbers is always divisible by their HCE.
. Ifdis the HCF of two natural numbers aand b, then there exist some integers x and y such that
d= a + yb.
For example, HCF (65, 117) = 13 (obtain it), then
13 = 2 x 65 + (-1) x 117
If anatural number n is divisible by two positive integers a and b, then n is also divisible by LCM
(a, b).

Some results based on Fundamental Theorem of Arithmetic

Theorem 1. Ifais any natural number and p is a prime number such that p divides a,
then p divides a.
The generalisation of the above theorem is:
If a, n are any natural numbers and p is a prime number such that p divides a", then p
divides a.
Theorem 2. If a and b are any natural numbers and p is a prime number such that p
divides ab then p divides a or p divides bor p divides both.

lustrative Examples
EXAMPLE 1. Check whether 4" can end with the digit 0 for any natural number n or not. (CBSE 2015)
SOLUTION. We know that every natural number which ends with the digit 0 is divisible
by 5.
Therefore, if 4"ends with the digit 0, it must bedivisible by 5
’ 5 divides 4" but 5 is prime
5divides 4 (using generalisation of theorem 1)
But 5 does not divide 4.
Hence, 4" cannot end with the digit 0 for any natural number n.
EXAMPLE2.Show that 21" cannot end with the digits 0, 2, 4,6or 8 for any natural number n.
SOLUTION. We know that every natural number which ends with the digits O, 2, 4, 6, or 8 is
divisible by 2.Therefore, if 21"ends with the digits 0, 2, 4, 6or 8,it must be divisible by 2
2divides 21" but 2 is prime
’ 2 divides 21. (using generalisation of theorem 1)
But 2 does not divide 21.
Hence, 21" cannot end with the digits 0, 2, 4, 6 or 8 for any natural number n.

Exercise 1.1
Very short answerlobjective questions (1 - 8):
1. Express each number as aproduct of its prime factors:
(1) 429 (CBSE 2019) (ii) 546 (ii) 3825
(iv) 5005 (0) 7429
2. Complete the following factor tree and find the composite numbers y, x :