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Real Number
CHAPTER
We are Starting from a Point but want to Make it a Circle of Infinite Radius
Basic Concepts
1.
Euclid’s division lemma
Given two positive integers a and b, there exist unique integers q and r satisfying
a = bq + r, 0  r < b.
Here a and b are called dividend and divisor respectively, q is called quotient and r the
remainder.
Euclid’s division algorithm is based on the above lemma.
2.
Method to find HCF of two positive integers a and b (a > b) using Euclid’s division algorithm
Step1. By Euclid’s algorithm, find q and r where
a = bq + r, 0  r < b.
Step2. If r = 0, the HCF is b. if r  0, apply the algorithm to b and r.
Step3. Continue the process till the remainder is zero. The divisor at this stage is the required
HCF.
3.
The Fundamental Theorem of Arithmetic
Every composite number can be expressed as a product of primes and this factorization is
unique, apart from the order in which the prime factors occurs.
4.
If a prime number p divides a2, then p also divides a where a is a positive integer.
5.
A rational number with terminating decimal expansion can always be expressed in the form
p
q
where p and q are coprime and the prime factorization of q is of the form 2m.5n where m and n
are non-negative integers and vice versa.
6.
If a rational number is of the form
p
and q is not of the form 2m.5n, then its decimal
q
representation is non-terminating but repeating.
7.
A number is called irrational if it cannot be expressed in the form
are q  0.
p
where p and q are integers
q
Example 1: Use Euclid’s division algorithm to find the HCF of 135 and 225.
Sol. HCF of 135 and 225;
Applying the Euclid’s lemma to 225 and 135, (where 225 > 135), we get 225 = (135 × 1) + 90,
since 90  0, therefore, applying the Euclid’s lemma to 135 and 90, we have 135 = (90 × 1) +
45 But 45  =

Applying Euclid’s lemma to 90 and 45, we get 90 = (45 × 2) + 0
Here, r = 0, so our procedure stops. Since, the divisor at the last step is 45.

HCF of 225 and 135 is 45.
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Example 2: Find the H.C.F. and L.C.M. of 25152 and 12156 by using the Fundamental Theorem
of Arithmetic.
Sol. The prime factors of 25152 and 12156 are given below :
25152 = 26 × 3 × 131
12156 = 22 × 3 × 1013
H.C.F. (25152, 12156) = Product of the smallest power of each common prime factors in the
number = 22 × 3 = 12.
We know that
H.C.F. (a, b) × L.C.M. (a, b) = a × b

H.C.F (25152, 12156) × L.C.M. (25152, 12156) = 25152 × 12156.
 12 L.C.M. (25152, 12156) = 25152 × 12156.
25152  12156
 L.C.M. (25152, 12156) =
12
= 25152 × 1013 = 25478976
EXERCISE 1.1
1.
Use Euclid’s division algorithm to find the HCF of :
(i) 4052 and 12576
(ii) 196 and 38220
(iii) 867 and 255
2.
Using Euclid’s Division Algorithm find the H.C.F. of 36575, 2223 and 1330.
3.
Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
4.
Find the LCM and HCF of 6 and 20 by the prime factorization method.
5.
Find the HCF of 96 and 404 by the prime factorization method. Hence, find their LCM.
6.
Find the HCF and LCM of 6, 72 and 120, using the prime factorization method.
7.
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF =
product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
8.
Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
9.
Find the L.C.M and H.C.F. of the following integers by applying the prime factorization
method.
(i) 275, 225, 175
(ii) 765, 510, 408
(iii) 19, 13, 7
10.
Find the L.C.M. and H.C.F. of the following integers by applying the prime factorization
method.
(i) 2800, 1680, 1200
(ii) 1335, 890, 534
(iii) 392, 343, 294
11.
Given that HCF (306, 657) = 9, find LCM (306, 657).
12.
Given HCF of (16, 100) = 4, write LCM of (16, 100)
13.
Given LCM of (26, 169) = 338, what is HCF of (26, 169)?
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Example 3 : Use Euclid's division lemma to show that the cube of any positive integer is
of the form 9m, 9m + 1 or 9m + 8.
Sol. Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
∴
a = 3q or 3q + 1 or 3q + 2
Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,
a3 = (3q)3 = 27q3 = 9(3q)3 = 9m,
Where m is an integer such that m = 3q3
Case 2: When a = 3q + 1,
a3 = (3q +1)3
a3= 27q3 + 27q2 + 9q + 1
a3 = 9(3q3 + 3q2 + q) + 1
a3 = 9m + 1
Where m is an integer such that m = (3q3 + 3q2 + q)
Case 3: When a = 3q + 2,
a3 = (3q +2)3
a3= 27q3 + 54q2 + 36q + 8
a3 = 9(3q3 + 6q2 + 4q) + 8
a3 = 9m + 8
Where m is an integer such that m = (3q3 + 6q2 + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
EXERCISE 1.2
1.
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.
2.
Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.
3.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some
integer.
4.
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.
5.
Prove that the square of any positive integer is of the form 4q or 4q + 1 for some integer q.
6.
Prove that the square of any positive integer is of the form 5q, 5q + 1, 5q + 4 for some integer q.
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Example 4: Check whether 6n can end with the digit 0 for any natural number n.
Sol. If any digit has last digit 10 that means it is divisible by 10 and the factors of
10 = 2 × 5.
So value 6n should be divisible by 2 and 5 both 6n is divisible by 2 but not divisible by 5 So it
can not end with 0.
Example 5: Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite
numbers.
Sol.
7 × 11 × 13 + 13
Taking 13 common, we get
13 (7 x 11 +1 )
13(77 + 1 )
13 (78)
It is product of two numbers and both numbers are more than 1 so it is a composite number.
7×6×5×4×3×2×1+5
Taking 5 common, we get
5(7 × 6 × 4 × 3 × 2 × 1 +1)
5(1008 + 1)
5(1009)
It is product of two numbers and both numbers are more than 1 so it is a composite number.
EXERCISE 1.3
1.
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
maximum number of barfis that can be placed in each stack for this purpose?
2.
An army contingent of 616 members is to march behind an army band of 32 members in
a parade. The two groups are to march in the same number of columns. What is the
maximum number of columns in which they can march?
3.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the
field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at
the same time, and go in the same direction. After how many minutes will they meet again at the
starting point?
4.
In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84 and 108,
respectively. Find the minimum number of rooms required if in each room the same number of
participants are to be seated and all of them being in the same subject.
5.
Three sets of English, Hindi and Mathematics books have to be stacked in such a way that all the
books are stored topic wise and he height of each stack is the same. The number of English
books is 96, the number of Hindi books is 240 and the number of mathematics books is 336.
Assuming that the books are of the same thickness, determine the number of stacks of English,
Hindi and Mathematics books.
6.
105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat
which will have to make many trips in order to do so. The lazy boatman has his own conditions
for transporting them. He insists that he will take the same number of animals in every trip and
they have to be of the same kind. He will naturally like to take the largest possible number each
time. Can you tell how many animals went in each trip?
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7.
15 pastries and 12 biscuit packets have been donated for a school fete. These are to be packed in
several smaller identical boxed with the same number of pastries and biscuit packets in each.
How many biscuit packets and how many pastries will each box contain?
8.
A mason has to fit a bathroom with square marble tiles of the largest possible size. The size of
the bathroom is 10 ft. by 8 ft. What would be the size inches of the tile required that has to be cut
and how many such tiles are required?
9.
What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each
case?
10. In a morning walk three persons step off together, their steps measure 80 cm, 85 cm and 90 cm
respectively. What is the minimum distance each should walk so that he can cover the distance in
complete steps?
11. Determine the number nearest to 100000 but greater than 100000 which is exactly divisible by
each of 8, 15 and 21.
12. Consider the numbers 4 n , where n is a natural number. Check whether there is any value of n for
which 4 n ends with the digit zero.
13. Check whether 5n can end with the digit 0 for any n  N.
14. Check whether 8n can end with the digit 0 for any n  N.
15. Check whether 12n can end with the digit 5 for any n  N.
16. Check whether 2n can end with digit 6 for any n  N.
17. Check whether 3n can end with the digit 1, n  1, n  N.
Example 6 : Prove that
Sol. Let

n is not a rational number, if n is not a perfect square.
n be a rational number.
n
 n
p
[p and q are co-prime]
q
p2
[Squaring both sides]
q2
 p2 = nq2
….(1)
2
 n divides p  n divides p
….(2)
2
2 2
Let p = nm  p = n m
[Squaring both sides]
2
Putting the value of p in (1), we get
n2m2 = nq2 
q2 = nm2  n divides q2  n divides q
….(3)
From (2), n divides p and from (3) n divides q. It means n is a common factor of both p and q.
This contradicts the assumption that p and q are co-prime. So, our supposition is wrong. Hence,
n cannot be a rational number.
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Example 7 : Prove that 3  2 5 is irrational.
Sol. Let take that 3 + 2√5 is a rational number.
So we can write this number as
3 + 2√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 3 both sides we get
2√5 = a/b – 3
2√5 = (a-3b)/b
Now divide by 2, we get
√5 = (a-3b)/2b
Here a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational
number But √5 is a irrational number so it contradicts.
Hence, 3 + 2√5 is a irrational number.
EXERCISE 1.4
Question based on proof of irrational number
1.
If a prime number p divides a2, then p also divides a where a is a positive integer.
2.
A number is called irrational if it cannot be expressed in the form
are q  0.
3.
Show that
2 is an irrational number.
4.
Show that
3 is not a rational number.
5.
Show that 5 –
6.
Show that 5  2 3 is an irrational number.
7.
Show that 3 5  1 is not a rational number.
8.
Show that 3 +
9.
Show that 7 3 is an irrational number.
10. Show that
p
where p and q are integers
q
3 is irrational.
2 is not a rational number.
1
is irrational.
3
11. Show that sum and product of two irrational numbers 7 +
numbers.
5 and 7 –
5 are rational
12. Prove that 3 2 is irrational.
13. Prove that
5 is an irrational number.
14. Prove that 5  3 is an irrational number.
15. Prove that
2
5 is irrational.
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
A rational number with terminating decimal expansion can always be expressed in the form
p
q
where p and q are coprime and the prime factorization of q is of the form 2m.5n where m and n
are non-negative integers and vice versa.

If a rational number is of the form
p
and q is not of the form 2m.5n, then its decimal
q
representation is non-terminating but repeating.
EXERCISE 1.5
Question based on Terminating and non-terminating decimals
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
23
(vi)
2.
129
2 257 7 5
(iv)
15
1600
(v)
29
343
111
125
(ii).
127
8
(iii).
9
455
(iv).
9
27
23
8
(ii)
125
441
(iii)
35
50
(iv)
77
210
(v)
129
2  57  717
2
Write down the decimal expansions of the following rational numbers by writing their
denominators in the form 2m × 5n, where m, n are non-negative integers:
(i)
5.
(vii)
64
455
6
(viii)
15
(iii)
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)
4.
17
8
Without actually performing long division sate whether the following rational numbers have a
terminating decimal or non-terminating recurring decimal expression :
(i)
3.
2 35 2
(ii)
3
8
(ii)
13
7
(iii)
125
80
(iv)
14588
625
(v)
129
22  57
What can you say about the prime factorizations of the denominators of the following rationals:
.
(i) 43.123456789 (ii) 43123456789
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ANSWER
Exercise 1.1
(1)
(i) H.C.F. = 4
(ii) 196
(iii) 51
(2)
1
(iv) 5 × 7 × 11 × 13
(3) (i) 22 × 5 × 7
(v) 17 × 19 × 23
(ii) 22 × 3 × 13
(iii) 32 × 52 × 17
(4) H.C.F. = 2, L.C.M. = 60
(5)
H.C.F. = 22, L.C.M. = 25 × 3 × 101
(6) H.C.F. = 2, L.C.M. = 23 × 32 × 5
(7)
(i) L.C.M. = 182, H.C.F. = 13
(iii) L.C.M. = 3024, H.C.F. = 6
(ii) L.C.M. = 23460, H.C.F. = 2
(8)
(i) L.C.M. = 420, H.C.F. = 3
(iii) L.C.M. = 1800, H.C.F. = 1
(ii) L.C.M. = 11339, H.C.F. = 1
(9)
(i) H.C.F. = 25, L.C.M. = 17325
(iii) H.C.F. = 1, L.C.M. = 1729
(ii) H.C.F. = 51, L.C.M. 6120
(10) (i) H.C.F. = 80, L.C.M. = 8400
(iii) H.C.F. = 49, L.C.M. = 8232
(ii) H.C.F. = 89, L.C.M. = 2670
(11) 22338
(13) H.C.F. = 13
(12) L.C.M. = 400
Exercise 1.2
(1)
H.C.F. = 10, So, the Sweetseller can make stack of 10 for both kind of barfi.
(2)
8 columns
(3) 36 minutes
(5)
48 stacked
(6) 35 Animal
(4) 12 room
Exercise 1.5
(1)
(i) terminating
(iv) Terminating
(vi) Terminating
(2)
(i) Terminating
(ii) Terminating
(iv) non terminating repeating
(3)
(i) Terminating
(ii) non terminating repeating
(iii) Terminating
(iv) non terminating repeating
(v) non terminating repeating
(4)
(i) Terminating
(iv) Terminating
(5)
(i) The denominators is of the form of 2m × 5n.
(ii) The denominator is not in the form of 2m × 5n.
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(ii) Terminating
(iii) non terminating repeating
(v) Non terminating repeating
(vii) non terminating repeating
(viii) Terminating
(iii) non terminating repeating
(ii) Terminating
(v) Terminating
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