2. What is a Real Number?
What is a Real Number?
We have studied about many types of numbers as:
●
Natural Numbers
●
Whole Numbers
●
Rational Numbers
●
Irrational Numbers
●
Fractions
●
Decimal Numbers
All these comprised together are called real
numbers.
4. EUCLID’S DIVISION LEMMA
EUCLID’S DIVISION LEMMA
STATEMENT:
For any two given positive integers a
and b, there exists unique integers q
and r satisfying
a = bq +r; 0 ≤ r < b.
e.g. let we have two numbers 455 and
42, then 455 = 42 X 10 + 35
5. Euclid’s Division Lemma is a technique to
compute Highest Common Factor (HCF) of two
given numbers.
It is based on the fact that
Dividend = Divisor X Quotient + Remainder.
6. If we compare both we find
a = bq +r; 0 ≤ r < b.
Dividend = Divisor X Quotient + Remainder.
Here a = Dividend
b = divisor
q = Quotient
r = remainder
Hence when r becomes 0 then last divisor
is the HCF of a and b.
7. EXAMPLE
EXAMPLE
To find HCF of 135 and 225
greater number is 225 so dividing it by 135 the
smaller number
225 = 135 X 1 + 90
Now 135 is to be divided by 90
135 = 90 X 1 + 45
again 90 is to be divided by 45
90 = 45 X 2 + 0
Last divisor = 45
So HCF = 45
8. Comparison between Euclid’s
division lemma and long division
method
Comparison between Euclid’s
division lemma and long division
method
To understand the similarity between Euclid’s
Division Lemma and Long Division Method for
finding HCF let us take an example.
●
To find the HCF of 12576 and 4052
9. 12576 = 4052 X3 + 420
4052 = 420 X9 + 272
420 =272 X 1+ 148
272 = 148 X 1+ 124
148 = 124X 1+ 24
124 = 24X5 + 4
24 = 4X 6 + 0
Dividend = Divisor X quotient + Remainder
HCF = 4
Both cases are exactly same with just a difference in method
of expression.
10. Similar Example
Similar Example
Find HCF of 196 and 38220
39220 = 196 X 195 + 0
Last divisor = 196
So, H.C.F = 196
●
Find HCF of 867 and 255
867 = 255 X 3 + 102
255 = 102 X 2 + 51
102 = 51 X 2 + 0
Last Divisor = H.C.F = 51
11. Show that any positive integer is of the form 2q+1,
2q+3 or 2q+5, where q is an integer
Show that any positive integer is of the form 2q+1,
2q+3 or 2q+5, where q is an integer
Rule 1: For any integer, 2X Integer = Even
integer.
Rule 2: Even integer + 1 = Odd Integer.
Case I: Let q is an integer
2q+1= 2X some integer + 1
= Even Integer + 1= Odd Integer
Hence any positive odd integer can be expressed
as 2q+1
12. Case II: Form 2q+3
2q+3 = 2q +2+1 = 2X Some Integer +1
= Even Integer +1 = Odd Integer.
Hence any positive odd integer can be
expressed as 2q+3
Case III: Form 2q+5
2q+5 = 2q+ 4+1= 2(q+2)+1=2X Some Integer +1
= Even Integer +1 = Odd Integer
Hence any positive odd integer can be
expressed as 2q+5
13. Use Euclid’s Division Lemma to prove that square of any
positive integer is of the form 3m or 3m+1
Use Euclid’s Division Lemma to prove that square of any
positive integer is of the form 3m or 3m+1
Let m is any positive integer
●
Case I: 3m
(3m)2 = 9m2. Here for any integer m; m2 is always
a positive integer.
Hence (3m)2 = 9m2 = 3X3m2= 3X Some integer=
3m
●
Case II: 3m+1
(3m+1)2 = (3m)2+ 2X3mX1 + 1= 9m2+6m+1
= 3X (3m2+2m)+1 = 3X integer +1
=3m+1
14. The Fundamental Theorem of Arithmetic
The Fundamental Theorem of Arithmetic
Every Composite number can be expressed as
product of prime numbers.
The prime factorization of natural numbers is
unique except for the order.
e.g. 12 = 2X2X 3
27 = 3X3X3
45 = 3X3X5 etc.
Here these numbers can be expressed in these
prime factors only and we can just change the
order.
15. H.C.F. and L.C.M of two numbers
H.C.F. and L.C.M of two numbers
HCF and LCM of two numbers can be calculated
using fundamental theorem of arithmetic.
H.C.F. of two numbers = Product of smallest
power of each common prime factor in the
numbers.
e.g. HCF of 12 and 45
12 = 2X2X3 = 22X 31
45= 3X3X5 = 32X51
Common factor =3 and lowest power is 1
So HCF = 31 =3
16. LCM of 12 and 45
12 = 2X2X3 = 22X 31
45= 3X3X5 = 32X51
LCM = Highest power of common factor X
Highest power of other factors.
= 32 X 22 X 51
= 9 X 4 X 5 = 180.
17. For any two numbers a and b
a X b = HCF (a,b) X LCM (a,b)
e.g Two numbers 12 and 45
HCF = 3
LCM = 180
I number X II number = 12 X 45 = 540
HCF X LCM = 3 X 180 = 540
So, I number X II number = HCF X LCM
18. IRRATIONAL NUMBERS
IRRATIONAL NUMBERS
To prove that √2 is an irrational number.
Here we shall use contradiction method to prove
√2 as irrational.
Hence Let √2 is a rational number.
If so then √2 = p/q; where
p and q are integers, q ≠ 0
and they have no common factor other than 1
19. As √2 = p/q
So √2 X q = p
Squaring both the sides
(√2 X q)2 = p2
2 X q2 = p2.
Hence 2 is a factor of p2
so 2 is a factor of p ---------------(1)
20. Now Let p = 2c where c is any real number.
Squaring both sides; p2 = (2c)2
4c2 = p2
but p2 = 2q2
Hence 4c2 = 2q2 which implies 2 X c2 = q2.
So, 2 is a factor of q2 which implies
2 is a factor of q ---------------(2)
From equations 1 and 2
2 is a common factor of p and q
21. But we assumed that there is no common
factor between p and q other than 1. Hence our
assumption is wrong.
So we can conclude that √2 is an irrational
number.
22. Operations on Irrational Numbers
Operations on Irrational Numbers
●
Rational Number + Irrational number = Irrational
Number.
●
Rational Number - Irrational number = Irrational
Number.
●
Rational Number X Irrational number = Irrational
Number.
●
Rational Number / Irrational number = Irrational
Number.
23. To prove 3+2√5 is irrational number.
To prove 3+2√5 is irrational number.
√5 is an irrational number and 2 is a rational
number.
2 X √5 = Rational X Irrational = Irrational
Hence 2√5 is an irrational number.
3+2√5 = Rational + Irrational = Irrational
Hence 3 + 2√5 is an irrational number.
24. Terminating and Non- terminating decimals
Terminating and Non- terminating decimals
Working Rule:
Step 1: Factorize the denominator of the fractions
into its prime factors.
Step 2: Check the factors of the denominators, if
the factors contain 2 or 5 or 2 & 5 only then it is
terminating decimal.
Step 3: If the factors of the denominator contain
any number other than 2 and 5 then it is non-
terminating decimal.
25. Example
Example
13/3125
Here denominator is 3125
3125 = 5X5X5X5X5
The prime factors contain only 5 so 13/ 3125 is a
terminating decimal.
●
64/455
Here denominator is 455
455 = 5 X 7 X 13
The prime factors contain numbers other than 5 &
2 so 64/455 is a non terminating decimal.
26. PRACTICE QUESTIONS
PRACTICE QUESTIONS
1. Using Euclid’s Division lemma find the HCF of (I)
510 and 92 (ii) 336 and 54.
2. Find HCF and LCM of 26 and 91 and hence verify
I No. X II No. = HCF X LCM.
3. Given that HCF (306, 657)= 9. Find the HCF.
4. Explain why 7X11X13+13 is a composite number.
5. Identify terminating and non terminating: (I)
15/1600. (ii) 23/40. (iii) 77/210.
27. Prepared By:
Dr. Vivek Naithani TGT Maths
k.v Jagdalpur
This work is licensed under a Creative Commons
Attribution-ShareAlike 3.0 Unported License.
It makes use of the works of Mateus Machado Luna.
Prepared By:
Dr. Vivek Naithani TGT Maths
k.v Jagdalpur
This work is licensed under a Creative Commons
Attribution-ShareAlike 3.0 Unported License.
It makes use of the works of Mateus Machado Luna.
