1.6 Real Numbers:
We have studied properties of Natural numbers, Whole Numbers, Integers, fractions,
irrational number such as_{} and _{}. We have also studied that every non zero number has a
negative number associated with it such that their sum is zero.
The combined set of rational numbers and irrational
numbers is called ‘real number’ and is denoted by R. Note
that a number can be either a rational number or an irrational number and it
can not be both.
Therefore
If Q is set of rational numbers and I_{r} is set of irrational numbers then we say Q_{} I_{r} =_{} (null set)
The relationship between various types of numbers
can be represented in a tree structure as follows:
If
N is Set of Natural numbers, W is set of Whole numbers,
Z is set of Integers ,Q is set of Rational numbers, R is set of Real numbers
and I_{r}_{ } is set of Irrational numbers
then
The above relationship can also be expressed using
a Diagram (called Venn diagram) as follows:
Notice that
N_{}W _{} Z _{} Q_{} R and I_{r}_{}R and Q_{}I_{r} = R
If a, b, c_{} R where R is the set
of real numbers then
No 
Relationships 
Name of the property 
1 
a=a 
Reflexive
property 
2 
If
a=b then b=a 
Symmetric
property 
3 
If
a=b and b=c then a=c 
Transitive
property 
4 
If
a=b then a+c =b+c, ac=bc 

5 
If
ac=bc and c _{} 0then a=b 

6 
a+b _{} R 
Closure
property of addition 
7 
ab _{} R 
Closure
property of subtraction 
8 
a*b
_{} R 
Closure
property of multiplication 
9 
a/b
_{} R provided b_{}0 
Closure
property of division 
10 
a+b = b+a 
Commutative
property of addition 
11 
a*b
= b*a 
Commutative
property of multiplication 
12 
(a+b)+c = a+(b+c) 
Associative
property of addition 
13 
a*(b*c)
= (a*b)*c 
Associative
property of multiplication 
14 
a*(b+c) = a*b + a*c, (b+c)*a = b*a+c*a 
Distributive
law 
15 
a+0
=0+a =a 
0
is additive identity 
16 
a*1=
1*a=a 
1
is multiplicative identity 
17 
a+
(a) = 0 
a,
the additive inverse exists for every
a 
18 
a*1/a
=1 provided a_{}0 
1/a, the multiplicative inverse exists for every
a 
If a, b and c are real
numbers then their order relations are:
1 
Either a=b or a<b or a>b 

2 
If a <b 
Then b>a 
3 
If a<b and b <c 
Then a<c 
4 
If a<b and for any value of c 
Then a+c
< b+c 
5 
If a<b 
Then ac< bc
if c>0 
Then ac > bc if c<0 
1.6 Problem 1: Solve (x3)/x^{2}+4
>= 5/x^{2}+4
Solution:
Solve means finding value of x
Multiplying both sides of the given statement by x^{2}+4
we get
(x3)>= 5(_{} x^{2}+4 >0)
_{}x >=5+3 (Add 3 to both sides)
_{}x >=8
Verification: By substituting value of x =8,9
notice that the the given statement is satisfied.
Lemma is a
proven statement which is used to prove another statement
We know that
Dividend = (Divisor*Quotient)+Reminder
with 0 _{}reminder< Divisor
This relationship is called
It is also stated as follows:
For any positive
integers a and b there exists unique integers q and r such that
a=b*q+r with 0 _{}r<b
Alternate way of finding HCF:
We have
learnt in earlier class finding of HCF
by factorisation method and division method.
Using
In this method, starting with small number as first divisor, we
successively divide divisor of each step by reminder of that step till reminder
becomes zero, Then the last divisor is HCF.
1.6 Problem 2: Find HCF of 305 and 793
Solution:
Divisor 
Division 
Quotient 
Reminder 
305 
305)793(2 610 
2 
183 
183 
183)305(1 183 
1 
122 
122 
122)183(1 122 
1 
61 
61 
61)122(1 122 
2 
0 
Thus 61 is HCF.
1.6 Problem 3: The length and breadth of a rectangular field
is 110m and 30m respectively. Calculate the length of the longest rod which can measure the length and breadth
of the field exactly.
Solution:
Divisor 
Division 
Quotient 
Reminder 
30 
30)110(3 90 
3 
20 
20 
20)30(1 20 
1 
10 
10 
10)20(2 20 
2 
0 
Thus HCF is 10
With a rod of length of 10M, we can measure the
length and breadth of the field exactly.
A natural number(>1) which is not
prime is called composite number
Note that 24= 2*2*2*3= 3*2*2*2 and 55=11*5= 5*11
Thus, every composite number can be expressed as product
of primes in a unique way ignoring the order of
the terms. This us called fundamental theorem of
arithmetic.
1.6 Problem 3: Find HCF and LCM of 18,81 and 108
Solution:
18=2*9= 2*3^{2}^{}
81=9*9=3^{4}
108 = 12*9 = 4*3*9 = 2^{2}*3^{3}
_{}HCF = 3^{2}=9
_{}LCM = 2^{2}*3^{4}= 4*81= 324
Note that HCF*LCM
of 3 numbers _{} product of 3 numbers
Theorem: If a prime
number p divides a^{2} then p divides a, where a is natural number > 1
Proof : Let a = p_{1}*p_{2}*p_{3}*
… *p_{n} where
p_{1},p_{2},p_{3}, … ,p_{n} are prime factors of a
Squaring both sides gives
a^{2}=( p_{1}*p_{2}*p_{3}*
… *p_{n})^{2}
Since it is given that p
divides a^{2}, p is a factor of p_{1}^{2}*p_{2}^{2}*p_{3}^{2}*
… *p_{n}^{2}
_{} p has to be one
among p_{1},p_{2},p_{3},
… ,p_{n} .
Hence the conclusion is
that p divides a
1.6 Problem 3: Prove that _{} where p is a prime
number is irrational
Solution:
Assume _{} is not rational, then _{}= a/b( simplest form)
_{}pb^{2}= a^{2}=( p_{1}*p_{2}*p_{3}* … *p_{n})^{2 }where p_{1},p_{2},p_{3}, …
,p_{n}
are prime factors of a.
This means that p divides
a( p is a factor of a) and hence
a=kp for some value of k
_{} pb^{2}=k^{2}p^{2}
_{} b^{2}= k^{2}p^{}
_{} p
is a factor of b
In the earlier step we
have concluded that p is a factor a. now we are concluding that p is a factor
of b also.
This means p divides both
a and b which is a contradiction as we
have said earlier that a/b is in
simplest form.
Note that
Rational number+
Irrational number = Irrational number( Ex. 3+_{})
Rational
number*Irrational number = Irrational number( Ex 4_{})
1.6 Summary of learning
No 
Points
studied 
1 2 3 4 
Real numbers and their
properties Relationship between real
numbers and other types of numbers. Sum of rational and irrational
number is irrational Product of rational and
irrational number is irrational 