1.7 Surds:
We have learnt to locate rational numbers on the
number line. Is it possible to locate any irrational number on a number line?
By using calculator we find that the value of _{} =1.41421356237310.
. .and _{ } = 2.23606797749979. . . .
The numbers such as_{},_{}, _{},_{} are not rational numbers. They have non recurring and non
terminating decimal points. They are called surds.
A ‘surd’ is
defined as an irrational root of a rational number and is of the form
_{} with a >0 and n>1.
‘a’ is called ‘radicand’
and ‘n’ is called ‘order’ . _{} is the root sign.
We note that every surd is an irrational number but
every irrational number need not be a surd (e.g. _{}, _{} are irrationals but not surds)
Surds can be expressed in exponential (index) forms
as shown below.
_{} = 2^{1/2}, _{}= 5^{1/3}, 8_{}= 8*7^{1/5}
Observations:
1. (2^{1/2})
*(2^{1/2}) = _{} *_{} = 2. We also know by first law of indices that (2^{1/2})
*(2^{1/2}) =2^{1/2+1/2} = 2^{1}= 2
2. (_{})* (_{})*(_{}) = _{} = 4 and by first law
of indices 4^{1/3}*4^{1/3}4^{1/3} = 4^{1/3+1/3+1/3}=
4^{3/3}= 4^{1}=4
Some times surds can be reduced to its simplest
form: For example:
_{} =(405)^{1/4} =
(81*5)^{1/4} = (3^{4}*5)^{1/4} = 3^{4*1/4}* 5^{1/4}
= 3^{1}*5^{1/4}= 3_{}
Surds which have 1 as its rational co efficient
are called ‘pure surds’. The examples are_{}, _{}, _{}.
Surds which have rational coefficients other than
1 are called ‘mixed surds’. The examples
are 5_{},8_{},4_{}(Their co –efficients are 5, 8, 4 respectively).
Surds whose order and radicands are same in their
simplest form are called ‘like surds’. The
examples are 5_{},7_{}.8_{}_{} ( Their order is two and radicand is 3).
Surds whose order or/and radicands are not the same
in their simplest form are called ‘unlike surds’.
The examples are
(i) _{} , _{}, _{} ( Though the order is
2, radicands are different)
(ii), _{} , _{} , _{}, 4 ( Though the
radicand is 4, orders are different)
From law of indices (Refer section 2.2), and
ordering of real numbers, we have following laws on surds
1.
(_{})^{n} =a
2.
_{}*_{} =_{}
3. _{}/_{} =_{}
4. If _{} =_{}then a=b
5. If _{} >_{}then a>b
6. If _{} <_{}then a<b
1.7 Problem 1 : Compare _{} and _{}
Solution:
Since the order of two surds(3 and 4) are different
we need to convert their orders in to same number.
The smallest common order is the LCM of
3 and 4 which is 12.
_{} = 4^{1/3}= 4^{4/12
}= (4^{4})^{1/12}=256^{1/12}=_{}
_{} =6^{1/4} = 6^{3/12}=
(6^{3})^{1/12}=216^{1/12}=_{}
Since 256>216 it follows that _{} >_{}
I.e. _{}> _{}
Note : For most of operation on surds, it is necessary that they
are converted first to same orders.
1.7.1 Representing
square root of numbers on number line:
We have learnt to represent integers and fractions
on a number line.
We have also studied earlier that by division
method (Section 1.5.2), we can find the value of _{} to the required number
of decimal places.
_{} to 5 decimal places is
=1.73205
With non repeating and non terminating decimal
value of_{}, we can not accurately represent _{} on a number line.
But, with the help of Pythagoras theorem we can
represent irrational number of the form _{} on a number line
accurately.
Note that
any number x, can be represented as
(_{})^{2}_{ }= x +1 = (_{})^{2}+1^{2 } ====è(1)
We know that in a right angled triangle, square of
hypotenuse^{ }is equal to sum of squares of other two sides (Proof
given in 6.11)
(Hypotenuse)^{2
} =(1^{st}
Side)^{2 }+(2^{nd} Side)^{2
}
1^{st} Side 
2^{nd} Side 
Equation 
Hypotenuse 

4 
3 
5^{2 }= 25 = 16 + 9 = 4^{2}+3^{2} 
5 

12 
5 
13^{2 }= 169 = 144+ 25 = 12^{2}+5^{2} 
13 

20 
15 
25^{2 }= 625 = 400+225 = 20^{2}+15^{2} 
25 
By observing equation (1), we can conclude that if
the sides of a right angled triangle are 1 and
_{} then the hypotenuse of that triangle = _{}
Sides of right angled triangle = 
Pythagoras theorem 
Hypotenuse= 

1,
1 
1^{2}+
1^{2}=(_{})^{2} 
_{} 

_{}, 1 
(_{})^{2}+ 1^{2}=3^{2} 
_{} 

_{},1 
(_{})^{2}+ 1^{2}=4^{2} 
_{} 

……… 
……. 
…… 

_{},1 
(_{})^{2}+ 1^{2}=(99)^{2} 
_{} 

In
general _{},1 
(_{})^{2}+ 1^{2}=(x+1)^{2} 
_{} 
Thus, if we can construct a right angled triangle
whose sides are_{}, 1 (say base of _{} and height of 1), then
the hypotenuse gives the value of_{}.
Let us find values of_{},_{}
1.7.1 Problem 1: Locate _{} on number line
Draw
a number line and mark O. Mark the point A, say at a distance of 1cm from O.
Therefore OA=1cm. At
A, draw a perpendicular to the number line. Draw an arc of 1cm from A to cut
this perpendicular line at B. Join AB. Therefore AB=1cm.Join
Thus,
we have a right angled triangle OAB whose sides OA
= AB = 1cm. _{} By Pythagoras
Theorem OB^{2 }=OA^{2}+AB^{2}
= 1^{2}+1^{2}= 1+1 =2 _{} With
_{} OP = _{} 

1.7.1 Problem 2: Locate _{} on the number line.
Solution:
At
P draw a perpendicular line to the number line. With P as center and 1cm as
radius draw an arc to cut this perpendicular line at C. Join
PC Therefore
PC=1cm. Join
OC _{} By Pythagoras
Theorem OC^{2 }=OP^{2}+PC^{2} = (_{})^{2}+1^{2} =2+1 _{} OC= _{} With
OC as radius draw an arc to cut the number line at Q. _{} OQ= _{} 

Observe the adjacent figure:
We
note that with O as center, we have concentric circles whose
radii ( In
the same way we can locate _{} for any positive
integer n after _{}has been located. This
is called Wheel of Theodorus named
after a Every
number (rational or irrational) is represented by a unique point on the
number line(refer later part of this section) Also,
every point on the number line represents a unique real number. 

1.7 Summary of learning
No 
Points studied 
1 
Surds,
laws on surds, Locating irrational
numbers on number line 
Additional points:
Representing
real numbers on the number line:
We have learnt how to represent rational numbers on
the number line in 1.1 and how to represent _{} on the number line in
this section.
We have also learnt that every irrational number
has a decimal representation.
Thus, if we could represent a decimal number on the
number line, then we can conclude that every irrational number can also be represented
on the number line.
1.7.1 Problem 2: Locate 3.1415 on the number line.
Solution:
We know that the given number 3.1415 lies in
between 3 and 4.
Draw a number line with a large scale.
Identify 3 and 4 on this line. Divide the portion
between 3 and 4 into 10 equal parts.
Identify 3.1 and 3.2 on this line. Again divide the
portion between 3.1 and 3.2 into 10 equal parts.
Identify 3.14 and 3.15 on this line. Again divide
the portion between 3.14 and 3.15 into 10 equal parts.
Identify 3.141 and 3.142 on this line. Again divide
the portion between 3.141 and 3.142 into 10 equal parts.
Identify 3.1415 on this line.
Thus we have located _{} on the number line to
an approximate value of 3.1415.
In this way we conclude that any irrational number
can be represented on the number line.
1.7.1 Problem 3: Locating_{} when x is any positive number (rational
or irrational) on the number line.
Solution:
We
know that any real number can be represented in decimal form. Let
x be the given number and we know how to represent this on the number line
(Refer 1.7.1 Problem 2) Similarly,
we can represent (x+1)/2 and (x1)/2 on the number line, as they are also
real numbers. Then
Construct a right angled triangle with base (AB) = (x1)/2 and hypotenuse
(AC) = (x+1)/2. By
Pythagoras theorem we know that AC^{2
}= AB^{2}+BC^{2} i.e. BC^{2}= AC^{2}AB^{2}
= {(x+1)/2}^{2}{(x1)/2}^{2} = {(x+1)^{2}(x1)^{2}}/4
= 4x/4 =x _{}BC = _{} With
BC as the length, we can draw an arc from 0 to cut the number line which then
represents_{} 

Addition/Subtraction
of Like Surds:
1.7.1 Problem 3: Simplify _{} +_{}_{}
Solution:
Since 50 =25*2 = 5^{2}*2
_{} = 5_{}
Since 32 =16*2 = 4^{2}*2
_{} = 4_{}
Since 72 =36*2 = 6^{2}*2
_{} = 6_{}
_{} _{} +_{}_{} = 5_{}+4_{}6_{} =(5+46)_{} = 3_{}
For multiplication and division of surds having same order follow the following rule: (Refer 1.7
Problem 1)
_{}*_{} =_{} and _{}/_{} =_{}
Note: If orders are different then find LCM
of orders to convert them to same order.
Rationalization:
The process of multiplication of two irrational
numbers to get their product as a rational number is called ‘rationalisation’.
The two irrational numbers are called rationalising factors of each other.
We have seen earlier that _{} is an irrational
number. What about the product of _{} and 3_{}?
_{}*3_{} = 3*2 = 6 which is a rational number.
Thus 3_{} is a rationalising factor of_{}. Similarly _{} is also a rationalising
factor of _{}.
Also note that 1/_{} is also a rationalizing factor of _{}.
Thus rationalising factor is not unique. In fact the product of a rational number
and the rationalising factor is also a rationalising factor.
1.7.1 Problem 3: Rationalize the denominator in 1/(_{}2)
Solution:
We are required to convert the denominator into a
rational number.
Let us multiply the numerator and the denominator
by (_{}+2)
_{}1/(_{}2) =(_{}+2)÷ {(_{}2)*(_{}+2)}
By expanding the terms in the denominator or by using
an identity (a+b)*(ab) = a^{2}b^{2} (refer to section 2.3)
We note that {(_{}2)*(_{}+2)} = 74 =3
_{}1/(_{}2) =(_{}+2)/3
Thus 1/(_{}+2) is the rationalising factor of the denominator.
If the product of two mixed surds a+_{}and a_{}(where a and b are rational numbers) is a rational number
then they are called conjugate surds and a+_{}and a_{}are conjugate to each other.