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.