
Q) √2=
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Q) (c+√d)(c√d)=
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Q) 5√3+10√3=
A) Show/hide Answer
Q) The product of a number and itself is 3969, then the number is
A) Show/hide Answer
Q) Principle square root of 1296 is
A) Show/hide Answer
Q) Find a rational number between 1/2 and √1
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Q) Is √16 rational? Explain
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Q) Evaluate √2 x 4√2
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Q) Express 2025 as a product of prime factors
A) Show/hide Answer
Q) Find the LCM of 12 and 18
A) Show/hide Answer
Q) What is the HCF of two consecutive numbers ?
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Q) State some irrational numbers.
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Q) Expand log 15
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Q) Evaluate log_{2} 512
A) Show/hide Answer
Q) Evaluate log_{x} √X
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Q) Evaluate log_{√2} 32
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Q) Evaluate log_{10} 0.01
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 Real Numbers
 Introduction
 Problems on Surds
 Euclid's Division Lemma Theorem
 Fundamental Theorem Of Arithmetic
 Examples
We have studied different types of numbers in earlier classes.
We have learnt about natural
numbers, whole numbers, integers, rational numbers and irrational numbers.
Let us recall a little
bit about rational numbers and irrational numbers.
Rational numbers are numbers which can be written in the form of `p/
q` where both p and q
are integers and q 0.
They are a bigger collection than integers as there can be many rational
numbers between two integers.
All rational numbers can be written either in the form of terminating
decimals or nonterminating repeating decimals.
Numbers which cannot be expressed in the form of `p/
q`
are irrational.
These include numbers like 2, 3, 5 and mathematical quantities like π.
When these are written as decimals, they are nonterminaing, nonrecurring.
For example, 2 = 1.41421356... and π = 3.14159...
These numbers can be located on the number line.
The set of rational and irrational numbers together are called real numbers.
We can show them in the form of a diagram:
In this chapter, we will see some theorems and the different ways in which we can prove them.
We will use the theorems to explore properties of rational and irrational numbers.
Finally, we will study about a type of function called logarithms (in short logs) and see how they are useful in science and everyday life.
But before exploring real numbers a little more, let us solve some questions.
1. Which of the following rational numbers are terminating and which are nonterminating, repeating in their deimenal form?
 `2/ 5`
 `17/ 18`
 `15/ 16 `
 `7/ 40`
 `9/ 11`
2. Find any rational number between the pair of numbers given below:
 `1/ 2` and 1
 3`(1/ 3)` and 3`(2/ 3)`
 `sqrt(4/9)` and `sqrt2`
Are all integers also in real numbers? Why?
Let us explore real numbers more in this section.
We know that natural numbers are also
in real numbers.
So, we will start with them.
In earlier classes, we have seen that all natural numbers, except 1, can be written as a
product of their prime factors.
For example, 3 = 3, 6 as 2 3, 253 as 11 23 and so on.
(Remember: 1 is neither a composite nor a prime).
Do you think that there may be a composite number which is not the product of the
powers of primes? To answer this, let us factorize a natural number as an example.
We are going to use the factor tree which you all are familiar with. Let us take some large
number, say 163800, and factorize it as shown :
So we have factorized 163800 as 2x 2 x 2 x3 x3 x 5 x5 7x 13. So 163800
= `2^3x3^2x5^2x7x13`when we write it as a product of power of primes.
Try another number, 123456789.
This can be written as 32 3803 3607. Of course,
you have to check that 3803 and 3607 are primes! (Try it out for several other natural numbers
yourself.)
This leads us to a conjecture that every composite number can be written as the
product of powers of primes.
Now, let us try and look at natural numbers from the other direction. Let us take any
collection of prime numbers, say 2, 3, 7, 11 and 23.
If we multiply some or all of these numbers,
allowing them to repeat as many times as we wish, we can produce infinitely many large positive
integers.
Let us list a few :
2 x 3 x 11 = 66 7
7 x 77 = 77
Now, let us suppose your collection of primes includes all the possible primes.
What is
your guess about the size of this collection? Does it contain only a finite number of primes or
infinitely many? In fact, there are infinitely many primes.
So, if we multiply all these primes in all
possible ways, we will get an infinite collection of composite numbers.
This gives us the Fundamental Theorem of Arithmetic which says that every composite
number can be factorized as a product of primes. Actually, it says more.
It says that given any
composite number it can be factorized as a product of prime numbers in a ‘unique’ way, except
for the order in which the primes occur.
For example, when we factorize 210, we regard 2 x 3
x5 x 7 as same as 3 any other possible order in which these primes are
written.
That is, given any composite number there is one and only one way to write it as a
product of primes, as long as we are not particular about the order in which the primes occur.
Let
us now formally state this theorem.
(Fundamental Theorem of Arithmetic) : Every composite number can be expressed (factorised) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
In general, given a composite number x, we factorize it as x = p_{1}
p_{2}
...p_{n}
, where p_{1}
, p_{2}
...,
p_{n}
are primes and written in ascending order, i.e., p_{1} ≤ p_{2} ≤... ≤pn
. If we use the same primes,
we will get powers of primes.
Once we have decided that the order will be ascending, then the
way the number is factorised, is unique. For example,
163800 = `2^3x3^2x5^2x7x13`
While this is a result that is easy to state and understand, it has some very deep and
significant applications in the field of mathematics.
Let us see two examples.
You have already learnt how to find the HCF (Highest Common Factor) and LCM
(Lowest Common Multiple) of two positive integers using the Fundamental Theorem of Arithmetic
in earlier classes, without realizing it! This method is also called the prime factorization method.
Let us recall this method through the following example.
Example1: Find the HCF and LCM of 12 and 18 by the prime factorization method.
Solution:
We have 12 = 2x2x3 = `2^2`x3
18 = 2x3x3 = 2x `3^2`
LCM (12, 18) =`2^2`x`3^2`
From the example above, you might have noticed that HCF (12, 18) x LCM (12, 18)
= 12 x 18. In fact, we can verify that for any two positive integers a and b, HCF (a,b) LCM
(a, b) = a x b.
We can use this result to find the LCM of two positive integers, if we have
already found the HCF of the two positive integers.
Example2: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?
For the number `4^n`
to end with digit zero for any natural number n, it should be
divisible by 5.
This means that the prime factorisation of 4n
should contain the prime number 5.
But it is not possible because `4^n`
= `(2)2^n` so 2 is the only prime in the factorisation of 4n
. Since 5
is not present in the prime factorization, so there is no natural number n for which `4^n`
ends with the
digit zero.
1.Check whether `5^n`
can end with the digit 0 for any natural number n.
2. Find the LCM and HCF of the following integers by the prime factorization method.
(i) 12, 16 and 31
(ii) 13, 23, and 39
(iii) 3, 9 and 25
(iv) 73 and 103
(v) 236 and 657
In this section, we are going to explore when their decimal expansions of rational numbers
are terminating and when they are nonterminating, repeating.
Let us consider the following terminating decimal forms of some rational numbers:
(i) 0.375
(ii) 1.04
(iii) 0.0875
(iv) 12.5
(v) 0.00025
Now let us express them in the form of `p/q` .
We see that all terminating decimals taken by us can be expressed as rational numbers
whose denominators are powers of 10.
Let us now prime factorize the numerator and denominator
and then express in the simplest rational form :
0.375 = `375/1000`=`(375)/((10)^3)`
Do you see a pattern in the denominators? It appears that when the decimal expression
is expressed in its simplest rational form then p and q are coprime and the denominator (i.e., q)
has only powers of 2, or powers of 5, or both.
This is because the powers of 10 can only have
powers of 2 and 5 as factors.
Let Us Conclude
Even though, we have worked only with a few examples, you can see that any rational
number which has a decimal expansion that terminates can be expressed as a rational number
whose denominator is a power of 10.
The only prime factors of 10 are 2 and 5.
So, when we
simplyfy the rational number, we find that the number is of the form p
q , where the prime
factorization of q is of the form `2^n`
`5^m`, and n, m are some nonnegative integers.
We can write our result formally :
Let x be a rational number whose decimal expansion terminates.
Then x
can be expressed in the form `p/
q `, where p and q are coprime, and the prime factorization
of q is of the form `2^n`
`5^m`, where n, m are nonnegative integers.
You are probably wondering what happens the other way round.
That is, if we have a
rational number in the form `p/q` , and the prime factorization of q is of the form `2^n`
`5^m`, where n, m
are nonnegative integers, then does `p/q` have a terminating decimal expansion?
So, it seems to make sense to convert a rational number of the form `p/q` , where q is of the form `2^n`
5m, to an equivalent rational number of the form
`a/b`, where b is a power of 10.
Let us go
back to our examples above and work backwards.
Ex:
`25/2`=`(5^3)/(2xx5)`=`125/10`=12.5
So, these examples show us how we can convert a rational number of the form `p/
q` ,
where q is of the form `2^n`
`5^m`, to an equivalent rational number of the form
`a/
b` , where b is a power
of 10.
Therefore, the decimal expansion of such a rational number terminates.
We find that a
rational number of the form `p/q` , where q is a power of 10, will have terminating decimal expansion.
So, we find that the converse of theorem 12 is also true and can be formally stated as :
Theorem Let x =
`p/
q` be a rational number, such that the prime factorization of q is
of the form `2^n`
`5^m`, where n, m are nonnegative integers.
Then x has a decimal expansion
which terminates.
Non  Terminating Recurring Decimals In Rational Numbers
Let us look at the decimal conversion of
`1/7 xx 1/7 `= 0.1428571428571 ..... which is a nonterminating and recurring
decimal.
Notice, the block of digits '142857' is repeating in the quotient.
Notice that the denominator here, i.e., 7 is not of the form `2^n`
`5^m`.
Theorem: Let x =
`p/
q `be a rational number, such that the prime factorization of q is
not of the form `2^n`
`5^m`, where n, m are nonnegative integers.
Then, x has a decimal
expansion which is nonterminating repeating (recurring).
From the discussion above, we can conclude that the decimal form of every rational
number is either terminating or nonterminating repeating.
Example1:
Using the above theorems, without actual division, state whether the following
rational numbers are terminating or nonterminating, repeating decimals.
(i)`
16/
125`
(i) `16/125` = `16/(5xx5xx5)` = `16/(5^3)`is terminating decimal.
Example 2:
Write the decimal expansion of the following rational numbers without actual division.
(i)`
35/
50`
`
35/
50`
= `(7xx5)/(2xx5xx5)`
`7/(10^1)`
= 0.7
Recall, a real number ("`Q^1`" or "S") is called irrational if it cannot be written in the form`p/q` , where p and q are integers and q≠ 0.
Some examples of irrational numbers, with which you
are already familiar, are `sqrt2`,`sqrt3`,`sqrt[15]`,π,0.101100111000................................
In this section, we will prove some real numbers are irrationals with the help of the
fundamental theorem of arithmetic. We will prove that`sqrt2,sqrt3,sqrt5`,and in general ,`sqrtp` is irrational and
where p is a prime.
Before we prove that `sqrt2` is irrational, we will look at a statement, the proof of which is
based on the Fundamental Theorem of Arithmetic.
Statement1 : Let p be a prime number.
If p divides `a^2`
, (where a is a positive integer),
then p divides a.
Proof : Let a be any positive integer. Then the prime factorization of a is as follows :
a = `p_1p_2`… `p_n`
, where `p_1`
, `p_2`
, …., `p_n`
are primes, not necessarily distinct.
Therefore `a^2`
= (`p_1`
`p_2`
… `p_n`
) (`p_1`
`p_2`
… `p_n`
) = `p_1^2`
`p_
2^2`
… `p^2
n`
.
Now, here we have been given that p divides `a^2`
. Therefore, from the Fundamental
Theorem of Arithmetic, it follows that p is one of the prime factors of `a^2`
.
Also, using the uniqueness
part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of `a^2`
are
`p_1` `p_2` … `p_n`.
So p is one of `p_1`, `p_2`, … `p_n`.
Now, since p is one of `p_1p_2` … `p_n`, it divides a.
We are now ready to give a proof that `sqrt2` is irrational. We will use a technique called
proof by contradiction.
Prove that `sqrt2` is irrational.
Proof : Since we are using proof by contradiction, let us assume the contrary, i.e., `sqrt2` is rational.
If it is rational, then there must exist two integers r and s (s≠ 0) such that `sqrt2`=`r/s`.
Suppose r and s have a common factor other than 1. Then, we divide by the common
factor to get `sqrt2` = `a/b`
where a and b are coprime.
So, b`sqrt2` = a,
On squaring both sides and rearranging, we get `2b^2`
= `a^2`
. Therefore, 2 divides `a^2`
Now, by statement 1, it follows that if 2 divides `a^2`
it also divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get `2b^2`,
= `4c^2`
, that is, `b2`
= `2c^2`
.
This means that 2 divides `b^2`
, and so 2 divides b (again using statement 1 with p= 2).
Therefore, both a and b have 2 as a common factor.
But this contradicts the fact that a and b are coprime and have no common factors other
than 1.
This contradiction has arisen because of our assumption that `sqrt2` is rational.
So, we
conclude that `sqrt2` is irrational.
In general, it can be shown that d is irrational whenever d is a positive integer which is
not the square of an integer.
As such, it follows that 6, 8, 15 24 etc.
are all irrational
numbers.
In earlier classes, we mentioned that :
 The sum or difference of a rational and an irrational number is irrational and
 The product or quotient of a nonzero rational and irrational number is irrational.
Example:3
Show that 5 – `sqrt3` is irrational.
Let us assume, to the contrary, that 5 – `sqrt3` is rational.
That is, we can find coprimes a and b (b ≠ 0) such that 5 – `sqrt3` = `a/b`
Therefore 5`a/b` = `sqrt3`
Rearranging this equation,
we get
`sqrt3`= 5 `a/b`
`(5ba)/b`
Since a and b are integers, we get 5`a/b` is rational so `sqrt3` is rational.
But this contradicts the fact that `sqrt3` is irrational.
This contradiction has arisen because of our incorrect assumption that 5 – `sqrt3` is rational.
So, we conclude that 5 – `sqrt3` is irrational.
Example:4
Show that 3 `sqrt2`is irrational.
Let us assume, the contrary, that 3 `sqrt2` is rational.
i.e., we can find coprimes a and b (b ≠ 0) such that 3 `sqrt2` =`a/b`.
Rearranging, we get `sqrt2` = `a/[3b]`
Since 3, a and b are integers, `a/[3b]` is rational, and so `sqrt2` is rational
So, we conclude that 3`sqrt2`is irrational.
In this section, we are going to learn about logarithms.
Logarithms are used for all sorts
of calculations in engineering, science, business, economics and include calcuating compound
interest, exponential growth and decay, pH value in chemistry, measurement of the magnitude of
earthquakes etc.
However, before we can deal with logarithms, we need to revise the laws of exponents
as logarithms and laws of exponents are closely related.
Exponents Revisted
We know that when 81 is written as 34
it is said to be written in its exponential form.
That
is, in 81 = `3^4`
, the number 4 is the exponent or index and 3 is the base.
We say that 
81 is the 4th power of the base 3 or 81 is the 4th power of 3. Similarly, 27 = `3^3`
.
Now, suppose we want to multiply 27 and 81; one way of doing this is by directly
multiplying.
But multiplication could get long and tedious if the numbers were much larger than 81
and 27.
Can we use powers to makes our work easier?
We know that 81 = 34
We also know that 27 = 33
Using the Law of exponents `a^m xx a^n `= `a^(m+n)` ,
we can write
`27 xx 81` = `3^3 xx 3^4` = `3^7`
Now, if we had a table containing the values for the powers of 3, it would be straight
forward task to find the value of `3^7`
and obtain the result of `81 xx 27 = 2187` .
Similaly, if we want to divide 81 by 27 we can use the law of exponents
= where m > n.
`(a^m)/(a^n)` = `a^(mn)`
Then, 81 ÷ 27 = `3^4`
÷ `3^3`
= `3^1`
or simply 3
Notice that by using powers, we have changed a multiplication problem into one involving
addition and a division problem into one of subtration i.e., the addition of powers, 4 and 3 and
the subtraction of the powers 4 and 3.
We know that 10000 =104 . Here, 10 is the base and 4 is the exponent. Writing a
number in the form of a base raised to a power is known as exponentiation.
We can also write
this in another way called logarithms as
log_{10} (10000) = 4.
This is stated as "log of 10000 to the base 10 is equal to 4".
We observe that the base in the original expression becomes the base of the logarithmic
form.
Thus,
10000=`10^4`
is the same as log(1010000)= 4.
In general, if an
= x; we write it as log_{a}
x = n where a and x are positive numbers
and a ≠ 1.
Let us understand this better through examples
Example5: Write i) 64 = `8^2 `
ii) 64 = `4^3`
in logarithmic form.
Solution:: (i) The logarithmic form of 64 = `8^2`
is log`8^2`
64 = 2.
(ii) The logarithmic form of 64 = `4^3`
is log`4^3`
64 = 3.
In this example, we find that log base 8 of 64 is 2 and log base 4 of 64 is 3.
So, the
logarithms of the same number to different bases are different.
Just like we have rules or laws of exponents, we have three laws of logarithms.
We will
try to prove them in the coming sections.
1.5.3a The first law of logarithms
Suppose x = `a^n`
and y = `a^m` where a>0 and `a_1`.
Then we know that we can write:
loga^{
x} = n and loga^{
y} = m .............. (1)
Using the first law of exponents we know that `a^m x a^n`=`a^(m+n)`
So, xy = `a^m x a^n`=`a^(m+n)`
xy = `a^(m+n)`
Writing in the logarithmic form, we get
loga^{
xy} = n+m .............. (2)
But from (1), n = loga^{
x} and m=loga
^{y}.
So, log_{a}
xy = log_{a}
x + log_{a}
y
So, if we want to multiply two numbers and find the logarithm of the product, we can do
this by adding the logarithms of the two numbers.
This is the first law of logarithms.
log_{a}
xy = log_{a}
x + log_{a}
y
1.5.3b The second law of logarithms states log_{a} `x/ y` = log_{a} x  log_{a} y
1.5.3c The third law of logarithms
Let x = `a^n `so log_{a}
x = n. Suppose, we raise both sides of x = `a^n`
to the power m, we get `x^m`
= (`a^n
)^m`
Using the laws of exponentsxm
= `a^nm`
If we think of `x^m` as a single quantity, the logarithmic form of it, is
log_{a}
`x^m` = nm
log_{a}
`x^m` = m log_{a}
x (`a^n` = x so log_{a}
x = n)
This is the third law.
It states that the logarithm of a power number can be obtained by
multiplying the logarithm of the number by that power.
log_{a}
`x^m` = m log_{a}
x
Example 6:Expand log15
Solution
As you know, log_{a}
xy = log_{a}
x + log_{a}
y.
So, log15 = log (3 × 5)
= log3 + log5.
Example7:Write 2log3 + 3log5 – 5log2 as a single logarithm.
Solution:
2log3 + 3log5 – 5log2
= log`3^2`
+ log`5^3` – log`2^5`
(since in m log_{a}
x=log_{a}
`x^m`)
= log9 + log125 – log32
= log (9 × 125) – log32 (Since log_{a}x + log_{a}y = log_{a}xy)
= log1125 – log32
= log`1125/32` (Since log_{a}x – log_{a}y = log_{a}`x/y `)
Standard Bases Of A Logarithm
There are two bases which are used more commonly than any others and deserve
special mention.
They are base 10 and base e
Usually the expression log x implies that the base is 10.
In calculators, the button
marked log is preprogrammed to evaluate logarithms to base ‘10’.
For example,
log 2 = 0.301029995664…
log 3 = 0.4771212547197…
The second common base is ‘e’.
The symbol ‘e’ is called the exponential constant.
This
is an irrational number with an infinite, nonterminating nonrecurring decimal expansion.
It is
usually approximated as 2.718.
Base ‘e’ is used frequently in scientific and mathematical
applications.
Logarithms to base e or loge
, are often written simply as ‘ln’.
So, “ln x” implies
the base is ‘e’. Such logarithms are also called natural logarithms.
In calculators, the button
marked ‘ln’ gives natural logs.
For example
ln 2 = 0.6931471805599…
ln 3 = 1.0986122886681…
Applications Of Logarithms
Let us understand applications of logarithms with some examples.
Examples:The magnitude of an earthquake was defined in 1935 by Charles Richer with the
expression M=log
`I/S `; where I is the intensity of the earthquake tremor and S is the intensity of
a "threshold earthquake".
(a) If the intensity of an earthquake is 10 times the intensity of a threshold earthquake, then
what is its magnitude?
(b) If the magnitude of an earthquake registers 10 on the Richter scale, how many times is
the intensity of this earthquake to that of a threshold earthquake?
Solution
(a) Let the intensity of the earthquake be I, then we are given
I = 10 S
The magnitude of an earthquake is given by M
= log `(10S)/S`
The magnitude of the Delhi earthquake will be M
= log
`I/S`
= log 10
= 1
(b) Let x be the number of times the intensity of the earthquake to that of a threshold
earthquake.
So the intensity of earthquake is I
= xS
We know that M
= log
`I/
S`
So, the magnitude of the earthquake is M
= log
`(xs)/
s`
or M = log x
We know that M = 10
So log x = 10
and therefore x = `(10)^(10)`
A number of the form p/q, where p (may be a positive or negative integer or zero) and q (taken as a positive integer) are integers prime to each other and q not equal to zero is called a rational number or commensurable quantity.
 For example, each of the numbers 7, 3/5, 0.73, `sqrt(25)` etc. is a rational number. Evidently, the number 0 (zero) is a rational number.
A number which cannot be expressed in the form `p/q` where p and q are integers and q ≠ 0, is called an irrational number or incommensurable quantity.
 For example, each of the numbers `sqrt7, root3(3)`, [Maths Processing Error] etc. is an irrational number.
A root of a positive real quantity is called a surd if its value cannot be exactly determined.
 For example, each of the quantities `sqrt3, root3(7)`, `root4(19)`, (16)^{2/5} etc. is a surd.
 From the definition it is evident that a surd is an incommensurable quantity, although its value can be determined to any degree of accuracy. It should be noted that quantities `sqrt9, root3(64), root4(256/625)` etc.
 expressed in the form of surds are commensurable quantities and are not surds (since `sqrt9 = 3, root3(64) = 4, root4(256/625) = 4/5` etc.). In fact, any root of an algebraic expression is regarded as a surd.
 Thus, each of `sqrtm, root3(n)`, [Maths Processing Error] etc. may be regarded as a surd when the value of m ( or n or x) is not given. Note that `sqrtm` = 8 when m = 64; hence, in this case `sqrtm` does not represent a surd. Thus, `sqrtm` does not represent surd for all values of m.
All surds are irrationals but all irrational numbers are not surds. Irrational numbers like Π and e, which are not the roots of algebraic expressions, are not surds.
1. State whether the following are surds or not with reasons:
(i) `sqrt5` × `sqrt10`
(ii) `sqrt8` × `sqrt6`
(iii) `sqrt27` × `sqrt3`
(iv) `sqrt16` × `sqrt4`
(v) `5sqrt8` × `2sqrt6`
(vi) `sqrt125` × `sqrt5`
(vii) `sqrt100` × `sqrt2`
(viii) `6sqrt2` × `9sqrt3`
(ix) `sqrt120` × `sqrt45`
(x) `sqrt15` × `sqrt6`
(xi) `root3(5)` ×`root3(25)`
Solution:
 `sqrt5 xx sqrt10`
= `sqrt5 xxsqrt5 xx sqrt2`
= `5sqrt2 (sqrt5 xx sqrt5 = 5)`
= `5sqrt 2`, which is an irrational number. Hence, it is a surd.  (ii) `sqrt 8 xx sqrt6`
= `sqrt 2 xx sqrt 2 xx sqrt 2 xx sqrt 2 xx sqrt3 `
=`2 xx 2 xx sqrt3( sqrt 2 xx sqrt 2 = 2)`
= `4sqrt3`,
= `4sqrt3`, which is an irrational number. Hence, it is a surd.  (iii) `sqrt 27 xx sqrt3`
= `sqrt3 xx sqrt 3 xx sqrt 3 xx sqrt3`
= `3 × 3( sqrt 3 xx sqrt3 = 3)`
= 9, which is a rational number. Hence, it is not a surd.  (iv) `sqrt 16 × sqrt4`
= `4 xx2`
= 2 `xx`2 `xx`2
= 2 × 2 × 2
= 8, which is a rational number. Hence, it is not a surd.  (v) `5sqrt 8 × 2 sqrt6 `
= `5 (sqrt 2 xx sqrt 2 xx sqrt 2 ) xx 2( sqrt 2 xx sqrt3 )`
= `10 × 2 × 2 × sqrt3`
= `40sqrt3`, which is an irrational number. Hence, it is a surd.  (vi) `sqrt 125 × sqrt5`
= `[sqrt 5 xx sqrt 5 xx sqrt5 xx sqrt5]`
= `[ sqrt 5 xx sqrt 5 = 5]`
= 5 × 5
= 25, which is a rational number. Hence, it is not a surd.  (vii) `sqrt 100 × sqrt2`
= `[sqrt 5 xx sqrt 5 xx sqrt 2 xx sqrt 2 xx sqrt2]`
`[sqrt 5 xx sqrt 5 =5, sqrt 2 xx sqrt 2=2]`
= 5 × 2 × `sqrt2`
= `10sqrt2`, which is an irrational number. Hence, it is a surd.  (viii) `6sqrt2` × `9sqrt3`
= `6 × 9 [sqrt 2 xx sqrt3]`
= 54 × `sqrt6`
= `54sqrt6`, which is an irrational number. Hence, it is a surd.  (ix) `sqrt 120 × sqrt45`
= `[sqrt 40 xx sqrt 3 xx sqrt 9 xx sqrt5]`
= `2 × 3 × 5 × sqrt6`
= `30sqrt 6`, which is an irrational number. Hence, it is a surd.  (x) `sqrt 15 × sqrt6`
= `[sqrt 3 xx sqrt 5 xx sqrt 2 xx sqrt3]`
= `3sqrt 10`, which is an irrational number. Hence, it is a surd.  (xi) `root3(5) ×root3(25)`
= `[root3(5) xx ∛5 xx root3(5)]`
= `[(root3(5))^3]`
= 5, which is a rational number. Hence, it is not a surd.
2. Expand `(2sqrt 2  sqrt 6)(2 sqrt 2 + sqrt6)`, expressing the result in the simplest form of surd:
Solution:
`(2sqrt2  sqrt6 )(2 sqrt 2 + sqrt6)`
=`(2 sqrt 2)^2  ( sqrt 6)^2 `,[Since,`(x+ y)(x y) = x^2  y^2`]
= 8  6
= 2
3. Fill in the blanks:
(i) Surds having the same irrational factors are called ____________ surds.
(ii) `sqrt50` is a surd of order ____________.
(iii) [Maths Processing Error] × [Maths Processing Error] = ____________.
(iv) `6sqrt5` is a ____________ surd.
(v) `sqrt18` is a ____________ surd.
(vi) `2sqrt 7 + 3 sqrt7` = ____________.
(vii) The order of the surd `3root4(5)` is a ____________.
(viii) `root3(4) ×root3(2)` in the simplest form is = ____________.
Solution:
(i) similar.
(ii) 2
(iii) [Maths Processing Error], [Since, we know, 10[Maths Processing Error] = 1]
(iv) mixed
(v) pure
(vi) `5sqrt7`
(vii) 4
(viii) 2
➢ Every rational number is not a surd.
➢ Every irrational number is a surd.
➢ A root of a positive real quantity is called a surd if its value cannot he exactly determined.
➢ `sqrt9, root3(64), root4(16/81)` etc. are rational numbers but not surds because `sqrt9 = 3, root3(64) = 4, root4(16/81) = 2/3` etc.
➢ `sqrta` × `sqrta` = a ⇒`sqrt5` ×`sqrt5` = 5
➢ The sum and difference of two simple quadratic surds are said to be conjugate surds or complementary surds to each other. Thus, `(4sqrt 7 + sqrt6)` and `(4sqrt 7  sqrt 6)` are surds conjugate to each other.
➢ To express in the simplest form, denominator must be rationalized.
➢ The method of convening a given surd into a rational number on multiplication by another suitable surd is called rationalization of surds. In this case the multiplying surd is called the rationalizing factor of the given surd and conversely.
➢ If a and b are both rationals and `sqrtx` and `sqrty` are both surds and `a + sqrt x= b + sqrt y` then a = b and x= y
➢ If `a  sqrt x= b  sqrt y `then a = b and x= y.
➢ If `a + sqrtx= 0`, then a = 0 and x= 0.
➢ If `a  sqrtx= 0`, then a = 0 and x= 0.
Given positive integers a and b,there exist unique integers q and r satisfying a = bq + r,
0 ≤ r < b.
Euclid's Elements: Euclid's division algorithm is based on this lemma HCF of two given positive integers.
➢ The HCF of two positive integers a and b is the largest positive integer d that divides both a and b.
➢ Let us see how the algorithm works, through an example first. Suppose we need
➢ To find the HCF of the integers 455 and 42. We start with the larger integer, that is455.
Then we use Euclid's lemma to get
455 = 42 × 10 + 35
➢ Now consider the divisor 42 and the remainder 35, and apply the division lemma to get
42 = 35 × 1 + 7
Now consider the divisor 35 and the remainder 7, and apply the division lemma to get
➢ 35 = 7 × 5 + 0
Notice that the remainder has become zero.
➢ We claim that the HCF of 455 and 42 is the divisor at this stage, i.e., 7. verify this by listing all the factors of 455 and 42.
➢ let us state Euclid's division algorithm clearly.
To obtain the HCF of two positive integers, say c and d, with c > d, follow
Step 1 : Apply Euclid's division lemma, to c and d. So, we find whole numbers, q and r such that
c = dq + r, 0 ≤ r < d.
Step 2: If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.
Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.
➢ This algorithm works because HCF (c, d) = HCF (d, r) where the symbol HCF (c, d) denotes the HCF of c and d, etc.
Use Euclid's algorithm to find the HCF of 4052 and 12576
Solution:
Step 1: Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get
12576 = 4052 × 3 + 420
Step 2 : Since the remainder 420 ≠ 0, we apply the division lemma to 4052 and 420, to get
4052 = 420 × 9 + 272
Step 3 : We consider the new divisor 420 and the new remainder 272, and apply the division lemma to get
420 = 272 × 1 + 148
➢ We consider the new divisor 272 and the new remainder 148, and apply the division lemma to get
272 = 148 × 1 + 124
➢ We consider the new divisor 148 and the new remainder 124, and apply the division lemma to get
148 = 124 × 1 + 24
➢ We consider the new divisor 124 and the new remainder 24, and apply the division lemma to get
124 = 24 × 5 + 4
➢ We consider the new divisor 24 and the new remainder 4, and apply the division lemma to get
24 = 4 × 6 + 0
➢ The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4
➢ Notice that 4 = HCF (24, 4) = HCF (124, 24) = HCF (148, 124)
= HCF (272, 148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576, 4052).
➢ Euclid's division algorithm is not only useful for calculating the HCF of very large numbers, but also because it is one of the earliest examples of an algorithm that a computer had been programmed to carry out.