Real Numbers

 Mind Maps

Class X - Maths: Real Numbers
Q) √2=

Q) (c+√d)(c-√d)=

Q) 5√3+10√3=

Q) The product of a number and itself is 3969, then the number is

Q) Principle square root of 1296 is

Q) Find a rational number between 1/2 and √1

Q) Is √16 rational? Explain

Q) Evaluate √2 x 4√2

Q) Express 2025 as a product of prime factors

Q) Find the LCM of 12 and 18

Q) What is the HCF of two consecutive numbers ?

Q) State some irrational numbers.

Q) Expand log 15

Q) Evaluate log2 512

Q) Evaluate logx √X

Q) Evaluate log√2 32

Q) Evaluate log10 0.01

• 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 non-terminating 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 non-terminaing, non-recurring.
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.

Exercise1.1

1. Which of the following rational numbers are terminating and which are non-terminating, repeating in their deimenal form?

1. 2/ 5
2. 17/ 18
3. 15/ 16
4. 7/ 40
5. 9/ 11

2. Find any rational number between the pair of numbers given below:

1. 1/ 2 and 1
2. 3(1/ 3) and 3(2/ 3)
3. sqrt(4/9) and sqrt2
Think-Discuss

Are all integers also in real numbers? Why?

Exploring Real Numbers

Let us explore real numbers more in this section.
We know that natural numbers are also in real numbers.

The Fundamental Theorem Of Arithmatic

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
= 2^3x3^2x5^2x7x13when 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.

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 = p1 p2 ...pn , where p1 , p2 ..., pn are primes and written in ascending order, i.e., p1 ≤ p2 ≤... ≤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 = 2^2x3
18 = 2x3x3 = 2x 3^2
LCM (12, 18) =2^2x3^2
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.

Exercise

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

Rational Numbers And Their Decimal Expansions

In this section, we are going to explore when their decimal expansions of rational numbers are terminating and when they are non-terminating, 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 non-negative 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 non-negative 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 non-negative 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 non-negative 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 non-terminating 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 non-negative integers.
Then, x has a decimal expansion which is non-terminating repeating (recurring).
From the discussion above, we can conclude that the decimal form of every rational number is either terminating or non-terminating repeating.

Example1:
Using the above theorems, without actual division, state whether the following rational numbers are terminating or non-terminating, 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 formp/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 thatsqrt2,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.
Statement-1 : 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 co-prime.
So, bsqrt2 = 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 co-prime 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 non-zero rational and irrational number is irrational.
We prove some particular cases here

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
(5b-a)/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 sqrt2is irrational.

Let us assume, the contrary, that 3 sqrt2 is rational.
i.e., we can find co-primes 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 3sqrt2is 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^(m-n)
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.

Writing Exponents As Logarithms

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 log10 (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 loga 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 log8^2
64 = 2.
(ii) The logarithmic form of 64 = 4^3
is log4^3
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.

Laws Of Logarithms

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, loga xy = loga x + loga 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.
loga xy = loga x + loga y

1.5.3b The second law of logarithms states loga x/ y = loga x - loga y

1.5.3c The third law of logarithms
Let x = a^n so loga 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 loga x^m = nm
loga x^m = m loga
x (a^n = x so loga 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. loga x^m = m loga x

Example 6:Expand log15
Solution
As you know, loga xy = loga x + loga y.
So, log15 = log (3 × 5)
= log3 + log5.

Example7:Write 2log3 + 3log5 – 5log2 as a single logarithm.
Solution:
2log3 + 3log5 – 5log2
= log3^2 + log5^3 – log2^5 (since in m loga x=loga x^m)
= log9 + log125 – log32
= log (9 × 125) – log32 (Since logax + logay = logaxy)
= log1125 – log32
= log1125/32 (Since logax – logay = logax/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 pre-programmed 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, non-terminating non-recurring 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)

Real Numbers
Rational Number
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.

Irrational 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.

Definitions of surd
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.
Note:

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.

Problems on 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 xx2 xx2
= 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

Rules of Surds

➢ 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.

Theorem (Euclid's Division Lemma)

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

The steps below

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.

Example :

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.

Remarks

1. Euclid's division lemma and algorithm are so closely interlinked that people often call former as the division algorithm also.

2. Although Euclid's Division Algorithm is stated for only positive integers, it can be extended for all integers except zero, i.e., b ≠ 0.

Euclid's division lemma/algorithm has several applications related to finding properties of numbers.

We give some examples of these applications below:

Example 1

Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer.

Solution

Let a be any positive integer and b = 2.
➢ Then, by Euclid's algorithm,a = 2q + r, for some integer q ≥ 0, and r = 0 or r = 1, because 0 ≤ r < 2. So,a = 2q or 2q + 1.If a is of the form 2q, then a is an even integer. Also, a positive integer can be either even or odd. Therefore, any positive odd integer is of the form 2q + 1.

Example 2

Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.

Solution

Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b = 4. Since 0 ≤ r < 4, the possible remainders are 0, 1, 2 and 3.

➢ That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient.
However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).
Therefore, any odd integer is of the form 4q + 1 or 4q + 3.

Theorem

sqrt2, is irrational.

Proof

Let us assume, to the contrary, that sqrt2, is rational.So, we can find integers r and 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 co-prime
➢ So, b sqrt2 = a.
➢ Squaring on both sides and rearranging, we get 2b^2 = a^2. Therefore, 2 divides a^2.
It follows 2 divides a.
So, we can write a=2c for some integer c.
substituting for a,we get 2b^2=4c^2,that is, b^2=2c^2.
Therefore 2 divides  b^2 and so 2 divides b
Therefore a and b have at least 2 as a common factor.
This contradiction has risen because of our incorrect assumption that sqrt2 is rational.
so, we conclude that sqrt2 is irrational

Example

A sweet seller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the number of that can be placed in each stack for this purpose?

Solution

This can be done by trial and error. But to do it systematically, we find HCF (420, 130).

➢ Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least.
➢ The area of the tray that is used up will be the least. Now, let us use Euclid's algorithm to find their HCF.
➢ We have : 420 = 130 × 3 + 30
130 = 30 × 4 + 10
30 = 10 × 3 + 0
So, the HCF of 420 and 130 is 10.
Therefore, the sweet seller can make stacks of 10 for both kinds of barfi.

Theorem (Fundamental Theorem of Arithmetic)

Every composite number can be expressed (factorised) as a product of primes,and this factorisation is unique,apart from the order in which the prime factors occur.

Proof

➢ The Fundamental Theorem of Arithmetic says that every composite number can be factorised as a product of primes. Actually it says more. It says that given any composite number it can be factorised as a product of prime numbers in a 'unique' way, except for the order in which the primes occur.

➢ 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. So, for example, we regard 2 × 3 × 5 × 7 as the same as 3 × 5 × 7 × 2, or any other possible order in which these primes are written.

This fact is also stated in the following form: The prime factorisation of a natural number is unique, except for the order of its factors.

➢ In general, given a composite number x, we factorise it as xx= p1p2 ... pn, where p1, p2 ,..., pn are primes and written in ascending order, i.e., p1 ≤ p2 ≤ . . . ≤ pn . If we combine the same primes, we will get powers of primes. For example, 32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 23 × 32 × 5 × 7 × 13

➢ Once we have decided that the order will be ascending, then the way the number is factorised, is unique.

Example

Consider the numbers 4n , where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero.

Solution :
If the number 4n , for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 4n would contain the prime 5.

➢ This is not possible because 4n = (2)2n ; so the only prime in the factorisation of 4n is 2.
So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4n. So, there is no natural number n for which 4n ends with the digit zero.
➢ You have already learnt how to find the HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic in earlier classes, without realising it! This method is also called the prime factorisation method.

The Fundamental Theorem of Arithmetic

Any natural number can be written as a product of its prime factors.
For instance, 2 = 2,
4 = 2 × 2,
253 = 11 × 23, and so on.

➢ Natural numbers from the other direction. That is, can any natural number be obtained by multiplying prime numbers.
➢ Let us see,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

we can produce a large collection of positive integers .

➢ 3 × 7 × 11 × 23 = 5313
➢ 2 × 3 × 7 × 11 × 23 = 10626
➢ 22 × 3 × 7 × 11 × 23 = 21252
We are going to use the factor tree with which you are all familiar. Let us take some large number, say, 32760, and factorise it as shown :

➢ we have factorised 32760 as 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 as a product of primes, i.e., 32760 = 23 × 32 × 5 × 7 × 13 as a product of powers of primes. Let us try another number, say, 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. In fact, this statement is true, and is called the Fundamental Theorem of Arithmetic because of its basic crucial importance to the study of integers. Let us now formally state this theorem.

Example 1

Find the LCM and HCF of 6 and 20 by the prime factorisation.method.
Solution :

We have : 6 = 21× 31 and 20 = 2 × 2 × 5 = 22 × 51.
You can find HCF(6, 20) = 2 and LCM(6, 20) = 2 × 2 × 3 × 5 = 60,
Note that HCF(6, 20) = 21 = Product of the smallest power of each common
prime factor in the numbers.
LCM (6, 20) = 22 × 31 × 51= Product of the greatest power of each prime factor,
involved in the numbers.
HCF(6, 20) × LCM(6, 20) = 6 × 20. In fact, we can verify that for any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × 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.

Example 2

Find the HCF of 96 and 404 by the prime factorisation method. Hence,
find their LCM.
Solution :
The prime factorisation of 96 and 404 gives :
96 = 25× 3, 404 = 22 × 101
Therefore, the HCF of these two integers is 22 = 4.

=(96**404)/4
=9696

Example 3

Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.
Solution :

We have 6 = 2 × 3,
➢ 72 = 23 × 32
➢ 120 = 23× 3 × 5

Here, 21 and 31 are the smallest powers of the common factors 2 and 3
HCF (6, 72, 120) = 21 × 31
= 2 × 3 = 6

➢ 23, 32and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively
➢ LCM (6, 72, 120) = 23 × 32 × 51 = 360

Remarks:

Notice 6* 72* 120 != HCF(6,72,120)*LCM(6,72,120).So the product of three numbers is not equal to the product of their HCF and LCM.

Formulae
• The Fundamental Theorem of Arithmetic states that every composite number can be expressed (factorized) as a product of its primes, and this factorization is unique, apart from the order in which the prime factors occur.
• If p is a prime and p divides a^2 , where a is a positive integer, then p divides a.
• Let x be a rational number whose decimal expansion terminates. Then we can express x in the form p/q , where p and q are coprime, and the prime factorization of q is of the form 2^n5^m, where n, m are non-negative integers.
• Let x = p/q be a rational number, such that the prime factorization of q is of the form2^n5^m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.
• Let x =p/q be a rational number, such that the prime factorization of q is of the form 2^n5^m, where n, m are non-negative integers. Then x has a decimal expansion which is non-terminating, repeating (recurring).
• We define loga x = n, if an = x, where a and x are positive numbers and a ≠ 1.
• laws of logarithms: (i) a^[log_a^N]=N
(ii)log_a^x + log_a^y = log_a^[xy]
(iii) log_a^x – log_a^y = log_a^[x/y]
(iv)log_a^[x^m] = m log_a^x
• A root of a positive real quantity is called a surd if its value cannot be exactly determined
• 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.
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