Real Numbers

 Mind Maps

Class X - Maths: Real Numbers
Q) 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 ________________?

Q) A root of a positive real quantity is called a _____________?

Q) State the LCM and HCF of 12, 15 and 21 by Prime Factorization Method.

Q) All surds are irrationals but all irrational numbers are not ____________.

Q) What is the decimal expansion of 17/8.

Q) Every______________ is a surd.

Q) State Fundamental Theorem of Arithmetic.

Q) Express 7429 as a product of its prime factors.

Q) The method of convening a given surd into a rational number on multiplication by another suitable surd is called______________.

Q) Find the LCM and HCF of 26 and 91.

Q) The HCF of 95 and 152, is?

Q) 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 _____ number

Q) 7√10 + √10 =?

Q) If p and q are co-prime numbers,then p2 and q2 are?

Q) Expand (2√2 - √6)(2√2 + √6), expressing the result in the simplest form of surd?

Q) Expand √4 x √5?

Q) Expand 5√8 x 2√6?

Q) Expand √27 x √3?

Q) Expand √2 x √7?

Q) Expand 4√3 x 7√3?

Q) 3√7 is a ____________ surd.

Q) LCM of 12, 18 is ?

Q) Conjugate of 3 - √5 is?

Q) If n is any natural number ,then 6n - 5n always ends with?

Q) Expand (2√5 + √7)(2√2 + √6), expressing the result in the simplest form of surd?

Q) Expand (11√5 + √7)(2√2 -√6), expressing the result in the simplest form of surd?

Q) Expand (11√5 - √19)(3√2 +√6), expressing the result in the simplest form of surd?

Q) Expand (4√7 - √17)(3√3 +√7) (9√2 +√5), expressing the result in the simplest form of surd?

Q) Expand (4√7+ √15)(3√3 -√7) (9√2 +√5), expressing the result in the simplest form of surd?

Q) If a = 23 3, b = 2 3 5, c = 3n and LCM(a,b,c) = 23 32 5, then n=

Q) The exponent of 2 in the prime factorisation of 144, is

Q) If HCF(26,169) = 13,then LCM(26,169) =?

Q) If n is a natural number ,then 92n - 42n is always divisible by

Q) If n= 23 ×34 ×54 ×7,then the number of consecutive zeros in n,where n is natural number,is

Q) The LCM of two numbers is 1200.Which of the following cannot be their HCF?

• Real Numbers
• Introduction
• Problems on Surds
• Euclid's Division Lemma Theorem
• Fundamental Theorem Of Arithmetic
• Examples
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
• 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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