Real Number

Mathematics

Mathematics Overview Polynomial Real Number Quadratic Equation Pair of Linear Equations in Two Variables Arithmetic Progression

Trigr as Classroom

R = Real Numbers:

All rational and irrational numbers are called real numbers.

I = Integers:

All numbers from (…-3, -2, -1, 0, 1, 2, 3…) are called integers.

Q = Rational Numbers:

Real numbers of the form p/q, q ≠ 0, p, q ∈ I are rational numbers.

All integers can be expressed as rational, for example, 5 = 5/1

Decimal expansion of rational numbers terminating or non-terminating recurring.

Q’ = Irrational Numbers:

Real numbers which cannot be expressed in the form p/q and whose decimal expansions are non-terminating and non-recurring.

Roots of primes like √2, √3, √5 etc. are irrational

N = Natural Numbers:

Counting numbers are called natural numbers. N = {1, 2, 3, …}

W = Whole Numbers:

Zero along with all natural numbers are together called whole numbers. {0, 1, 2, 3,…}

Even Numbers:

Natural numbers of the form 2n are called even numbers. (2, 4, 6, …}

Odd Numbers:

Natural numbers of the form 2n -1 are called odd numbers. {1, 3, 5, …}

Why can’t we write the form as 2n+1?

Remember this!

All Natural Numbers are whole numbers.

All Whole Numbers are Integers.

All Integers are Rational Numbers.

All Rational Numbers are Real Numbers.

Prime Numbers:

The natural numbers greater than 1 which are divisible by 1 and the number itself are called prime numbers, Prime numbers have two factors i.e., 1 and the number itself For example, 2, 3, 5, 7 & 11 etc.

1 is not a prime number as it has only one factor.

Composite Numbers:

The natural numbers which are divisible by 1, itself and any other number or numbers are called composite numbers. For example, 4, 6, 8, 9, 10 etc.

Note: 1 is neither prime nor a composite number.

Euclid’s Division lemma:

Euclid’s division algorithm, as the name suggests, has to do with divisibility of integers. Stated simply, it says any positive integer a can be divided by another positive integer b in such a way that it leaves a remainder r that is smaller than b.

In other words,Given two positive integers a and b, there exist unique integers q and r satisfying a = b q + r, 0 ≤ r ≤ b.

Notice this. Each time ‘r’ is less than b. Each ‘q’ and ‘r’ is unique.

Many of you probably recognise this as the usual long division process. Although this result is quite easy to state and understand, it has many applications related to the divisibility properties of integers. We touch upon a few of them, and use it mainly to compute the HCF of two positive integers.

Application of lemma:

Euclid’s Division lemma is used to find HCF of two positive integers. Example: Find HCF of 24 and 56

Apply lemma to 24 and 56.
Take bigger number and locate ‘b’ and ‘r’. 56 = 24 × 2 + 8
Since 8 ≠ 0, consider 24 as the new dividend and 8 as the new divisor. 24 = 8 × 3 + 0
Since the remainder is zero, divisor (8) is HCF.
Although Euclid’s Division lemma is stated for only positive integers, it can be extended for all integers except zero, i.e., b ≠ 0.

II. Constructing a factor tree

Write the number as a product of prime number and a composite number
Example:
Factorize 72
Repeat the process till all the primes are obtained
∴ Prime factorization of 72 = 23 x 32

Fundamental Theorem of Arithmetic:

The Fundamental Theorem of Arithmetic, on the other hand, has to do something with multiplication of positive integers. You already know that every composite number can be expressed as a product of primes in a unique way—this important fact is the Fundamental Theorem of Arithmetic. In other words, Every composite number can be expressed as a product of primes, and this expression is unique, apart from the order in which they appear.

Again, while it is a result that is easy to state and understand, it has some very deep and significant applications in the field of mathematics. We use the Fundamental Theorem of Arithmetic for two main applications.

Applications:

I. To prove the irrationality of numbers.

The sum or difference of a rational and an irrational number is irrational.
The product or quotient of a non-zero rational number and an irrational number is irrational.

II. To determine the nature of the decimal expansion of rational numbers

Let x = pq, p and q are co-primes, be a rational number whose decimal expansion terminates. Then the prime factorization of q’ is of the form 2m5n, m and n are non-negative integers.
Let x = pq be a rational number such that the prime factorization of ‘q’ is not of the form 2m5n, ‘m’ and ‘n’ being non-negative integers, then x has a non-terminating repeating decimal expansion.

QUESTIONS: SHORT ANSWER TYPE I

Find the largest number that will divide 398, 436 and 542 leaving remainders 7, 11, and 15 respectively.

Sol. 398 – 7 = 391, 436 – 11 = 425, 542 – 15 = 527

HCF of 391, 425, 527 = 17

Express 98 as a product of its primes.

Sol. 2 723

If the HCF of 408 and 1032 2 + 480 p, then find the value of p.

Sol. HCF of 408 and 1032 is 24.

1032 × 2 + 408 × (p) = 24

408p = 24 – 2064

p = -5

HCF and LCM of two numbers are 9 and 459 respectively. If one of the numbers is 27, find the other number.

Sol. We know,

1st number × 2nd number = HCF × LCM

⇒ 27 × 2nd number = 9 × 459

⇒ 2nd number = 9 45927

= 153

Find HCF and LCM of 13 and 17 by prime factorisation method.

Sol. 13 = 1 × 13; 17 = 1 × 17

HCF = 1 and LCM = 13 × 17 = 221

Find LCM of numbers whose prime factorization are expressible as 352 and 32 72.

Sol. LCM(352, 3272) = 92549

= 11025

What are the possible values of remainder r, when a positive integer a is divided by 3?

Sol. According to Euclid’s division lemma

a = 3q + r, where 0 r < 3 and r is an integer.

Therefore, the values of r can be 0, 1 or 2.

Without actually performing the long division, find 98710500 will have terminating or non.terminating repeating decimal expansion. Give reason for your answer.

Sol. 98710500 = 47500, and 500 = 2253

So, it has terminating decimal expansion.

Can the number 4n, n be a natural number, end with the digit 0? Give a reason.

Sol. If 4n ends with 0, then it must have 5 as a factor. But, (4)n = (22)n = 22n i.e., the only

prime factor of 4n is 2. Also, we know from the fundamental theorem of arithmetic

that the prime factorisation of each number is unique. Therefore, 4n can never end

with a zero.

The product of two consecutive integers is divisible by 2. Is this statement true or false? Give a reason.

Sol. True

because n(n + 1) will always be even, as one out of the n or n+ 1 must be even.

What is the least number that is divisible by all the numbers from 1 to 10?

Sol. Clearly, the required number will be the LCM of numbers 1, 2, 3, …,10.

LCM of 1, 2, 3, …,10 = 2520

Find LCM 96 and 360 using fundamental theorem of Arithmetic.

Sol. 96 = 25 3

360 = 23 32 5

LCM = 25 32 5

= 1440

Find the largest number which divide 70 and 125 leaving the remainder 5 and 8 respectively.

Sol. It is given that by dividing 70 by the required number, there is a remainder 5.

This means 70-5=65, will be exactly divisible by the required number.

Similarly, 125-8=117, will also be exactly divisible by the required number.

Hence, the required number i sth HCF of 65 and 117.

65 = 5 13

117 = 32 13

HCF = 13

13 is the required number.

Complete the following factor tree and find the number x and y.

Sol. y = 5 13 = 65

x = 3 195 = 585

Explain why 3 5 7 + 7 is a composite number.

Sol. 3 5 7 + 7 = 7(35 +1)

= 7 16

which has more than two factors.

That is why the given number is a composite number.

“The product of three consecutive positive integers is divisible by 6”. Is this statement true or false? Justify your answer.

Sol. The given statement is true because n(n+1)(n+2) will a;ways be divisible by 6, as at least one of the factors will be divisible by 2 and at least one of the factors is divisible by 3.

There is a circular path around a sports field. Rahul takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point, at the same time and go in the same direction. After how many minutes will they meet again at the starting point?

Sol. To find the time after which they meet again at the starting point, we have to find the LCM of 18 and 12 minutes.