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REAL NUMBERS

TOPIC:

Presenting By-Swapnil Yadav , X

rational numbers

Whole numbers are a subset of integers and counting numbers are a subset of whole numbers.

integers

whole numbers

counting numbers

If we express a new set of numbers as the quotient of two integers, we have the set of rational numbers

This means to divide one integer by another or “make a fraction”

rational numbers

There are numbers that cannot be expressed as the quotient of two integers. These are called irrational numbers.

integers

whole numbers

counting numbers 2

irrational

numbers

The rational numbers combined with the irrational numbers make up the set of real numbers.

REAL NUMBERS

Real Number System

Ir ra t ion a l N u m be rsp i, sq r t 7

N o n in te g er R a tion a l N um be rs-1 4 /5 , 9 /1 0 , 30 ,13

N eg a tiv e In te ge rs-2 0 , -1 3 , -1

Z e ro0

N a tu ra l N u m b e rs o rP os it iv e In te ge rs

1 , 16 , 1 70

W h o le Nu m b e rs0 , 2 , 5 6 , 1 98

In te ge rs-1 0 , 0 , 8

R a tio n a l N u m b e rs-3 5 , -7 /8 , 0 , 5 , 27 /11

R e al N u m b e r S ys tem-1 8 , -1 /2 , 0 , sq r t 2 , p i , 47 /10

RATIONAL NUMBERS

IRRATIONAL NUMBERS

Theorem 1.1 (Euclid’s Division Lemma) : Given positive integers a and b , there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.• Example :

Prove that every positive even integer is of the form 2m and every positive odd integer is of the form 2m + 1, where m is any integer. Solution: Let a be any positive integer and let b = 2. According to Euclid’s division lemma, there exist two unique integers m and r such that a = bm + r = 2m + r, where 0 ≤ r < 2. Thus, r = 0 or 1 If r = 0, i.e., if a = 2m, then the expression is divisible by 2. Thus, it is an even number . If r = 1, i.e., if a = 2m + 1, then the expression is not divisible by 2. Thus, it is an odd number . Thus, every positive even integer is of the form 2m and every positive odd integer is of the form 2m + 1.

Euclid’s division algorithm To obtain the HCF of two positive integers, say c and d, where c > d, follow the steps below:Step 1 : Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q andr 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 willbe the required HCF.

Example : Find the HCF of 48 and 88.

Solution : Take a = 88, b = 48 Applying Euclid’s Algorithm, we get 88 = 48 × 1 + 40 (Here, 0 ≤ 40 < 48) 48 = 40 × 1 + 8 (Here, 0 ≤ 8 < 40) 40 = 8 × 5 + 0 (Here, r = 0) Therefore , HCF (48, 88) = 8

For any positive integer a, b, HCF (a, b) × LCM (a, b) = a × b

Theorem 1.2 (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.• The Fundamental Theorem of Arithmetic says that every

composite number can be factorised as a product of primes.

• It also says that 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.

Example:Find the HCF of 300, 360 and 240 by the prime factorisation method. Solution: 300 = 22 × 3 × 52 360 = 23 × 32 × 5 240 = 24 × 3 × 5 Therefore , HCF (300, 360, 240) = 22 × 3 × 5 = 60

Theorem 1.3 : Let p be a prime number. If p divides a₂, then p divides a, where a is a positive integer.

*Proof : Let the prime factorisation of a be as follows : a = p₁ p₂ . . . Pn, where p₁, p₂, . . ., pn are primes, not necessarily distinct. Therefore, a² = (p1p₂ . . . pn)(p₁p₂ . . . pn) = p²₁ p²₂. . . p²n. Now, we are given that p divides a ₂. Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a². However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a₂ are p₁, p₂, . . ., pn. So p is one of p₁, p₂ , . . ., pn. Now, since a = p₁ p₂ . . . pn, p divides a.

Theorem 1.4 : √2 is irrational.• Proof : Let us

assume, to the contrary, that √2 is rational.So, we can find integers r and s (≠ 0) such that √ 2 =r/s.Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get √ 2 a/b,where a and b are coprime.So, b √ 2 = a.Squaring on both sides and rearranging, we get 2b² = a². Therefore, 2 divides a².Now, by Theorem 1.3, it follows that 2 divides a.So, we can write a = 2c for some integer c. Substituting for a, we get 2b² = 4c², that is, b² = 2c² This means that 2 divides b², and so 2 divides b (again using Theorem 1.3 with p = 2). Therefore, a and b have at least 2 as a common factor. But this contradicts the fact that a and b have no common factors other than 1. This contradiction has arisen because of our incorrect assumption that √ 2 is rational. So, we conclude that √ 2 is irrational.

Theorem 1.5 : 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 2n5m, where n, m are non-negative integers.

• Example

Theorem 1.6 : Let x = p/q be a rational number, such that the prime factorization of q is of the form 2n5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

• Example

Theorem 1.7 : Let x =p/q be a rational number, such that the prime factorization of q is not of the form 2n5m, where n, m are non-negative integers. Then, x has a decimal expansion which is non-terminating repeating (recurring).Thus, we can conclude that the decimal expansion of every rational number is either terminating or non-terminating repeating.

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