Tiêu đề 01-Real Numbers(1).pdf ✅

REAL NUMBERS 1 CHAPTER CONTENTS • Number System • Euclid's Division Lemma or Euclid's Division Algorithm • The Fundamental Theorem of Arithmetic • The Fundamental Theorem of Arithmetic to Find H.C.F. and L.C.M. ➢ NUMBER SYSTEM  Natural Numbers : The simplest numbers are 1, 2, 3, 4....... the numbers being used in counting. These are called natural numbers.  Whole numbers : The natural numbers along with the zero form the set of whole numbers i.e. numbers 0, 1, 2, 3, 4 are whole numbers. W = {0, 1, 2, 3, 4....}  Integers : The natural numbers, their negatives and zero make up the integers. Z = {....–4, –3, –2, –1, 0, 1, 2, 3, 4,....} The set of integers contains positive numbers, negative numbers and zero.  Rational Number : (i) A rational number is a number which can be put in the form q p , where p and q are both integers and q  0. (ii) A rational number is either a terminating or non-terminating and recurring (repeating) decimal. (iii) A rational number may be positive, negative or zero.  Complex numbers : Complex numbers are imaginary numbers of the form a + ib, where a and b are real numbers and i = –1 , which is an imaginary number.  Factors : A number is a factor of another, if the former exactly divides the latter without leaving a remainder (remainder is zero) 3 and 5 are factors of 12 and 25 respectively.  Multiples : A multiple is a number which is exactly divisible by another, 36 is a multiple of 2, 3, 4, 9 and 12.  Even Numbers : Integers which are multiples of 2 are even number (i.e.) 2,4, 6, 8............... are even numbers.  Odd numbers : Integers which are not multiples of 2 are odd numbers.  Prime and composite Numbers : All natural number which cannot be divided by any number other than 1 and itself is called a prime number. By convention, 1 is not a prime number. 2, 3, 5, 7, 11, 13, 17 ............. are prime numbers. Numbers which are not prime are called composite numbers.  The Absolute Value (or modulus) of a real Number : If a is a real number, modulus a is written as |a| ; |a| is always positive or zero.It means positive value of ‘a’ whether a is positive or negative |3| = 3 and |0| = 0, Hence |a| = a ; if a = 0 or a > 0 (i.e.) a  0 |–3| = 3 = – (–3) . Hence |a| = – a when a < 0 Hence, |a| = a, if a > 0 ; |a| = – a, if a < 0
 Irrational number : (i) A number is irrational if and only if its decimal representation is non-terminating and non- repeating. e.g. 2 , 3 , ................ etc. (ii) Rational number and irrational number taken together form the set of real numbers. (iii) If a and b are two real numbers, then either (i) a > b or (ii) a = b or (iii) a < b (iv) Negative of an irrational number is an irrational number. (v) The sum of a rational number with an irrational number is always irrational. (vi) The product of a non-zero rational number with an irrational number is always an irrational number. (vii) The sum of two irrational numbers is not always an irrational number. (viii) The product of two irrational numbers is not always an irrational number. In division for all rationals of the form q p (q  0), p & q are integers, two things can happen either the remainder becomes zero or never becomes zero. Type (1) : Eg : 8 7 = 0.875 70 64 60 56 40 40 × 8 0.875 This decimal expansion 0.875 is called terminating.  If remainder is zero then decimal expansion ends (terminates) after finite number of steps. These decimal expansion of such numbers terminating. Type (2) : Eg : 3 1 = 0.333.......... = 0.3 10 9 10 9 1...... 3 0.33...... or 7 1 = 0.142857142857..... = 0.142857 10 7 30 28 20 14 60 56 40 35 50 49 1.... 7 0.14285.... In both examples remainder is never becomes zero so the decimal expansion is never ends after some or infinite steps of division. These type of decimal expansions are called non terminating. In above examples, after Ist step & 6 steps of division (respectively) we get remainder equal to dividend so decimal expansion is repeating (recurring). So these are called non terminating recurring decimal expansions. Both the above types (1 & 2) are rational numbers. Types (3) : Eg :The decimal expansion 0.327172398......is not ends any where, also there is no arrangement of digits (not repeating) so these are called non terminating not recurring. These numbers are called irrational numbers. Eg. : 0.1279312793 rational terminating 0.1279312793.... rational non terminating or 0.12793 & recurring 0.32777 rational terminating 0.327 or rational non terminating 0.32777....... & recurring
0.5361279 rational terminating 0.3712854043.... irrational non terminating non recurring 0.10100100010000 rational terminating 0.10100100010000.... irrational non terminating non recurring. Rational no. Irrational no. If remainder = 0 If remainder  0 Terminating & Non repeating Non terminating & repeating (recurring) & rem. = devidend If remainder  0 & rem.  any devidend Non terminating non repeating Real Numbers Eg : 3.6 5 18 = Eg : 0.33.... 3 1 = .. Eg : 0.671234..... = 0.3 Eg : 1.343634003908...... ❖ EXAMPLES ❖ Ex.1 Insert a rational and an irrational number between 2 and 3. Sol. If a and b are two positive rational numbers such that ab is not a perfect square of a rational number, then ab is an irrational number lying between a and b. Also, if a,b are rational numbers, then 2 a + b is a rational number between them.  A rational number between 2 and 3 is 2 2 + 3 = 2.5 An irrational number between 2 and 3 is 23 = 6 Ex.2 Find two irrational numbers between 2 and 2.5. Sol. If a and b are two distinct positive rational numbers such that ab is not a perfect square of a rational number, then ab is an irrational number lying between a and b.  Irrational number between 2 and 2.5 is 2 2.5 = 5 Similarly, irrational number between 2 and 5 is 2 5 So, required numbers are 5 and 2 5 . Ex.3 Find two irrational numbers lying between 2 and 3 . Sol. We know that, if a and b are two distinct positive irrational numbers, then ab is an irrational number lying between a and b.  Irrational number between 2 and 3 is 2  3 = 6 = 61/4 Irrational number between 2 and 61/4 is 1/ 4 2 6 = 21/4 × 61/8 . Hence required irrational number are 61/4 and 2 1/4 × 61/8 . Ex.4 Find two irrational numbers between 0.12 and 0.13. Sol. Let a = 0.12 and b = 0.13. Clearly, a and b are rational numbers such that a < b. We observe that the number a and b have a 1 in the first place of decimal. But in the second place of decimal a has a 2 and b has 3. So, we consider the numbers c = 0.1201001000100001 ...... and, d = 0.12101001000100001....... Clearly, c and d are irrational numbers such that a < c < d < b. Theorem : 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 = p1p2.....pn, where p1,p2,.....pn are primes, not necessarily distinct. Therefore, a 2 = (p1p2.....pn) (p1p2 ..... pn) = 2 2 2 p1 p ..... 2 pn . Now, we are given that p divides a2 . Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a 2 . 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. We are now ready to give a proof that 2 is irrational. The proof is based on a technique called ‘proof by contradiction’. Ex.5 Prove that (i) 2 is irrational number (ii) 3 is irrational number Similarly 5, 7, 11 ...... are irrational numbers. Sol. (i) Let us assume, to the contrary, that 2 is rational. So, we can find integers r and s ( 0) such that . s r 2 = Suppose r and s not having a common factor other than 1. Then, we divide by the common factor to get , b a 2 = 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 it following that 2 divides a. So, we can write a = 2c for some integer c. Substituting for a, we get 2b2 = 4c2 , that is, b 2 = 2c2 . This means that 2 divides b2 , and so 2 divides b (again using Theorem 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. (ii) Let us assume, to contrary, that 3 is rational. That is, we can find integers a and b ( 0) such that b a 3 = . Suppose a and b not having a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime. So, b 3 = a . Squaring on both sides, and rearranging, we get 3b2 = a2 . Therefore, a2 is divisible by 3, and by Theorem, it follows that a is also divisible by 3. So, we can write a = 3c for some integer c. Substituting for a, we get 3b2 = 9c2 , that is, b 2 = 3c2 . This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem with p = 3). Therefore, a and b have at least 3 as a common factor. But this contradicts the fact that a and b are coprime. This contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that 3 is rational. So, we conclude that 3 is irrational. Ex.6 Prove that 7 − 3 is irrational Sol. Method I : Let 7 − 3 is rational number  q p 7 − 3 = (p, q are integers, q  0)  3 q p 7 − =  q 7q p 3 − = Here p, q are integers  q 7q − p is also integer