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Real number system full

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Real number system full

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Real number system full

  1. 1. CHAPTER 2 :Real Number System The real number system evolved over time by expanding the notion of what we mean by the word “number.” At first, “number” meant something you could count, like how many sheep a farmer owns. These are called the natural numbers, or sometimes the counting numbers. 2.1 Real Number Counting Number / Natural Number N = { 1 , 2 , 3 , 4 , … } • The use of three dots at the end of the list is a common mathematical notation to indicate that the list keeps going forever. Whole Numbers • Natural Numbers together with “zero” W = { }0 , 1 , 2 , 3 , 4 , ... Integer Number (I) • Whole numbers plus negatives { }I ..., 3 , 2 , 1 , 0 , 1 , 2 , 3 ,...= − − − { }I 0 , 1 , 2 , 3 ,...= ± ± ± Rational Number (Q) a Q = x x = ; a I , b I and b 0 b   ∈ ∈ ≠    • All numbers of the form a b , where a and b are integers (but b cannot be zero) • Rational numbers include what we usually call fractions
  2. 2. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 2 • Notice that the word “rational” contains the word “ratio,” which should remind you of fractions. • The bottom of the fraction is called the denominator. Think of it as the denomination—it tells you what size fraction we are talking about: fourths, fifths, etc. • The top of the fraction is called the numerator. It tells you how many fourths, fifths, or whatever. • RESTRICTION: The denominator cannot be zero! (But the numerator can) • Fractions can be numbers smaller than 1, like 1/2 or 3/4 (called proper fractions : In the numerator is smaller than the denominator) • Fractions can be numbers bigger than 1, like one-and-a-quarter, which we could also write as 5/4 (called improper fractions : the numerator is bigger than the denominator) • One way of expressing the improper fraction 5/4 is as the mixed number 1 1 4 , which is read as “one and one-fourth.” • The step for write Mixed Number to Improper Fraction ; 1. Multiply the integer part with the bottom of the fraction part. 2. Add the result to the top of the fraction. • The step for write Improper Fraction to Mixed Number 1. Do the division to get the integer part 2. Put the remainder over the old denominator to get the fractional part •••• Equivalent Fractions Equivalent fractions are fractions that have the same value, for example 1 2 3 4 2 4 6 8 = = = etc. • All integers can also be thought of as rational numbers, with a denominator of 1: for example 3 3 1 = , 5 5 1 − − = • This means that all the previous sets of numbers (natural numbers, whole numbers, and integers) are subsets of the rational numbers. Now it might seem as though the set of rational numbers would cover every possible case, but that is not so. There are numbers that cannot be expressed as a fraction, and these numbers are called irrational because they are not rational. • Integer 2 , 5, 0 , 3 , 7− − • Fraction 2 11 , 3 17
  3. 3. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 3 • Periodic Decimal 0.2 , 0.3 , 0.47& Irrational Number (Q′ ) • Cannot be expressed as a ratio of integers. • As decimals they never repeat or terminate (rationals always do one or the other) Examples: Rational (terminates) 2 = 0.66666 3 & Rational (repeats) 5 = 0.45454545 11 & & Rational (repeats) 5 = 0.714285714285 7 & & Rational (repeats) Irrational (never repeats or terminates) Irrational (never repeats or terminates) The Real Numbers(R) •••• Rationals + Irrationals •••• All points on the number line • Or all possible distances on the number line When we put the irrational numbers together with the rational numbers, we finally have the complete set of real numbers. Any number that represents an amount of something, such as a weight, a volume, or the distance between two points, will always be a real number. The following diagram illustrates the relationships of the sets that make up the real numbers.
  4. 4. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 4 Imaginary Number a+bi ; a,b R and b 0∈ ≠ 2+3i , 4 5i− , 1 3i− , 2i Union of Real Number and Imaginary Number called Complex Number. Properties for real number system 1) Reflexive Property ; • If a R∈ then a = a • Example ; 3 = 3 , 2 = 2π π , 2x = 2x , 1+2x = 1+2x etc. 2) Symetric Property ; • If a = b then b = a • Example ; If 5 2x= then 2x = 5 If 12 4 3x= − then 4 3x = 12− If 5 2 + 3= then 2 + 3 = 5 etc. 3) Transitive Property ; • If a = b and b = c then a = c • Example ; If 2x + 1 = 10 + 12 and 10 + 12 = 22 then 2x + 1 = 22 If 2x 5 = x + 1− and x + 1 = 7 then 2x 1 = 7− etc. 4) Addition by the equal number ; • If a = b then a + c = b + c • Example ; If a = 5 then a + 3 = 5 + 3 If ( ) ( ) ( )2x + 5 = 10 then 2x + 5 + 5 = 10 + 5− − etc.
  5. 5. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 5 5) Multiplication by the equal number ; • If a = b then ac = bc • Example ; If a = 3 then a 2 = 3 2× × If ( ) ( )2x = 8 then 3 2x = 3 8 etc. 6) If a + c = b + c then a = b 7) If ac = bc and c 0≠ then a = b Addition Property for real number system 1) Closure Property • If Ra∈ , Rb∈ then Rba ∈+ 2) Commutative Property • If Ra∈ , Rb∈ then abba +=+ 3) Associative Property • If RcandRb,Ra ∈∈∈ then ( ) ( )cbacba ++=++ • Example ; 1. consider ( )532 ++ and ( ) 532 ++ Hence ( ) 1082532 =+=++ and ( ) 1055532 =+=++ so ( )532 ++ = ( ) 532 ++ etc. 4) Identity (0) • There is R0∈ for all Ra∈ such that a0aa0 =+=+ • Example ; 0 + 2 = 2 + 0 = 2 20220 =+=+ etc. 5) Inverse ( a− ) • For all Ra∈ there exists Ra ∈− such that ( ) ( ) 0aaaa =+−=−+ • Example ; 1. Consider addition inverse of 2 Hence ( ) ( ) 02222 =+−=−+ so addition inverse of 2 is 2− 2. Consider addition inverse of 5− Hence ( ) ( )( ) ( )( ) ( ) 05555 =−+−−=−−+− so addition inverse of 5− is ( )5−− , and ( ) 55 =−− etc. Cancallation Property ( ก ก)
  6. 6. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 6 Multiplication Property for real number system 1) Closure Property • If Ra∈ , Rb∈ then Rab∈ 2) Commutative Property • If Ra∈ , Rb∈ then baab = 3) Associative Property • If RcandRb,Ra ∈∈∈ then ( ) ( )bcacab = 4) Identity (1) • There is R1∈ for all Ra∈ such that ( ) ( ) a1aa1 == • Example ; ( ) ( ) 21221 == ( ) ( )( ) 21221 == etc. 5) Inverse ( 1 a− ) • For all Ra∈ and 0a ≠ there exists Ra 1 ∈− such that ( ) ( ) 1aaaa 11 == −− • multiplication inverse of a is 1 a− and a 1 a 1 =− hence ( ) ( ) 1a a 1 a 1 a =      =      so multiplication inverse of a is a 1 • Example ; 1. Consider multiplication inverse of 2 hence ( ) ( ) 12 2 1 2 1 2 =      =      so multiplication inverse of 2 is 2 1 2. Consider addition inverse of 5 1 Hence ( ) ( ) 1 5 1 55 5 1 =      =      so multiplication inverse of 5 1 is 5 6) Distributive Property • If RcandRb,Ra ∈∈∈ then ( ) bcabcba +=+ and ( ) bcaccba +=+
  7. 7. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 7 Example : Given a and b are real number a b = a + b 5⊕ − Find 1) Identity 2) Inverse of 10 Find Identity Let x is identity x a⊕ = a x + a 5− = a x = 5 Find Inverse of 10 Let y is Inverse of 10 y 10⊕ = 5 y + 10 − 5 = 5 y = 0 Example : Change the number to fraction 1. 0.252525... Let x = 0.252525...  (1) (1) × 100 ; 100x = 25.252525...  (2) (2) – (1) ; 99x = 25 x = 25 99 Therefore0.252525... = 25 99 2. 0.45326& & Let x = 0.45326326326  (1) (1)×100 ; 100x = 45.326326326  (2) (1)×100,000 ; 100,000x = 45326.326326326...  (3) (3) – (2) ; 99900x = 45326 45− 99900x = 45281 x = 45281 99900 Therefore0.45326326326... = 45281 99900
  8. 8. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 8 3. 2.354& Let x = 2.35444...  (1) (1)×100 ; 100x = 235.444...  (2) (1)×1,000 ; 1,000x = 2354.444...  (3) (3) – (2) ; 900x = 2119 x = 2119 900 Therefore 2.354& = 2119 900 Example : Given a * b = a + b + 4 where a , b ∈ R. Find the property of operation * And find inverse of 5 1. Closure Property If a ∈ R and b ∈ R then a * b = a + b + 4 ∈ R 5 ∈ R and 7 ∈ R , 5 7∗ = 5 7 4+ + = 16 ∈ R 2. Commutative Property a * b = a + b + 4 = b + a + 4 = b * a 3. Associative Property (a * b) * c = (a + b + 4) * c a * (b * c) = a * (b + c + 4) = (a + b + 4) + c + 4 = a + (b + c + 4) +4 = a + b + 4 + c + 4 = a + b + c + 4 + 4 = a + b + c + 8 = a + b + c + 8 (a * b) * c = a * (b * c) 4. Identity Let x is identity x * a = a x + a + 4 = a x = 4− Therefore identity is 4−
  9. 9. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 9 5. Inverse Let y is inverse of 5 y * 5 = 4− ( identity) y + 5 + 4 = 4− y = 13− Therefore inverse of 5 is 13− 2.2 Fundamental Theorem For Real Number System Theorem 2.2.1 ; cancallation property for addition (กLMNOPQQกRSLTMOUกLMUVก) Given a , b , c ∈ R (1) if a + c = b + c then a = b ( 2) If a + b = a + c then b = c Theorem 2.2.2 ; cancallation property for multiplication (กLMNOPQQกRSLTMOUกLMWXY) Given a , b , c ∈ R (1) if ac = bc and c ≠ 0 then a = b ( 2) If ab = ac and a ≠ 0 then b = c Theorem 2.2.3 ; multiplied by 0 (กLMWXYPZV[ 0) If a ∈ R then ( ) ( )= =a 0 0 a 0 Theorem 2.2.4 ; multiplied by −1 (กLMWXYPZV[ − 1) If a ∈ R then a(−1) = (−1)a = −a Theorem 2.2.5 ; 89:;<=>?@กBC 0 Given a , b∈R , If ab = 0 then a = 0 or b = 0 Theorem 2.2.6 ; Given a ∈R , If a ≠ 0 then 1 a− ≠ 0
  10. 10. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 10 Theorem 2.2.7 ; ( inverse of the inverse ) (1) If a ∈R then ( ) aa =−− (2) If a ∈R and a ≠ 0 then ( ) aa 11 = −− Theorem 2.2.8 ; Given a , b ∈R , where a ≠ 0 and b ≠ 0 then ( ) 11111 baabab −−−−− == Theorem 2.2.9 ; Given a , b ∈R then ; (1) ( )ba− = ab− (2) ( )ba − = ab− (3) ( )( )ba −− = ab Subtraction And Division For Real Number Subtraction for real number Definetion ; Given a , b ∈ R then ( )− −a b = a + b Theorem 2.2.10 ; distributive property for subtraction (กLM]ก]^RSLTMOUกLM_U) Given a , b , c ∈R then ; (1) ( )−a b c = −ab ac (2) ( )−a b c = −ac bc (3) ( )( )− −a b c = −ab + ac Theorem 2.2.11 ; cancallation property for subtraction (กLMNOPQQกRSLTMOUกLM_U) Given a , b , c ∈ R (1) if − −a b = a c then b = c (2) If − −a b = c b then a = c
  11. 11. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 11 Devision for real number Definetion ; Given a , b ∈ R , where b ≠ 0 then ( )−1a = a b b Theorem 2.2.12 ; Given a , b , c ∈R , where b ≠ 0 and c ≠ 0 then ( ) − − − −− = =1 1 1 11 ab b a a b Theorem 2.2.13 ; Given a , b , c ∈R , where b ≠ 0 and then ( )a ca c = b b Theorem 2.2.14 ; Given a , b ∈R , where b ≠ 0 then − − − a a a = = b b b Theorem 2.2.15 ; Given a , b , c ∈R , where b ≠ 0 and c ≠ 0 then ( )a ab = c bc Theorem 2.2.16 ; Given a , b , c , d ∈ R , where b ≠ 0 and d ≠ 0 then + a c ad + bc = b d bd Theorem 2.2.17 ; Given a , b , c , d ∈R , where b ≠ 0 and d ≠ 0 then − − a c ad bc = b d bd
  12. 12. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 12 Theorem 2.2.18 ; Given a , b , c , d ∈R , where b ≠ 0 and d ≠ 0 then ( )( )a c ac = b d bd Theorem 2.2.19 ; Given a , b ∈R , where a ≠ 0 and b ≠ 0 then ( ) −1 a b = b a Theorem 2.2.20 ; Given a , b , c , d ∈R , where b ≠ 0 , c ≠ 0 and d ≠ 0 then ( ) ( ) a adb = c bc d Theorem 2.2.21 ; Given a , b , c ∈ R , where b ≠ 0 then (1) If a = c b then a = bc (2) If a = bc then a = c b Theorem 2.2.22 ; Given a , b , c ∈ R , where b ≠ 0 then a = c b if and only if a = bc Theorem 2.2.23 ; Given a , b , c ∈R , where b ≠ 0 and c ≠ 0 then ( ) a ac = b b c
  13. 13. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 13 2.3 Solving polynomial equations of one variable polynomial equations of one variable is an equation of the form − −+ + + + =n n 1 n n 1 1 0a x a x ... a x a 0 , where na , −n 1a , … , 1a , 0a are real numbers , which are constants , x is variable and n is an integer or zero. Solving equations by factorization(ก@GHกIJKก@GLMNก@GHNกOBPQGRกSC) Example 1 : Find the solution set of equation − − + =3 2 4x 3x 64x 48 0 Solution Since − − +3 2 4x 3x 64x 48 = 0 ( ) ( )− − −3 2 4x 3x 64x 48 = 0 ( ) ( )− − −2 x 4x 3 16 4x 3 = 0 ( )( )− −2 x 16 4x 3 = 0 ( )( )( )+ − −x 4 x 4 4x 3 = 0 So −x 4 = 0 or +x 4 = 0 or −4x 3 = 0 So x = 4 or x = −4 or x = 3 4 The solution set is { }− 3 4 , , 4 4 Solving equations by remainder theorem (ก@GHกIJKก@GLMN>TUVWC>=XU=Y9ZS) P(x) is polynomial − −+ + + + =n n 1 n n 1 1 0a x a x ... a x a 0 , which n is a positive integer , na , −n 1a , … , 1a , 0a are real numbers , which na ≠ 0 . If division p(x) by −x c , which c is real number then remainder is equal to p(c) Example 2 : Find the remainder when division + −3 9x 4x 1 by − 1 x 2 Solution Given p(x) = + −3 9x 4x 1 By remainder theorem ; when division p(x) by − 1 x 2 then the remainder is p( )1 2
  14. 14. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 14 p( )1 2 = ( ) ( )+ − 3 1 1 9 4 1 2 2 = 17 8 Therefore the remainder is 17 8 Factor Theorem (>TUVWC>OBPQGRกSC) P(x) is polynomial − −+ + + + =n n 1 n n 1 1 0a x a x ... a x a 0 , which n is a positive integer , na , −n 1a , … , 1a , 0a are real numbers , which na ≠ 0 . −x c is factor of p(x) if and only if p(c) = 0 RM`abOcdNQdกLM[กNOVaMeกQUbQ^fT`dLg p(x) hP[ijZklmnoUkpqmpT_rQidกMYodoc kSLPO^doc 1. TLNOVaMeกQU c bQ^ 0a koskSLiTZ p(c) = 0 2. dSL cx − taTLMfT`dLg p(x) u_TLM]epavdfT`dLgPoกMoNsSLกVwLPoกMobQ^ p(x) Q[Xw 1 3. xZLu_TLMidbZQ 2 [O^goPoกMoRX^กVwLRQ^_eRLgLMx[กNOVaMeกQUNwQtatPZQoกกy[ก NOVaMeกQUNLgbOcdNQdidbZQ 1 _e 2 NwxZLu_TLMgoPoกMoRQ^]eijZVz{o[กNOVaMeกQU NLgkospMo[dgL_ZV Example 3 : Show that −x 2 is factor of − + +3 2 x 5x 2x 8 Solution Given p(x) = − + +3 2 x 5x 2x 8 p(2) = − + +3 2 2 5(2) 2(2) 8 = 8 20 4 8− + + = 0 So −x 2 is factor of − + +3 2 x 5x 2x 8 From the example , −x 2 is factor of − + +3 2 x 5x 2x 8. When divide − + +3 2 x 5x 2x 8 by −x 2 ,
  15. 15. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 15 2 3 2 3 2 2 2 x 3x 4 x 2 x 5x 2x 8 x 2x 3x 2x 3x 6x 4x 8 4x 8 − − − − + + − − + − + − + − + So − + +3 2 x 5x 2x 8 = ( )( )2 x 2 x 3x 4− − − And 2 x 3x 4− − = ( )( )x 1 x 4+ − Therefore − + +3 2 x 5x 2x 8= ( )( )( )x 2 x 1 x 4− + − Rational Factor Theorem (>TUVWC>OBPQGRกSCOGGกNR) P(x) is polynomial − −+ + + + =n n 1 n n 1 1 0a x a x ... a x a 0 , where n is a positive integer , na , −n 1a , … , 1a , 0a are integer , where na ≠ 0 . If m k x − is factor of p(x), where m and n are integer , where m ≠ 0 and gcd. of m and k is 1 then na divided by m perfect and 0a divided by k perfect . Note : gcd = greatest common divisor ( T.M.g. = TLMMwVggLก) RM`abOcdNQdกLM[กNOVaMeกQUbQ^fT`dLg p(x) PO^doc 1. TL m k hP[fz]LMYL m _e k ]LกNOVaMeกQUbQ^ na _e 0a NLg_SLPOU _e T.M.g. bQ^ m _e k pkwLกOU 1 2. kPRQUVwL p( m k ) pkwLกOU 0 TMrQtgw xZL p( m k ) = 0 ]etPZ m k x − pavdNOVaMeกQU bQ^ p(x) idกMYokostgwgo m k koskSLiTZ p( m k ) = 0 RP^VwL fT`dLg p(x) tgwgo NOVaMeกQUkospavdfT`dLgPoกMoTd|s^idMXa m k x −
  16. 16. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 16 3. dSL m k x − }|s^pavdNOVaMeกQUbQ^fT`dLg p(x) taTLMfT`dLg p(x) u_TLM]epavd fT`dLgPoกMoNsSLกVwLPoกMobQ^ p(x) 4. xZLu_TLMidbZQ 3 [O^goPoกMoRX^กVwLRQ^_eRLgLMx[กNOVaMeกQUNwQtatPZQoก กy[กNOV aMeกQUNLgbOcdNQdidbZQ 1 , 2 _e 3 NwxZLu_TLMgoPoกMoRQ^]eijZVz{o[กNOVaMeกQU NLgkospMo[dgL_ZV Example 4 : Factorise of 35x16x12x 23 −−+ Solution Given p(x) = 35x16x12x 23 −−+ Since integers which divide 3− perfect are 1± , 3± (]SLdVdkosTLM 3− _^NOV) And integers which divide 12 perfect are 1± , 2± , 3± , 4± , 6± , 12± (]SLdVdkosTLM 12 _^NOV) So k m which p( m k ) = 0 is the number of the following numbers 1± , 3± , 1 2 ± , 3 2 ± , 1 3 ± , 1 4 ± , 3 4 ± , 1 6 ± , 1 12 ± ( ]SLdVdpT_wLdoc NOVpqm WrQ]SLdVdkospavdNOVaMeกQUbQ^ 3− _eNOVRwVd WrQ ]SLdVd kospavdNOVaMeกQUbQ^ 12 hP[kos T.M.g. bQ^NOVpqm_eNOVRwVd WrQ 1) Consider p( 1 2 ) = 3 21 1 1 12( ) 16( ) 5( ) 3 2 2 2 + − − = 12 16 5 3 8 4 2 + − − = 0 So 1 x 2 − is a factor of p(x) Devide 35x16x12x 23 −−+ by 1 x 2 − then quotient is 2 12x 22x 6+ + So 35x16x12x 23 −−+ = 21 (x )(12x 22x 6) 2 − + + = 21 (x )(2)(6x 11x 3) 2 − + + = 2 (2x 1)(6x 11x 3)− + + = (2x 1)(3x 1)(2x 3)− + +
  17. 17. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 17 Example 5 : Solve the equation 35x16x12x 23 −−+ Solution Given p(x) = 35x16x12x 23 −−+ From example 4 ; 35x16x12x 23 −−+ = (2x 1)(3x 1)(2x 3)− + + Since 35x16x12x 23 −−+ = 0 (2x 1)(3x 1)(2x 3)− + + = 0 So 2x 1− = 0 or 3x 1+ = 0 or 2x 3+ = 0 x = 1 2 or x = 1 3 − or x = 3 2 − Example 6 : If 2 is real number .Prove that 2 is irrational number. Prove 2 is the solution of equation x = 2 So 2 is the solution of equation 2 x = 2 or 2 x 2 = 0−  (1) From Rational Factor Theorem ; the solution of equation (1) is in set { 1± , 2± } but 2 ∉ { 1± , 2± }. So 2 is not rational number. And since 2 is real number so 2 is irrational number. Example 7 : If 2 5+ is real number .Prove that 2 5+ is irrational number. Prove 2 5+ is the solution of x = 2 5+ So 2 5+ is the solution of x 2 = 5− Or 2 x 2 2 x 2 = 5− + ([กกSL_O^RQ^kOc^RQ^bZL^) 2 x 3 = 2 2x− 4 2 2 x 6x 9 = 8x− + ([กกSL_O^RQ^kOc^RQ^bZL^) 4 2 x 14x 9 = 0− +  (1) From Rational Factor Theorem ; the solution of equation (1) is in set { 1± , 3± , 9± } but 2 5+ ∉ { 1± , 3± , 9± }. So 2 5+ is not rational number. Since 2 5+ is real number so 2 5+ is irrational number.
  18. 18. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 18 2.4 Inequality Definition ; 1) a > b if and only if −a b > 0 2) a < b if and only if −a b < 0 3) ≥a b means a > b or a = b 4) a b≤ means a < b or a = b 5) a < b < c means a < b and b < c 6) ≤ ≤a b c means ≤a b and ≤b c 7) < ≤a b c means <a b and ≤b c 8) ≤a b < c means ≤a b and b < c 9) ≠a b if and only if ( )− 2 a b > 0 Properties of inequality I1 : Trichotomy Property (RgUONztNMVz€LW) If a and b are real numbers then a = b , a < b and a > b , it is actually only one. I2 : Transitive Property (RgUONzกLMxwL[kQP) Given a , b and c are real numbers. (1) a < b and b < c then a < c (2) ≤a b and ≤b c then ≤a c (3) a > b and b > c then a > c (4) ≥a b and ≥b c then ≥a c I3 : property of number when compared with 0 (RgUONzbQ^]SLdVdpgrsQpaMo[Upko[UกOU 0) Given a is real number. (1) a is positive number if and only if a > 0. (2) a is negative number if and only if a < 0. (3) a is not positive number if and only if ≤a 0. (4) a is not negative number if and only if ≥a 0.
  19. 19. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 19 I4 : RgUONzกMtgwpavd]SLdVd_U_eกLMpavd]SLdVdUVก (1) If a is real number then ≥2 a 0 (2) If a is real number and a ≠ 0 then >2 a 0 I5 : Addition by the equal number (RgUONzกLMUVกPZV[]SLdVdpkwLกOd) Given a , b and c are real numbers. (1) If a < b then a + c < b + c (2) If ≤a b then ≤a + c b + c (3) If a > b then a + c > b + c (4) If ≥a b then ≥a + c b + c I6 : cancallation property for addition (กLMNOPQQกRSLTMOUกLMUVก) Given a , b and c are real numbers. (1) If a + c < b + c then a < b (2) If ≤a + c b + c then ≤a b (3) If a + c > b + c then a > b (4) If ≥a + c b + c then ≥a b I7 : RgUONzกLMWXYPZV[]SLdVdpkwLกOdkospavd]SLdVdUVก Given a , b and c are real numbers. (1) If a < b and c > o then ac < bc (2) If ≤a b c > o then ≤ac bc (3) If a > b c > o then ac > bc (4) If ≥a b c > o then ≥ac bc I8 : RgUONzกLMWXYPZV[]SLdVdpkwLกOdkospavd]SLdVd_U Given a , b and c are real numbers. (1) If a < b and c < o then ac > bc (2) If ≤a b c < o then ≥ac bc (3) If a > b c < o then ac < bc (4) If ≥a b c < o then ≤ac bc
  20. 20. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 20 I9 : RgUONzกLMNOPQQกbQ^]SLdVdUVกbQ^กLMWXY Given a , b and c are real numbers. (1) If ac < bc and c > o then a < b (2) If ≤ac bc c > o then ≤a b (3) If ac > bc c > o then a > b (4) If ≥ac bc c > o then ≥a b I10 : RgUONzกLMNOPQQกbQ^]SLdVd_UbQ^กLMWXY Given a , b and c are real numbers. (1) If ac < bc and c < o then a > b (2) If ≤ac bc c < o then ≥a b (3) If ac > bc c < o then a < b (4) If ≥ac bc c < o then ≤a b More Summaries 1) No Reflexive (tgwgoRgUONzกLMpkwLกOd) 2) No Symmetric (tgwgoRgUONzRggLNM) 3) If a b< and c d< then a c b d+ < + 4) If a b< and c d< then a d b c− < − 5) If 0 a b< < and 0 c d< < then ac bd< 6) 0a b< < and 0c d< < then ac bd> 7) If 0 a b< < and 0 c d< < then a b d c < 8) If 0a b< < and 0c d< < then a b d c > 9) If 0 a b< < then 2 2 a b< 10) If 0a b< < then 2 2 a b> 11) If 0 a b< < then 1 1 a b > 12) If 0a b< < then 1 1 a b > 13) If 0ab > and a b< then 1 1 a b >
  21. 21. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 21 2.5 Interval Given a R∈ , b R∈ and a b< 1) ( ),a b = { }x a x b< < 2) [ ],a b = { }x a x b≤ ≤ 3) ( ],a b = { }x a x b< ≤ 4) [ ),a b = { }a x<bx ≤ 5) ( ),a ∞ = { }x x a> 6) [ ),a ∞ = { }x x a≥ 7) ( ),a−∞ = { }x x a< 8) ( ],a−∞ = { }x x a≤ 9) ( ),−∞ ∞ = R Example 1 : Given A = ( )1,7 , B = [ ]3,4− , C = [ )0,6 1) A B∪ = [ )3,7− 2) B C∪ = [ )3,6− 3) A B C∪ ∪ = [ )3,7− 4) A B∩ = ( ]1,4 5) B C∩ = [ ]0,4 6) A B C∩ ∩ = ( ]1,4 7) ( )A B C∪ ∩ = [ )0,6 8) ( )A B C∩ ∪ = ( )1,6 9) A B− = ( )4,7 10) B C− [ )3,0− 11) ( )A B C− − = [ )6,7 12) ( )A B C∪ − j [ ) [ )3,0 6,7− ∪ 0
  22. 22. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 22 Example 2 : Given 1 ,2nA n n   = −    ; n I∈ . Find ( )1 2 3A A A∪ − ( ]1 = 1,2A − 2 1 = ,4 2 A   −    3 1 = ,6 3 A   −    Therefore ( )1 2 3 1 = 1, 3 A A A   ∪ − − −    . Example 3 : Given 2 ,3nA n n   = −    ; n I∈ . Find ( )3 5 2A A A∩ − ( ]1 = 1,6A − 2 2 = ,9 3 A   −    3 2 = ,15 5 A   −    Therefore ( ) [ ]3 5 2 = 6,9A A A∩ − . Example 4 : Given 5 10 and 3 8x y< < ≤ ≤ 1) 8 < < 18x y+ 2) 3 < < 7x y− − 3) 15 < < 80x y⋅ 4) 5 10 < < 8 3 x y
  23. 23. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 23 Example 5 : Given 5 2 and -3 1x y− < < − ≤ ≤ − 1) 8 < < -3x y− + 2) 4 < < 1x y− − 3) 2 < < 15x y⋅ 4) 2 < < 5 3 x y Example 6 : Given 3 1 and 1 3x y− < < < < 1) -2 < < 4x y+ 2) 6 < < 0x y− − 3) 9 < < 3x y− ⋅ 4) 3 < < 1 x y − 5) 2 2 0 < 9 ; 1 < < 9x y≤ 6) 2 2 9 < < 8x y− − Inequality solving Example 7 : Solve inequality 5 7 2 11x≤ − ≤ . SOLUTION 5 7 2 11x≤ − ≤ 7 5 2 11 7x− + ≤ − ≤ − 2 2 4x− ≤ − ≤ ( ) ( ) 1 1 2 4 2 8 x     − − ≤ ≤ −        -2 1x≤ ≤ Therefore solution set is { }2 1x x− ≤ ≤ or [ ]2,1− . Example 8 : Solve inequality 9 2 < 4 3 6x x x− − ≤ + . SOLUTION 9 2 < 4 3x x− − and 4 3 6x x− ≤ + 4 2 < 3 9x x− − − − 4 6 3x x− ≤ + 6 < 12x− − 3 9x ≤ ( ) ( ) 1 1 6 > 12 6 6 x     − − − −        ( ) 1 1 3 9 3 3 x     ≤        > 2x 3x ≤ 32 Therefore solution set is { }2 3x x< ≤ or ( ]2,3 .
  24. 24. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 24 Example 9 : Solve inequality 2 4 3 > 0x x− + . SOLUTION 2 4 3 > 0x x− + ( )( )1 3 > 0x x− − CASE 1 1 > 0x − and 3 > 0x − CASE 2 1 < 0x − and 3 < 0x − > 1x and > 3x < 1x and < 0x > 3x < 1x Therefore solution set is { }1 or 3x x x< > or ( ) ( ),1 3,−∞ ∪ ∞ . Example 10 : Solve inequality 2 2 5 3 0x x+ − ≤ . SOLUTION 2 2 5 3 0x x+ − ≤ ( )( )2 1 3 0x x− + ≤ CASE 1 2 1 0x − ≥ and 3 0x + ≤ CASE 2 2 1 0x − ≤ and 3 0x + ≥ 1 2 x ≥ and 3x ≤ − 1 2 x ≤ and 3x ≥ − No solution 1 3 2 x− ≤ ≤ Therefore solution set is 1 3 2 x x   − ≤ ≤    or 1 3, 2   −    . Example 11 : Solve inequality 2 2 5 > 0x x+ + . SOLUTION 2 2 5 > 0x x+ + ( )2 2 1 +4 > 0x x+ + ( ) 2 1 +4 > 0x + ; x R∈ Therefore solution set is R . Example 12 : Solve inequality 2 6 5 < 0x x+ − . SOLUTION 2 6 5 < 0x x+ − ( )2 6 9 14 < 0x x+ + − ( ) ( ) 22 3 14 < 0x + − ( )( )3 14 3 14 < 0x x+ − + + Therefore solution set is ( )3 14 , 3 14− − − + . -+ + -3+ 14-3- 14
  25. 25. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 25 Example 13 : Solve inequality 2 8 10 0x x− + ≥ . SOLUTION 2 8 10 0x x− + ≥ ( )2 8 16 6 0x x− + − ≥ ( ) ( ) 22 4 6 0x − − ≥ ( )( )4 6 4 6 0x x− − − + ≥ Therefore solution set is ( ), 4 16 4 6 , −∞ − ∪ + ∞  . Example 14 : Solve inequality 2 10 25 > 0x x+ + . SOLUTION 2 10 25 > 0x x+ + ( ) 2 5 > 0x + and 5x R x∈ ≠ − Therefore solution set is { }5x x ≠ − or { }5R − − . Example 15 : Solve inequality 2 4 4 0x x− + ≤ . SOLUTION 2 4 4 0x x− + ≤ ( ) 2 2 0x − ≤ 2x = Therefore solutionset is { }= 2x x or { }2 . Example 16 : Solve inequality 2 10 25 < 0x x− + . SOLUTION 2 10 25 < 0x x− + ( ) 2 5 < 0x − Therefore no solution . Example 17 : solve the following inequality 1) ( )( )( )2 3 1 < 0x x x− − + SOLUTION ( )( )( )2 3 1 < 0x x x− − + ( )( )( )2 3 1 < 0x x x− − + - 2-1 -+ + 3 Solution set is ( ) ( ), 1 2,3−∞ − ∪ . -+ + 4+ 64- 6
  26. 26. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 26 2) ( )( ) ( ) 2 1 2 3 0x x x− − − ≥ SOLUTION ( )( ) ( ) 2 1 2 3 0x x x− − − ≥ ( )( ) ( ) 2 1 2 3 0x x x− − − ≥ Times ( ) 2 1 2x − ; ( )( )1 3 0x x− − ≥ 2-1 -+ + 3 Solution set is ( ] { } [ ),1 2 3,−∞ ∪ ∪ ∞ . 3) ( ) ( ) ( ) ( ) ( ) 2 3 4 5 6 2 3 4 0 5 6 x x x x x − − − ≤ − − SOLUTION ( ) ( ) ( ) ( ) ( ) 2 3 4 5 6 2 3 4 0 5 6 x x x x x − − − ≤ − − Times ( ) ( ) ( ) ( ) ( ) 5 6 2 3 4 5 6 2 3 4 x x x x x − − − − − ; ( )( )3 5 0x x− − ≤ 542 -+ + 3 Solution set is { } [ )2 3,5∪ . 4) 2 1 2 3 4 x x x − < + − SOLUTION 2 1 2 < 0 3 4 x x x − − + − ( )( ) ( )( ) ( )( ) 4 2 1 3 2 < 0 3 4 x x x x x − − − + + − ( )( ) 2 2 9 4 2 6 < 0 3 4 x x x x x − + − − + − ( )( ) 2 2 11 2 < 0 3 4 x x x x − − + − ( )( ) 2 11 1 2 < 0 3 4 x x x x − − + −
  27. 27. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 27 ( )( ) 2 11 121 137 2 16 16 < 0 3 4 x x x x   − + −    + − ( )( ) 22 11 137 4 4 < 0 3 4 x x x    − −       + − ( )( ) 11 137 11 137 4 4 4 4 < 0 3 4 x x x x    − + − −      + − Solution set is 11 137 11 137 3, 4, 4 4    − + − ∪           . 5) 5 2 3 2x x ≥ + − SOLUTION 5 2 0 3 2x x − ≥ + − ( ) ( ) ( )( ) 5 2 2 3 0 3 2 x x x x − − + ≥ + − ( )( ) 5 10 2 6 0 3 2 x x x x − − − ≥ + − ( )( ) 3 16 0 3 2 x x x − ≥ + − ( )( )( )3 16 3 2 0x x x− + − ≥ Solution set is [ ] 16 3,2 , 3   − ∪ ∞    .
  28. 28. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 28 2.6 Absolute Value When we want to talk about how “large” a number is without regard as to whether it is positive or negative, we use the absolute value function. The absolute value of a number is the distance from that number to the origin (zero) on the number line. That distance is always given as a non-negative number. • If a number is positive (or zero), the absolute value function does nothing to it: • If a number is negative, the absolute value function makes it positive: Definition ; ; 0 = 0 ; 0 ; 0 x x x x x x >  = − < 3 = 3 5 = 5− , ( ) 5 = 5 4 = 4 = 4− − − , 0 = 0 ; 0 = ; 0 x x x x x ≥  − < Absolute Value Property ; 1) =x x− 2) =x y y x− − 3) =xy x y 4) = ; 0 xx y y y ≠ 5) 2 2 =x x 6) = 0 if and only if 0x x = 7) = ; 0 if and only if orx a a x a x a> = = − 8) 2 2 = if and only ifx y x y= 9) 2 2 < if and only ifx y x y< 10) x y x y+ ≤ + 11) = if and only if 0x y x y xy+ + ≥ 12) < if and only if 0x y x y xy+ + < 13) x y x y− ≥ −
  29. 29. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 29 14) = if and only if 0x y x y xy− − ≥ 15) > if and only if 0x y x y xy− − < 16) < ; 0 if and only ifx a a a x a> − < < 17) ; 0 if and only ifx a a a x a≤ > − ≤ ≤ 18) > ; 0 if and only if orx a a x a x a> < − > 19) ; 0 if and only if orx a a x a x a≥ > ≤ − ≥ 20) 1 ; 0 = 1 ; 0 xx xx >  − < Example 1 : Solve the following equations 1) 2 6 = 0x x− − SOLUTION 2 6 = 0x x− − ( )( )3 2 = 0x x− + = 3 , 2x − Solution set is { }3, 2− . 2) 2 3 = 15x − SOLUTION 2 3 = 15x − 2 3 = 15 , 15x − − 2 = 18 , 12x − = 9 , 6x − Solution set is { }9, 6− . 3) 2 1 = 3x x− + SOLUTION ( ) ( ) 2 2 2 1 = 3x x− + 2 2 4 4 1 = 6 9x x x x− + + + 2 3 10 8 = 0x x− − ( )( )3 2 4 = 0x x+ − 2 = , 4 3 x − Solution set is 2 ,4 3   −    .
  30. 30. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 30 4) 2 1 = 2x x− + SOLUTION ( ) ( ) 2 2 2 1 = 2x x− + 2 2 4 4 1 = 4 4x x x x− + + + 2 3 8 3 = 0x x− − ( )( )3 1 3 = 0x x+ − 1 = , 3 3 x − Solution set is 1 ,3 3   −    . Example 2 : Solve the following equations 1) 2 2 2 15 = 2 15x x x x− − − − SOLUTION 2 2 15 0x x− − ≥ ( )( )5 3 0x x− + ≥ Solution set is ( ) [ ), 3 5,−∞ − ∪ ∞ . 2) 2 2 6 = 6x x x x− − + − SOLUTION ( )2 2 6 = 6x x x x− − − − − 2 6 0x x− − ≤ ( )( )3 2 0x x− + ≤ Solution set is [ ]2,3− . 3) = 1x x − SOLUTION If 0 then 1x x x≥ = − No solution If 0 then - 1x x x< = − 1 = (F) 2 x Solution set is φ .
  31. 31. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 31 4) = 1x x + SOLUTION If 0 then 1x x x≥ = + No solution If 0 then - 1x x x< = + 1 = (T) 2 x − Solution set is 1 2   −    . 5) 2 = 3x x− + SOLUTION If 2 0 then 2 3x x x− ≥ − = + No solution If 2 0 then 2 3x x x− < − + = + 1 = (T) 2 x − Solution set is 1 2   −    . 6) 2 1 = 3x x+ − SOLUTION If 2 1 0 then 2 1 3x x x+ ≥ + = − = 4 (F)x − If 2 1 0 then 2 1 3x x x+ < − − = − 2 = (F) 3 x Solution set is φ . 7) 1 + 2 = 10x x+ − SOLUTION ( ) ( )If 1 0 and 2 0 then 1 2 10x x x x+ ≥ − ≥ + + − = 11 2 and (T) 2 x x≥ = ( ) ( )If 1 0 and 2 0 then 1 2 10x x x x+ ≥ − < + − − = No solution ( ) ( )If 1 0 and 2 0 then - 1 2 10x x x x+ < − ≥ + + − = No solution ( ) ( )If 1 0 and 2 0 then 1 2 10x x x x+ < − < − + − − = 9 < 1 and (T) 2 x x− = − Solution set is 11 9 , 2 2   −    .
  32. 32. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 32 2.7 Absolute Value Inequality Example 1 : Solve the following inequality. 1) 2 1 3x x− ≤ + SOLUTION 2 1 3x x− ≤ + ( ) ( ) 2 2 2 1 3x x− ≤ + 2 2 4 4 1 6 9x x x x− + ≤ + + 2 3 10 8 0x x− − ≤ ( )( )3 2 4 0x x+ − ≤ Solution set is 2 ,4 3   −    . 2) 3 2 < 2 3x x− + SOLUTION If 3 2 0 then 3 2 2 3x x x− ≥ − < + 2 and 5 3 x x≥ < 2 < 5 3 x≤  (1) If 3 2 0 then 3 2 2 3x x x− < − + < + 2 1 < and 3 5 x x > − 1 2 < < 5 3 x−  (2) Solution set is 1 2 2 1 , ,5 = ,5 5 3 3 5       − ∪ −           . 3) 2 1 + 3 > 10x x− + SOLUTION ( ) ( )If 2 1 0 and 3 0 then 2 1 3 10x x x x− ≥ + ≥ − + + > 1 and 3 2 x x≥ ≥ − 8 3 x > 8 > 3 x  (1) ( ) ( )If 2 1 0 and 3 0 then 2 1 3 10x x x x− ≥ + < − − + > 1 and 3 2 x x≥ < − Oppose
  33. 33. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 33 ( ) ( )If 2 1 0 and 3 0 then 2 1 3 10x x x x− < + ≥ − − + + > 1 < and 3 2 x x ≥ − 6x < − No solution ( ) ( )If 2 1 0 and 3 0 then 2 1 3 10x x x x− < + < − − − + > 1 < and 3 2 x x < − 4x < − < 4x −  (2) Solution set is ( ) 8 , 4 , 5   −∞ − ∪ ∞    . 4) 2 3 13x − ≤ SOLUTION 13 2 3 13x− ≤ − ≤ 10 2 16x− ≤ ≤ 5 8x− ≤ ≤ Solution set is [ ]5,8− . 5) 3 1 > 11x − SOLUTION 3 1 11x − < − or 3 1 11x − > 3 < 10x − 3 > 12x 10 < 3 x − > 4x Solution set is ( ) 10 , 4, 3   −∞ − ∪ ∞    .
  34. 34. C h a p t e r 2 : R e a l N u m b e r S y s t e m P a g e | 34 English vocabulary for Mathematic Addition Property = RgUONzกLMUVก Associative Property = RgUONzกLMpa_os[dTgXw Cancallation Property = ก ก Closure Property for addition = RgUONza•PกLMUVก Complex Numbers = ]SLdVdpjz^}ZQd Commutative Property = RgUONzกLMR_OUkos Counting Numbers = ]SLdVddOU Decimal = kqdz[g Distributive Property = RgUONzกLM]ก]^ Fraction = pqmRwVd Identity = pQก_OกmY‚ Imaginary Number = ]SLdVd]zdN€Lf Integer numbers = ]SLdVdpNyg Inverse = NOVuกuOd Irrational Numbers = ]SLdVdNMMก[e multiplication = กLMWXY Multiplication Property = RgUONzกLMWXY Natural Numbers = ]SLdVddOU Periodic Decimal = kqdz[g}cSL Rational Numbers =]SLdVdNMMก[e Reflexive Property = RgUONzกLMRekZQd Subtraction = กLM_U Square = MXaRospT_os[g]ON`MOR Struction = hWM^RMZL^ Symetric Property = RgUONzกLMpkwLกOd therefore = PO^dOcd Transitive Property = RgUONzกLMxwL[kQP Whole Number = ]SLdVdkOc^TgP

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