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Tran Viet Dung LECTURE ON MATHEMATICS I For HEDSPI Students Hanoi University 2008 COMTENTS CHAPTER I SYMBOLIC LOGICS Propositions Logical operations 1.1 Negation operator NOT 1.2 Conjunction operator AND (  ) 1.3 Disjunction operator OR (  ) 1.4 Implication operator IMP ( 1.5 Equivalence operator IFF (  ) 1.6 Tautologies, contradictions 10 Generation of operators 11 4.1 Binary XOR operator ( ) 11 4.2 Binary operator NOR () 12 4.3 Binary Operator NAND () 13 Propositions with quantifiers , 13 CHAPTER II ) SETS 16 Sets and elements 16 1.1 Some definitions and notations .16 1.2 The ways to determine a set 16 1.3 Subsets 18 Set operations 18 2.1 Intersection and union of sets 18 2.2 Difference of sets, compliment of a subset 19 2.3 Properties 20 2.4 The Cartesian product of sets 21 Some properties of finite sets 21 CHAPTER III MAPPINGS 23 Basic concepts .23 Injective,surjective, bijective mappings .24 Composition of maps, inverse maps 25 Restriction, characteristic functions 26 Substitutions 27 Collections 29 6.1 Collection of sets 29 6.2 Collection of maps 30 CHAPTER IV RELATIONS 32 On relation Concepts 32 Order relation .33 2.1 Concepts on order relation 33 2.2 Lexicographical order 35 Equivalence relation .36 3.1 Definitions and examples 36 3.2 Equivalence classes 36 3.3 Partitions induced by maps 38 CHAPTER V ALGEBRAIC STRUCTURES 39 Binary operators 39 1.1 Definitions and examples 40 1.2 Properties of binary operators 41 Groups 43 2.1 Semigroups 43 2.2 Concepts on groups .43 2.3 Some properties 44 Subgroups, normal subgroups 45 3.1 Subgroups .45 3.2 Normal subgroups 46 Rings and fiels .48 4.1 Rings 48 4.2 Fields 48 4.3 Ring of integers 50 4.4 Euclidean Algorithm 51 4.5 Presentation of integers 54 CHAPTER VI FIELD OF COMPLEX NUMBERS 56 Concepts on complex numbers 56 1.1 Canonical form of complex numbers .56 1.2 Operations in canonical form 562 1.3 Modulus and conjucgate of complex numbers 562 Polar form of complex numbers 59 2.1 Definitions and examples 59 2.2 Some operations of complex numbers in the polar form 60 2.3 n-roots of a complex number 62 Quadratic equations on C 64 3.1 Quadratic equations of real coefficients 64 3.2 Quadratic equations of complex coefficients 65 Polynomials of complex variables 66 Chapter I SYMBOLIC LOGICS Propositions Definition A Proposition is a statement which is either true or false, although we may not know which Propositions are denoted by lower letters as p, q, r… The truth or falsity is called truth value of the proposition The truth value of the proposition p is denoted by V(p) If p is true then V(p) = or T If p is false then V(p) = or F Example The proposition p is given by p = “ The sun rises in the east “ and the proposition q is given by q = “ The sun rises in the west” Then V(p) = and V(q) = However, for several propositions we not know the truth values Example The proposition r = “ There exists life outside the earth.” Up to now we can not know the truth value of the statement r Logical operations Negation operator NOT Definition 2.1 The negative proposition of a proposition p is the proposition p defined by its truth table as follows p p 0 Table 1.NOT truth table  Note : For a proposition p we have V(p) = V p Example Given a proposition p = “ Hanoi is the capital of Vietnam “ Then NOT p is the proposition p = “ Hanoi is not the capital of Vietnam” Example The proposition q is “ The equation f(x) = has solutions ”, then the negative proposition is “ The equation f(x) = has no solution “ Conjunction operator AND (  ) Definition 2.2 Given two propositions p and q The proposition p  q is true only both p and q are true propositions The AND operator is defined by a truth table which lists of possible combinations of the truth values of p andq : p q pq 1 1 0 0 0 TABLE AND truth table Example If p = “ Pigs are mammals” and q = “Pigs fly “ , then p  q is interpreted as “ Pigs are flying mammals “ Disjunction operator OR (  ) Definition 2.3 For propositions p and q the proposition p  q is false only when both p and q are false The OR operator is defined by the following truth table p Q p q 1 1 1 0 Table OR Truth Table Theorem ( De Morgan ’s Law ) For any two propositions p and q we have 1) V( p  q ) = V( p )  V( q ), 2) V( p  q ) = V( p )  V( q ), Corollary The disjunction operator OR may be defined by NOT and AND operators: V( p  q ) = V ( p  q ) Theorem 3.(Distributive Laws ) For any three propositions p, q, r, we have 1) V(p(qr)) = V((pq)(pr)) 2) V(p(qr ) = V(( pq)(pr)) Theorem 4.( commutative and associative Laws) For propositions p, q, r, we have 1) V(pq) = V(qp), 2) V(pq) = V(qp), 3) V((pq)r)=V(pq(r)), 4) V((pq)r))= V(p(qr)) Implication operator IMP ( ) Definition 2.4 For two propositions p, q, the proposition p q is defined by its truth table p q P q 1 0 1 1 TABLE IMP truth table Thus, the proposition p q is false only when p is true and q is false Assertion If p, q are propositions then we have 1) the proposition p (pq) is true, 2) The proposition (pq) p is true Theorem The implication operator IMP may be built from the negation operator NOT and the conjunction operator AND: V(p q) = V( p  q ) Equivalence operator IFF (  ) Definition 2.5 Given two propositions p, q The proposition pq is true only when the truth values of p and q are the same p q Pq 1 1 0 0 TABLE IFF Truth Table Theorem For propositions p, q we have V( pq) = V((p q)(q p)) Theorem The equivalence operator may be built from the negation operator NOT and the conjunction operator AND: V(pq) = V(( p  q )( q  p )) Tautologies, contradictions Definition 3.1 A proposition composite by atomic propositions is called a tautology if it is always true regardless truth values of atomic components Example a)The proposition (pq) p is a tautology Actually, we have the truth table of this proposition p q pq (Pq) p 1 1 0 1 0 b)(( pq) p ) q is a tautology The truth table is p q pq (Pq) p ((Pq) p ) q 1 1 1 1 1 0 0 Definition 3.2 10 Presentation of integers We usually integers as form 10-adic For example, 2139 = 2.103 + 1.102 + 3.101 + 9.100 Remark Given a positive integer b For a natural number n we have the expression n = akbk + ak-1 bk-1 + + a1.b + a0,  aj  b , ak  (*) Then the presentation (*) is said to be the expansion of n by base b Denote n = ( ak ak-1 a1 a0)b If b = the presetation (*) is called the binary expansion of n Example 21 a) The binary expansion of 35 is 35 = 125 +024 + 023 +022 + 121 +120 35 = (100011)2 b) The expansion of 135 by base is 135 = 243 + 042 + 141 + 340 135 = (2013 )4 Algorithm for expantion of n by base b on the step k we find the numbers qk, ak satisfy qk-1 = b.qk + ak,  ak  b Step n = b.q0 + a0 Step q0 = b.q1 + a1 Step q1 = b.q2 + a2 54 The last step is Then qm-1 =b.qm + am if qm = n = ( am am-1 a1 a0 )b Example 22 Represent 1397 by base 1397 = 174 + 174 = 8.21 + 21 = 8.2 + = 8.0 + Hence, 1397 = (2565)8 55 Chapter VI FIELD OF COMPLEX NUMBERS Concepts on complex numbers Canonical form of complex numbers Definition Let R be the fiel of real numbers Put C = RR We define addtion and multiplication operators on C as follows For z1 = ( a, b ), z2 = (c, d) in C, z1 + z2 = ( a + c, b + d ), z1 z2 = (ac – bd, ad + bc ) Proposition The set C with the above addition and multiplication operators is a field called the fiedl of complex numbers Proof We can check following properties 1) ( z1 + z2 ) + z3 = z1 + ( z2 + z3 ), 2) z1 + z2 = z2 + z1, 3) There is an identity element = (0,0 ) for addition , + z = z + = z for z  C 4) For z =( a, b ) , -z = ( -a, -b ), 5) z1(z2z3 ) = (z1z2 )z3, 6) z1.z2 = z2.z1 7) z1( z2 + z3 ) = z1z2 + z1z3 56 8) The identity for multiplication is e = ( 1, ), ez = ze = z z = ( a, b ) ≠ , the inverse element is z-1 = ( 9) a b , ), a  b a  b2 z.z-1 = z-1.z = e We say that an element of C is a complex number Note For complex numbers of form (a, ), we have ( a, ) + (c, ) = ( a + c, ) (a, )(c, ) = ( ac, ) We can identify a complex number (a, 0) as the real number aR (a ,0)  Then the identity e = ( 1, )  Put i = ( 0, ) We have i2 = ( 0, )(0, 1) = (-1, ) = -1 This ellement i is called the imaginary unit For z = ( a, b ) , it can be expressed as z = ( a, b ) = a + b.i, a.b  R Definition For a complex number z  C, z = a + bi , ( 1.1) where a, b R, and i2 = -1 The above form of z is called the canonical form of z The real number a is called the real part of z and denoted by a = Re(z) The real number b is called the imaginary part of z denoted by b = Im(z) Note that for z1 ,z2  C, z1 = z2 iff Re(z1)=Re(z2), Im(z1)= Im(z2) (1.2) 57 Operations of complex numbers in canonical form Addition : ( a + b.i ) + ( c + d.i ) = ( a + c ) + (b + d )i, Substraction: ( a+ b.i ) - (c + d.i) = ( a - c ) + ( b – d )i, Multiplication : ( a + b.i ) ( c + d.i ) = ( ac - bd) +( ad + bc)i, Division is defined by z1 = z1 z2-1 z2 for z ≠ a  bi ac  bd bc  ad = + i c  di c  d c  d Example1 a) ( + 3i ) + ( -1 +4i ) = + 7i, b) ( +3i ) – ( + i ) = (-3) +2i, c) (2 +3i )(1 + i ) = -1 + 5i, d)  3i (2  3i )(1  i ) 5i = = 1i (1  i )(1  i ) = i + , 2 e) ( + 3i )3 = -46 + 9i Modulus, conjugation of complex numbers Definition For a complex number z = a + bi, where a, b  R Modulus  z of z is given z= a  b , (1 3) 58 Conjugate of z = a + bi is defined to be the complex number a – bi usualy denoted by z* or z z = a - bi (1 4) Proposition For complex numbers we have 1)  z ≥ 0, and  z = iff z = 0, 2)  z + w ≤  z + w, 3)  z.w =  z  w 4)  1 = 1, z z = , w w 5) z  w = z + w , z.w = z w , z z 6)   = , w w z = z, 7) z = z if and only if z is a real number, 8) z = - z if and only if z is purely imaginary, 9) Re(z) = 10)  z =  z  ,  z2 = z z 1 ( z + z ) , Im(z) = ( z - z ), 2i Polar form of complex numbers Definitions and examples A complex number z can be wiewed as a point or a vector in twodimension cartesean coordinate system called the complex plane The number z = x + yi can be considered as the poin z(x, y ) in the plane Oxy and Oxy is called the complex plane 59 So the modolus of z is the distance from O to z, the number x is x-coordinate of z and y is the y-coordinate of z If z = x + 0i then z is a real number and it lies on x-axis Therefore, Ox is called the real axis If z = + yi then z is purely imaginary and lies on Oy Then Oy is called the imaginary axis In the complex plane , for a point z = (x, y ) , we put r = z , and  the angle beween Ox and Oz We have z = x + yi = r (cos + i.sin) (2.1) The expantion (2.1) is called the polar form ( or trigonometric form ) of z Here we recall that r is the modulus of z The value  is called argument of z The argument of z is unique modulo 2 Put ei = cos + isin Then z = r(cos + isin ) = r ei (2.2) The form (2.2) is called the exponential form of z Some operations of complex numbers in the polar form We can show the following formulas Multiplication: If z1 = r1 (cos1 + isin1 ) and z2 = r2 (cos2 + isin2 ) then z1.z2 = r1.r2(cos (1 + 2 ) + isin(1 + 2 )) (2.3) Division: If z1 = r1 (cos1 + isin1 ) and z2 = r2 (cos2 + isin2 ), z2 ≠ 0, then z1 r1 = (cos(1 - 2 ) + isin(1 - 2 )) z r2 (2.4 ) 60 Exponentiation: If z = r( cos + i.sin) then zn = rn( cosn + i.sinn) (2.5) Here we pove the formula (2.3 ): z1.z2 = r1 (cos1 + isin1 ) r2 (cos2 + isin2 ) = = r1r2( cos1 cos2 - sin1 sin + i ( cos1 sin2 + sin1 cos2 ) = r1r2 ((cos (1 + 2) + isin(1 + 2 )) Moiver’s formula : ( cos + i.sin)n = ( cosn + i.sinn) (2.6) Example Express the complex number ( + i )99 in the canonical form Solution z = (1 + i ) = ( cos zn =( )99( cos99 = ( )99( cos3    + i sin + i sin99 + i sin3    ), )= ) = 249( -1 + i ) = - 249 + 249.i Example Express sinnx, cosnx in term of cosx sinx Solution We have (cos + i.sin)n = ( cosn + i.sinn) On the other hand, by binomial formula (cos + i.sin)n = n C k n (cos x ) k (i sin x ) n k k 0 here C nk = n( n  1) (n  k  1) k ( k  1) 2.1 61 Then express (cos + i.sin)n = A + Bi , where A, B are polynomials of cosx, sinx It follows A = cosnx, B = sinnx That are formulas we need Example ( cosx + isinx )3 = cos 3x + isin3x, Other hand, (cosx + isinx )3 = cos 3x + 3cos 2x.(isinx) + 3cosx(isinx) 2+ (isinx) = = (cos 3x – 3cosx sin x) + i(3cos 2x sinx - sin x ) cos3x = cos 3x – 3cosx sin x, Hence, sin3x = 3cos 2x sinx - sin x n-roots of a complex number Definition If c is a complex number, n is a positive integer, then any complex number z satisfying the formula zn = c is called an n-th root of the coplex number c Example a) The numbers 1, -1 are square roots of 1, b) The numbers 1, -1, i, -i are fouth roots of c) 2 i is a square root of i 2 For c = the n-th root of is For a complex number c , the set of all n-th roots of c is denoted by n c Thus, n c = { z C  zn = c } Proposition 62 If c = r( cos + i.sin) is nonzero then the set of all n-th roots contains n elements zk given by n zk = n where r (cos   2k   2k +i.sin ), k =1,2, ,n-, n n (2.7) r represents the usual (positive) n-th root of the positive number r Proof Let z = (cos + isin) be an n-th root of c Then zn = n (cosn + isinn) = r ( cos + i.sin) We have n = r ;  =n r ,  = n =  + 2k , k Z It follows   2k , kZ n However, we have only n distint values of z corresponding to n that denoted by zk = n r ( cos values of k =0, 1, , n-1 i.sin   2k + n   2k n The proposition is proved Example Find the set of all n-th roots of Solution We have = ( cos0 + isin0) By the for mula (2.7), there are n values of n-th roots that are k = cos 2k 2k + i.sin , where k = 0, 1, , n-1 n n Example The 6th roots of are 0 = 1,  = 1 = cos   + isin , 3 63 2 = cos2     + i sin2 , 3 = cos3 + isin3 , 3 3 4 = cos4     + isin4 , 5 = cos5 + isin5 3 3 Quadratic equations on C Quadratic equations of real coefficients Let us consider quadratic equations ax2 + bx + c = 0, (3.1) where a, b, c  R, a ≠ and x is the variable Recall that to find the real solutions of the equation (3.1) we have If  = b2 - 4ac  then the equation has two real root 1) x1=  b  b  4ac  b  b  4ac , x2 = 2a 2a (3.2) If  = b2 - 4ac = then the equation (3.1) has a double 2) solution x1 = x2 = b , 2a (3.3) 3) If  = b2 - 4ac  then the equation has no real solutions To find complex solutions of the equation (3.1) we have 1) If  ≥ solutions as in (3.2), (3.3) 2) If   the equation has two conjugate complex roots x1,2 = b i  2a (3.4) Example Solve the equation x2 - 2x + 10 = 64 Solution complex roots  = b2 - 4ac = - 36  We have two conjugate x1 = + 3i, x2 = – 3i Quadratic equations of complex coefficients Let us consider the equation ax2 + bx + c = 0, (3.5) where a, b, c are complex coefficients , a ≠ The equation always has complex roots  b  b  4ac x1,2 = 2a where , (3.6) b  4ac in the sense of square roots of a complex number So b  4ac contains two values Remark that when consider the equations in C we can see the case of real coefficients as special case of complex coefficients Example x2 + 2ix - 10 = Solve the equation Solution x1 = -i + 3, x2 = - i – Example 10 Solve the equation x2 – 2(1 + i )x – 14i = 0, Solution x1,2 =(1+i) + = (1+i) + i (1  i )  14i = (1+i) + 16i = ( 1+i)  ( 2 + i) 2 Thus, x1 = (1+2 ) + ( 1+2 )i ; x2 = ( - 2 ) + ( 1- 2 )i 65 Polynomials of complex variables Consider a polynomial of degree n on C p(x) = a0 + a1x + a2x2 + + anxn , (4.1) where a0, a1, , an  C, an ≠ and x is complex variable Definition A complex number  is said to be a root of the polynomial p(x) = a0 + a1x + a2x2 + + anxn if p( ) = i.e a0 + a1 + a22 + + ann = Proposition Every polynomial of degree n on C has exactly n complex roots (counting multiple roots according to their multiplicity ) Proposition If x1, x2, , xn are roots of the polynomial p(x) = a0 + a1x + a2x2 + + anxn then p(x) can be represented in the form p(x)= an( x - x1 )( x - x2 ) ( x - xn ) (4.2) Now we consider the polinomial of real coefficients p(x) = a0 + a1x + a2x2 + + anxn , (4.3) where a0, , an are real numbers Proposition If  is a complex root of p(x) in (4.3) then the conjugate complex number  of  is also a root of p(x) 66 Proof From p() = we have p() = a0 + a1 + a22 + + ann = and p( ) = a  a1   a n n = Hence, a0 + a1  +a2  2+ + an  n = Thus, p(  ) = and  is a root of p(x) The proof is complete Proposition Let p(x) be a polynomial of real coefficients Then p(x) can be expressed as a product of binomials and quadratic polynomials of negative discriminant Proof Let p(x) = a0 + a1x + a2x2 + + anxn ,  R Suppose that x1, , xn are n complex roots of p(x) Then p(x) = an( x - x1 )( x - x2 ) ( x - xn ) (4.4) If xk is a real root then ( x – xk) is a real binomial factor of p(x) If xk is a complex root ( noreal) then x k is also a root of p(x) We have (x - xk)( x - x k ) = x2 – (xk + x k )x + xk x k is a factor of p(x) Moreover, (xk + x k ) and xk x k are real numbers So this factor is a quadratic polynomial that has no real roots and therefore, the discrimant  of this factor is negative 67 68