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Unformatted text preview: PARTS 1 & 2 COMBINED COVER PRINCIPLES FOR BASIC, INTERMEDIATE AND COLLEGE COURSES SET THEORY NOTATION . { } braces indicate the beginning and end of a set notation; when listed, el
ements or members must be separated by commas. EX: A = {4, 8, 16}; sets
are finite (ending, or having a last element) unless otherwise indicated.  indicates continuation of a pattern. EX: B = {5, 10, 15, , 85, 90} 0 at the end indicates an infinite set, that is, a set with no last element.
EX: C = {3, 6, 9, 12, ...} 0  is a symbol which literally means "such that." 0 E means “is a member of” OR “is an element of.” EX: IfA = {4, 8, 12}
then 12 E A because 12 is in set A. 0 65 means “is not a member of” OR “is not an element of.” EX: If
B = {2, 4, 6, 8} then 365B because 3 is not in set B. 0% means empty set OR null set; a set containing no elements or mem
bers, but which is a subset of all sets; also written as { }. 0 C means “is a subset of;” also may be written as Q 0 ¢ means “is not a subset of;” also may be written as @  AC B indicates that every element of set A is also an element of set B.
EX: IfA = {3, 6} and B = {1, 3, 5, 6, 7, 9} then ACB because the 3 and 6
which are in set A are also in set B. 0 2“ is the number of subsets of a set when n equals the number of elements
in that set. EX: If A = {4, 5, 6} then set A has 8 subsets because A has 3 elements and 23 = 8.
. OPERATIONS  U means unlon. 0 AU B indicates the union of set A with set B; every element of this set is
either an element of set A OR an element of set B; that is, to form the
union of two sets, put all of the elements of both sets together into one set
making sure not to write any element more than once. EX: If A = {2,4} and B = {4, 8, 16} then A U B = {2, 4, 8,16}.
0 0 means intersection. 0 AﬂB indicates the intersection of set A with set B; every element of this
set is also an element of BOTH set A and set B; that is, to form the in tersection of two sets, list only those elements which are found in BOTH O P E RAT I O N S O F R EA L N U M B E RS _ ofthe two sets. EX: IfA = {2,4} and B = {4, 8, 16} then A n B = {4}. ' A indicates the complement of set A; that is, all elements in the univer
sal set which are NOT in set A. EX: If the Universal set is the set of
Integers and A = {0, 1, 2, 3, ...} then A {1, 2, 3, 4,...}. A I) A = Q. PROPERTIES A = B means all of the elements in set A are also in set B and all ele
ments in set B are also in set A, although they do not have to be in the
same order. EX: IfA = {5, 10} and B = {10, 5} then A = B. 0 n(A) indicates the number of elements in set A. EX: If A = {2, 4, 6} then
n(A) = 3. 0 ~ means “is equivalent to”; that is, set A and set B have the same number of el
ements although the elements themselves may or may not be the same. EX: If
A = {2, 4, 6} and B = {6, 12, 18} then A ~ B because n(A) = 3 and n(B) = 3. 0 A (I B = Q indicates disjoint sets which have no elements in common. SETS OF NUMBERS ' Natural or Counting numbers = {1, 2, 3, 4, 5, ..., 11, 12, ...} 'Whole numbers = {0, 1, 2, 3,..., 10, 11, 12, 13, ...} ' Integers = {... , 4, 3, 2, 1, 0, 1, 2, 3, 4, ...} ' Rational numbers = {p/q  p and q are integers, q at 0}; the sets of Nat
ural numbers, Whole numbers, and Integers, as well as numbers which
can be written as proper or improper fractions, are all subsets of the set of
Rational numbers. ° Irrational numbers : { x  x is a Real number but is not a Rational num
ber}; the sets of Rational numbers and Irrational numbers have no ele
ments in common and are therefore disjoint sets. 'Real numbers = {x  x is the coordinate of a point on a number
line}; the union of the set of Rational numbers with the set of Irra
tional numbers equals the set of Real numbers. ° Imaginary numbers = {ai  a is a Real number and i is the number
whose square is 1}; i2 = 1; the sets of Real numbers and Imaginary
numbers have no elements in common and are therefore disjoint sets. ° Complex numbers = {a + bi  a and b are Real numbers and i is the number
whose square is 1}; the set of Real numbers and the set of Imaginary num
bers are both subsets of the set of Complex numbers. EXs: 4 + 7i ; 3  2i ~___ I W  Part 1_ PROPERTIES OF REAL NUMBERS ... I mudelﬂ:puLﬁJﬂﬁlm.fm FOR ANY REAL NUMBERS a, b, AND G FOR ADDITION FOR MULTIPLICATION
a+bisaReal number ab isaReal number
a+b=b+a ab=ba
(a+b)+c=a+(b+c) (ab)c=a(bc)
0+a=aanda+0=a a‘1=aandl'a=a
a+(a)=0and a1/,=1and
(a)+a=0 ‘/a°a=1ifa¢0 Distributive Property a(b + c) = ab + ac; a(b  c) = ab  ac PROPERTIES OF EQUALITY
FOR ANY REAL NUMBERS a, b, AND c Reﬂexive: a = a
Symmetric: If a = b then b = a
Transitive: If a = b and b = c then a = c
Addition Property: If a = b then a + c = b + c
Multiplication Property: If a = b then ac = bc
Multiplication Property of Zero: a 0 0 = 0 and 0 0 a = 0
Double Negative Property: (a) = a PROPERTIES OF INEQUALITY
FOR ANY REAL NUMBERS a, b, AND c Trichotomy: Either a>b, or a=b, or a<b
Transitive: If a<b, andb<cthena<c Ifa<bthenac<b+c Ifa>bthenac>b+c Ifc¢0andc>0,anda>bthenac>bc;
also, if a<bthen ac<bc lfc¢0andc<0,anda>bthenac<bc;
also, ifa<bthen ac>bc PROPERTY Closure
Commutative
Associative Identity VNBd Inverse Addition Property of Inequalities: Multiplication Property of Inequalities: ABSOLUTE VALUE
x = x if X is zero or a positive number; x = X if x is a negative number;
that is, the distance (which is always positive) of a number from zero on the
number line is the absolute value of that number. EXs:   4 =  ( 4) = 4;
29=29; 0=0;43=(43)=43
ADDITION
If the signs of the numbers are the same: add the absolute values of the numbers; the sign of the answer is the same as the signs of the original two
numbers. EXs: 11 + 5 = 16 and 16 + 10 = 26
If the signs of the numbers are different: subtract the absolute values of the numbers; the answer has the same sign as the number with the larger
absolute value. EXs: 16 + 4 = 12 and 3 + 10 = 7 S U BTRACTI O N a  b = a + (b); subtraction is changed to addition of the opposite number:
that is, change the sign of the second number and follow the rules of addition
(never change the sign of the ﬁrst number since it is the number in back of the
subtraction sign which is being subtracted; 14  6 ¢ 14 +  6).
EXs:1542= 15+(42)=27; 245=24+(5)=29; 13 (45)=
13 + (+45) = 32;  62  (20) =  62 + (+20) =  42 MULTIPLICATION
The product of two numbers which have the same signs is positive;
EXs: (55)(3) = 165; ( 30)( 4) = 120; ( 5)( 12) = 60
The product of two numbers which have different signs is negative no matter
which number is larger. EXs: ( 3)(70) =  210; (21)( 40) =  840; (50)(3) =  150 DIVISION
(DIVIsons no NOT EQUAL ZERO) The quotient of two numbers which have the same sign is positive.
EXs: ( 14) / (7) = 2; (44) / (11) = 4; ( 4) / ( 8) = .5
The quotient of two numbers which have different signs is negative no
matter which number is larger.
EXs: (24) / (6) = 4; (40) / (8) =  5; (14) / (56) =  .25 DOUBLE NEGATIVE  ( a) = a; that is, the negative sign changes the sign of the contents of the
parentheses. EXs:  (4) = 4;  (17) = 17 VNSdov vuadov ALGEBRAIC TERMS COMBINING LIKE TERMS
ADDING on SUBTRAGTING a + a = 2a; when adding or subtracting terms, they must have exactly
the same variables and exponents, although not necessarily in the
same order; these are called like terms. The coefficients
(numbers in the front) may or may not be the same. ° RULE: combine (add or subtract) only the coefficients of like terms and
never change the exponents during addition or subtraction. EXs: 4xy3 and
7y3x are like terms and may be combined in this manner: 4xy3 + 7y3x =
3xy3. Notice only the coefficients were combined and no exponent changed. 15a2bc and 3bca4 are not like terms because the exponents of the a are not the same in both terms, so they may not be combined. MULTIPLYING
PRODUCT RULE FOR EXPONENTS (a"’)(a") = am"; any terms may be multiplied, not just like terms.
The coefficients m the variables are multiplied which means the
exponents also change. ' RULE: multiply the coefﬁcients and multiply the variables (this means you have to add the exponents of the same variable).
EX: (4a2c)(12a3b2c) = 48a5b2c2, notice that 4 times 12 became 48, a2
times a3 became as, c times c became c2, and the b2 was written down.
DISTRIBUTIVE PROPERTY FOR POLYNOMIALS
° Type 1:a(c + d) = ac + ad. EX: 4x3(2xy + y2) = 8x4y + 4x3y2
'Type2:(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd
EX: (2x + y)(3x  5y) = 2x(3x  5y) + y(3x  5y) =
6x2  10xy + 3xy  5y2 = 6x2  7xy  5yz
' Type 3:(a + b)(c2 + cd + d2) = a(c2 + cd + d2) + b(c2 + cd + d2) = ac2 +
acd + ad2 + be2 + bcd + bd2
EX: (5x + 3y)(x2  6xy + 4y2) = 5x(x2 6xy+4y2) +3y(x2  6xy + 4y2) =
5x330x2y+20xy2+3x2y18xy2+12y3=5x327x2y+2xy2+12y3
“FOIL” METHOD FOR PnODucTs 0F BINOMIALs ' This is a popular method for multiplying 2 terms by 2 terms only. FOIL
means ﬁrst times ﬁrst, outer times outer: inner times inner: and last times last.
EX: (2x + 3y)(x + 5y) would be multiplied by multiplying first term
times first term, 2x times x = 2x2; outer term times outer term, 2x times
5y = 10xy; inner term times inner term, 3y times x = 3xy; and last term
times last term, 3y times 5y = 15y2; then combining the like terms of 10xy
and 3xy gives 13xy with the final answer equaling 2x2 + 13xy + 15y2. SPECIAL PRODUCTS 'Type 1: (a+b)2=(a+b)(a+b)=a2+2ab+b2 'Type2: (ab)2=(ab)(ab)=a22ab+b2 'Type3: (a+b)(ab)=a2+0abb2=azb2 EXPONENT RULES ' RULE 1: (a“')11 = a‘"'11 ; (am)n means the parentheses contents are multi
plied n times and when you multiply you add exponents; EX: (2m“n2)3 = (2m4n2) (2m4n2) (2m4n2) = 8m12n6, notice the
parentheses were multiplied 3 times and then the rules of regular multi
plication of terms were used; ' SHORTCUT RULE: when raising a term to a power, just multiply expo
nents; EX: (2m4n2)3 = 23m12n6, notice the exponents of the 2, m4 and n2
were all multiplied by the exponent 3, and that the answer was the same as the
example above. CAUTION: am at (a)“; these two expressions are different.
EXs: 4yz2 at (4yz)2 because (4yz)2 = (4yz) (4yz) = 16yzz2 while 4yz2
means 4  y ' z2 and the exponent 2 applies only to the z in this situation. ° RULE 2: (ab)m = am b’“. EX: 6x3 y)2 = 62 x6 y2 = 36x6 y2
BUT (6x3 + y)2 = (6x3 + y) (6x + y) = 36x6 + 12x3y + yz; because there
is more than one term in the parentheses the distributive property for
polynomials must be used in this situation. m m 2
'RULE 3: 3 =Lwhen b at 0; EX: ‘4X2y =16x4y2
b 1"” 5z 25z2 ' RULE 4: Zero Power a0 = 1 when a at 0 DIVIDING ' QUOTIENT RULE: 27:11:?“ ; any terms may be divided, not just like terms; the coefficients ﬂi the variables are divided which means the
exponents also change. RULE: Divide coefficients and divide variables (this means you have
to subtract the exponents of matching variables). EX: (20x5yzz) / (5x22) = 4x3y2, notice that 20 divided by 5 became 4,
x5 divided by x2 became x3, and 1 divided by z became one and therefore
did not have to be written because 1 times 4x3y2 equals 4x3y . ' NEGATIVE EXPONENT: a' '1 = 1 / a'1 when a 1: 0; EXs: 2'1 = 1I2;
(41 '3y2) / (3ab'1) = (4y2b1) / (3az3). Notice that the 4 and the 3 both stayed
where they were because they both had an invisible exponent of positive 1;
the y remained in the numerator and the a remained in the denominator be
cause their exponents were both positive numbers; the z moved down and the
b moved up because their exponents were both negative numbers. STEPS FOR SOLVING A FIRST DEGREE
EQUATION WITH ONE VARIABLE ° FIRST, eliminate any fractions by using the Multiplication Property of
Equality. EX: 1I2 (3a + 5) = 2/3 (7a  5) + 9 would be multiplied on both sides
of the = sign by the lowest common denominator of 1I2 and 2I3, which is 6;
the result would be 3(3a + 5) = 4(7a  5) + 54, notice that only the 1/2, the
2/3, and the 9 were multiplied by 6 and not the contents of the parentheses;
the parentheses will be handled in the next step which is distribution. ° SECOND, simplify the left side of the equation as much as possible by us
ing the Order of Operations, the Distributive Property, and Combining
Like Terms. Do the same to the right side of the equation. EX: Use dis
tribution first, 3(2k  5) + 6k  2 = 5  2(k + 3) would become 6k  15 +
6k  2 = 5  2k  6 and then combine like terms to get 12k  17 = 1  2k. ' THIRD, apply the Addition Property of Equality to simplify and organize all
terms containing the variable on one side of the equation and all terms which do
not contain the variable on the other side. EX: 12k  17= 1  2k would become
2k + 12k  17 + 17= 1 + 17  2k + 2k and then combining like terms, 14k = 16. ' FOURTH, apply the Multiplication Property of Equality to make the co
efficient of the variable 1. EX: 14k = 16 would be multiplied on both sides
by 1/14 (or divided by 14) to get a 1 in front of the k so the equation would
become 1k = 16/14 or simply k = 11/7 or 1.143. ' FIFTH, check the answer by substituting it for the variable in the origi
nal equation to see if it works. NOTE: 1. Some equations have exactly one solution (answer). They are conditional
equations. EX: 2k = 18 2. Some equations work for all real numbers. They are identities. EX: 2k = 2k 3. Some equations have no solutions. They are inconsistent equations.
EX: 2k+3=2k+7 STEPS FOR SOLVING A FIRST DEGREE
INEQUALITY WITH ONE VARIABLE ADDITION PROPERTY OF INEQUALITIES For all real numbers a, b, and c, the inequalities a < b and a + c < b + c
are equivalent; that is, any terms may be added to both sides of an inequal
ity and the inequality remains a true statement. This also applies to a > b
and a + c > b + c. MULTIPLIOATION PROPERTY OF INEOUALITIES ' For all real numbers a, b, and c, with c at 0 and c > 0, the inequalities a >
b and ac > be are equivalent and the inequalities a < b and ac < be are
equivalent; that is, when c is a positive number the inequality symbols stay
the same as they were before the multiplication. EX: If 8 > 3 then multi
plying by 2 would make 16 > 6, which is a true statement. ° For all real numbers a, b, and c, with c at 0 and c < 0, the inequalities a > b and
ac < be are equivalent and the inequalities a < b and ac > be are equivalent; that
is, when c is a negative number the inequality symbols must be reversed from
the way they were before the multiplication for the inequality to remain a true
statement. EX: If 8 > 3 then multiplying by 2 would make 16 > 6, which is
false unless the inequality symbol is reversed to make it true, 16 < 6. STEPS FOR SOLVING ' FIRST, simplify the left side of the inequality in the same manner as an
equation, applying the order of operations, the distributive property, and
combining like terms. Simplify the right side in the same manner. ' SECOND, apply the Addition Property of Inequality to get all terms which
have the variable on one side of the inequality symbol and all terms which
do not have the variable on the other side of the symbol. ' THIRD, apply the Multiplication Property of Inequality to get the
coefficient of the variable to be a 1; (remember to reverse the in
equality symbol when multiplying or dividing by a negative number, this
is NOT done when multiplying or dividing by a positive number). ‘FOURTII, check the solution by substituting some numerical values of
the variable in the original inequality. ORDER OF OPERATIONS  FIRST, simplify any enclosure symbols: parentheses ( ), brackets [ ],
braces { } if present:
1.Work the enclosure symbols from the innermost and work outward.
2. Work separately above and below any fraction bars since the entire top of
a fraction bar is treated as though it has its own invisible enclosure sym
bols around it and the entire bottom is treated the same way.  SECOND, simplify any exponents and roots, working from left to
right; Note: The J— symbol is used only to indicate the positive root,
except that «m = 0.  THIRD, do any multiplication and division in the order in which they oc
cur, working from left to right; Note: If division comes before multiplica
tion then it is done first, if multiplication comes first then it is done first. ° FOURTH, do any addition and subtraction in the order in which they oc
cur,working from left to right; Note: If subtraction comes before addition in
the problem then it is done first, if addition comes first then it is done first. FACTORING Some algebraic polynomials cannot be factored. The following are meth
ods of handling those which can be factored. When the factoring process
is complete the answer can always be checked by multiplying the factors
out to see if the original problem is the result. That will happen if the
factorization is a correct one. A polynomial is factored when it is written as a product of polynomials
with integer coejﬁcients and all of the factors are prime. The order of the
factors does not matter. FIRST STEP  'GCF' Factor out the Greatest Common Factor (GCF), if there is one. The GCF
is the largest number which will divide evenly into every coefficient togeth
er with the lowest exponent of each variable common to all terms. EX: 15a3c3 + 25azc4d2  10a2c3d has a greatest common factor of 5a2c3 be
cause 5 divides evenly into 15, 25, and 10; the lowest degree of a in all three
terms is 2; the lowest degree of c is 3; the GCF is 5a2c3 ; the factorization is
Sazc3 (33 + Scd  2d). SECOND STEP  CATEGORIZE AND FACTOR Identify the problem as belonging in one of the following categories. Be
sure to place the terms in the correct order first: highest degree term to
the lowest degree term. EX: 2A3 + A4 + 1 =A4  2A3 + 1 CATEGORY FORM OF PROBLEM FORM OF FACTORS ax2+bx+c
(aaé0) Ifa= 1: (x+h)(x+k)wherehk=cand
h + k = b; h and k may be either
positive or negative numbers. Ifa ¢ 1: (mx + h)(nx + k) where m  n
=a,hk=c,andhn+mk=b; m,
h, n and k may be either positive or
negative numbers. Trial and error
methods may be needed. TRINOMIALS
(a rEnMs) (see Special Factoring Hints at right)
x2 + 2cx + c2 (x + c) (x + c) = (x + c)2 where c may be @e'fe“ 5‘1"!" 9) either a positive or a negative number a2x2 _ bzyz
(difference of 2 squares) (ax + by)(ax  by) a2x2 + bzyz
(sum of 2 squares)
asxs + bsys
(sum of 2 cubes) asxs _ bays
(difference of 2 cubes) BN0MAL5 PRIME — cannot be factored! (2 TERMS)
(ax + by) (azx2  abxy + bzyz) (ax  by) (azx2 + abxy + bzyz)
(see Special Factoring Hints at right) PERFECT
cUBEs
(4 TERMS) a3);3 + 3a2bx2  3ab2x + b3 (ax + b)3 = (ax + b)(ax + b) (ax + b) a3x3  3a2bx2  3ab2x  b3 (ax  b)3 = (ax  h)(ax  b)(ax  b) ax + ay + bx + by
(2  2 grouping) X2+20X+02y2
(3  I grouping) a(x+y)+b(x+y)=(x+y)(a+b) GROUPING (x + c)2  y2 = (x + c + y)(x + c  y) yzxzZcxc2 y2(x+c)2=(y+x+c)(yxc) (1  3 grouping) SPECIAL FACTORING HINTS
TRI NOMIALS
WHERE THE cOEFFIcIENr or THE HIGHEST DEGREE TERM :5 NOT 1 The first term in each set of parentheses must multiply to equal the first
term (highest degree) of the problem. The second term in each set of
parentheses must multiply to equal the last term in the problem. The
middle term must be checked on a trial and error basis using: outer
times outer plus inner times inner; ax2 + bx + c = (mx + h)(nx + k)
where mx times nx equals axz, h times k equals c, and mx times k plus
h times nx equals bx. EX: To factor 3x2 + 17x  6 all of the following are possible correct factor
izations, (3x + 3)(x  2); (3x + 2)(x  3); (3x + 6)(x  1); (3x + 1)(x  6). How
ever, the only set which results in a 17x for the middle term when applying
“outer times outer plus inner times inner” is the last one, (3x + 1)(x  6). It
results in 17x and +17): is needed, so both signs must be changed to get the
correct middle term. Therefore, the correct factorization is (3x  1)(x + 6). BINOMIALS
'rHE SU...
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 Real Numbers, ex

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