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Unformatted text preview: CHAPTER 1 Elementary Inequalities 1. Taking squares of or square roots of positive numbers preserves the order:
O<a<b=>a2<ab<b2and0<¢5<¢3 2. Triangle inequality:
ﬂ7+yl S l$+lyl; may 6R. Equality holds when 3:, 3,; have the same sign. 3. AGM inequality: Mary 5 a: J; y for may 2 0 real numbers. Equality holds when a: x y. Corollary:
22:3; 33+y sci—y Sx/wys 2 . In order to prove these inequalities we have used the properties of
ordered fields (page 17). When proving inequalities, you must justify each step by using other
(accepted) inequalities; you may be required to avoid ”wordy" proofs. 1 Sets 0 Deﬁnition, notation (listing, set—builder notation); elements, mem
bers; m E A, A g B. 0 Must use 6, g correctly (example: A E B is incorrect)
0 Empty set, proper subset, power set. a Equality of sets: A z: B — for ﬁnite sets, can compare the elements and verify if they are the
same — for inﬁnite sets: A g B and B Q A implies A a B 0 Set notation: 7 even numbers; odd numbers
— closed1 open intervals — maximum and minimum of a set — Cartesian product of sets — k—tuples; ordered pairs 0 Set operations: union, intersection, complement, difference; disjoint
sets. You must be able to use the deﬁnitious of these operations to
prove equality of sets. 0 Properties of set operations: (can be found in exercises) — Commutativity of unions and intersections: AUBzBUA;AﬂB:BﬂA — Associativity of unions and intersections: (AUB)UC = AU(BUC)
(AnBjnO = Aﬂ(BﬂC)
— Distributivity:
Au(BnO) = (AuB)n(AUC)
Aﬂ(BUC) = (AnB)U(AnO) — De Morgan’s laws: (AUBY AcﬂBc
(AoB)° : ACLJBc o Venn diagrams can be used to depict sets and set operations; accepted
way of showing set relations. K c To show equality of sets: A D (B — C) % (A H B) i (A 0 C) — draw Venn diagrams, clearly indicating each set used in the equal— '
ity. — use statements about memberships: that is, let SC 6 A H (B — C) ,
use the deﬁnitions of the operations involved, show that LU E
(A F] B)—(A D C). This would show that Aﬂ(B — 0) g (A H B)—
(A n C). Likewise, show that any 3: in (A F] B) —— (A (1 C) also be— longs to Am (B i 0). Sometimes, we can simply reverse the steps
to show the other inclusion. — use the properties of set operations (if suitable) 
e ' _ C Q
AMBnC) w k: (mgn ﬁne) — transform into} logical statément and use a truth table. (MM—(mo = (Mewtrmc) H \ ‘a
: (msynmc'ucLV—dx Gem wx
t (WaistbﬁﬁmmnﬂéWWW? {(A0 RC) MS) U fmtbnc") <— asmdatmlv
((5 . and stmuicdm’l) 5 = d3 U Arnie«Q
‘7:— ‘Pvﬁ(B—Q> Functions 0 Deﬁnition 0 Domain; image: Im(f) = {f(a:) 3: E D (f)}, range 0 Finding the image of a function: from the expression of the function,
make a guess on what the image is and call it 5'; then Show lm (f) = S. — To show that lm(f) Q 8’, let y = f(m) (e.g. y : x/(l + [3], show
that y (E S, that is x/(l 4— [22D 6 S). — To show that S g1m(f), let 3; E S, Show there exists a: E D (f)
such that y = f (3:) Solve for :1: in terms of y (in order to show
that such an :3 exists), and use that y E S to validate some of the
steps. Show then that the expression you found for a: does lead to f($)=y — If you use a different reasoning to show that Im ( f) = 8, make
sure that your solution is actually valid and is not'incomplete. — Graphical proofs won’t be accepted as the sole proof; graphs can
be used to guide your answer. o Well—deﬁned functions: meaning __ .to show that a piecewise function is welldeﬁned: (:51), a: 6 I1
“5"”) ={ ice), :1: e 12 — g and it must be deﬁned on 11 and I2 respectively — if [1 F] 12 # (25 (the intervals have an overlap), then on 11 H 15,?
must equal h — graphical proofs will not be accepted “Carﬁll Mil ‘ o Bounded set, bounded function; unbounded set and function e when proving boundedness, you must use justiﬁed inequalities; a
wordy argument will not receive full credit.
0 Increasing, nondecreasing, decreasing, nonincreasing, monotone.
— when proving these preperties, you must use the deﬁnition, not graphical or differential calculus methods. 0 Identity function; ﬁxed point of a function 0 Equal functions o Inverse image of an element: for f : A a» B , the inverse image of y E B
is: . 13(9) = {56 6 Al f(33) = y} o Inverse image of a set: me) = {x e A : HS?) 6 g}. o For functions f : A x B —> 0, the inverse image of a point c E C is
called a level set; We) : C — used to ﬁnd largest values for 3: and y satisfying inequalities of the
form: UiD'Wéjlr)
l$ + yl S M { lasyl S N — this method can be used to iind maxima for :c and y, if a: + y and
any are bounded. (problem 1.49 e) CHAPTER 2: LANGUAGE AND PROOFS 0 (Logical) Statement: sentence that can be classiﬁed as true or false: P(:c) : 3: 6 S
S is the universe, :1: is the variable 0 V2: 6 S, P(1L‘)Z a: is universally quantiﬁed. Negation: _' (V1: 6 S, P(m)) is Elm E 3, 113(23). Finding such an 93 means
to ﬁnd a counterexample. o 33: E 3, 13(55): :1: is existentially quantiﬁed. Negation: j (EWx E S, P(:::)) is V2: 6 S, "P(:c). This usually involves
saying: let x E S, and showing that 3: satisﬁes P(x). 0 Other expressions for universal and existential quantiﬁers: table on
page 29. 0 Order of quantiﬁers:
Va; 6 S, Ely E T, P(a:,y) This means that 3: is initially arbitrary, but for each choice of ac, there
exists a y that works. The value of each variable is chosen indepen
dently of subsequently quantiﬁed variables. 9 3:1: 6 S, Vy E T, P($,y) This means that there is an m in S, for which every 3; works. 0 When can one switch the order in a quantiﬁed statement? Dealing with
V$ E S, Vy E T, P($,y) and 3:1: 6 S, Ely E T, P(:1:, y) 0 Must know how convert statements to logical statements and negate
them both in logical notation and in plain English. 10 Compound Statements Logical connectives: 7, V, Y A, =>,<=> Truth table for each of them P 22> Q: hypothesis, conclusion, condfitional, converse, contrapositive vacuously true; quantifying over empty sets logically equivalent; tautology a conditional is logically equivalent to its contrapositive; other logical
equivalences Remark 2.20 Logical connectives and memberships in sets. 11 ...
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 Spring '07
 GHEORGHICIUC
 Math

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