頁二第脹四第
日五十月一十年申戊感婆 WAH KIU YAT PO
報日僑華
五期星
日三月一年九六九一曆公年八十
英文中學會考
【題預習 |
represent the statements and is a symbol
for 'and', a convenient way to determine the
現代數科學
(九)·李
statements is denoted by pɅq, where p, q
truth value of a compound statement is hy the use of a truth table.
Truth table of a conjunction paqi
一九六九率中文中学會考試題預習
鄧炳恩
物理科(元)
MODERN MATHEMATICS (9))
16. Solution
(1) or 158 T.V, viewers questioned, 94 watched
BBC-1 and 85 watched BBC-2, How many watched both?
Solution:
D
Venn diagram of the problēm
Let I, Y be the 2 sets of people. Than nYur)n(K) + n(i) - năni) 15894 85 = n(XNY)
n(X(Y) - 179-158
21
Answer! 21 of them watched both,
(2) in a form of 36 boys, every boy carried a pen.
20 had blue pens, 28 had black pens, and 5 had red pens. 7 of them carried pens of any two colors.
How many of them carried pans of all three colors?
Solution:
Let X 1, and Z be the three sets. Anon n(XUYUZ)-n(X)+n(Y)+n{2) — 3.7+x(XNINZ)
36 20 28 + 5 - 21
36
32. (INYNZ)
4 * n[XNYNZ) ;
(XOINZ)
Answer: There are 4 of them who carried pens
of all three colors.
(3) Statistics show that more than 70% of men drink
and more than 60% of më moke. What can you say about the percentage who do both? Solution:
Let D and S be the sets of people who drink and
smoke respecitvely. Then
n(DUS) = n(D) + n(S) – n(SAP)
100-70 60 – n(SND).
n(SAD) - 130 – 100 . ·
30
inswer: At: most 30% do both.
Bing
Boolean Algebra Set
Boolean Algebra defined
Propositions, and
ical
A mathematical system is a system consists of a sát of objects, and one or two operations. satisfying certain conditions. A Boolean Algebra 13, any mathematical system which consists of a set of
set of elementa, closed under two operations denoted by and U, satisfying the following
conditions
(1) Both operations are commulative (2) Both operations are associative
(3) Each operation is distributive over the other. (4) For each operation there is en identity element.
For each element A, there is an element A, such that AflA - Ø and AVA - if, where § is the empty set, and MA is the universal set.
Under this topic, you are to learn how Boolean bagra applies to suts, to propositions, and to electrical circuitings. Since we have had our lesson. on the algebra of sets, we are not going to study it in detail. However we shall consider in detail the algebra of prepositions and algebra of electrical cirauitings.
Basic propositions
Statements propositions).
Verbal assertions
BOL statement is a verbal aggartion, sxpressing a true or a false idea but not both. The truthfulness or falsity of a statement is called its truth value.
Trees are animals,
False statement.
True statement
not
Human beings are mortal. Mathematics is easy to learn.
Watatement.
(4) You must put on your cont.... not a
statement,
The last two are not statements because they
neither true nor false.
Some statements are composed of substatements and various connectives. Thesë are called compound or composite statements. An important and fundamental property of a comound statement is that its truth value depends upon the truth or falsity ofsits Substatements, and the connmotives used in the connection of the substatements. In the algebra of propositions we shall deal with the modes of forming propositions. Basically there are five kinds of propositions which will be: discussed subsequently. They are conjunction, disjunction, negation, conditional, and bi- conditional,
(b) Conjunction
When two statements are joined by the word and we have a compound statement called conjunction.
Examples:
(1) She is a teacher and I am a student, (2) The sky is clear and the moon is full,
Symbolically the conjunction of two
T
F
T
The table infor
us PA is true when
both p and q are true, otherwise it is false. The symbol 'A ! can be used to define the intersection of two sets A and B. That is.
AÏB * { XIX E A ALEB}
It is instructive for us to compare the
truth table of pag with the membership table of
AnB so as to show how Boolean Algebra is applicable to the algebra of propositions.
Membership table of AB
Wed Whed
The table shows that if any element x belongs to
both A and B it belongs to ANB, otherwise it is not in AB. This is the parallelism between the conjunction of two statements and the intersection of two seta.
(e) Disjunction
When two statements p and q are connected by the connective fort, we have a disjunction. It is denoted by pq. In mathematics or
means one or the other or both. That is or has the meaning and/or in mathematical use.
Truth table of a disjunction p V q:
P
The table informs us that the disjunction. is true when at least one of its substatements is true, and that it is falso when both substatements are false
Notice that the symbol V can be used to define the union of two sets. That is AUB -{X}Z E AVLEB}
Membership table of AUB:
B
AUB
www.ww
wwwas
Obviously there is a parallelism between.
the disjunction of two statements and the union of two sets.
In some occasions we need a symbol for the exclusive use of 'or!. That is we need a symbol for the idea to mean one or the other but not
both. The symbol for this purpose is pq, which
is a symbol for an exclusive disjunction.
Truth table of an exclusive disjunction:
Examples:
T
PX9
(1) Are you apt to get an A in mathematics or English?
..or here is used inclusively, meaning pVq. (2) Do you want ketcheep or mustard on your hamburger?
For here is used exclusively, meaning pg.
Assignment for the week.
In each of the following, construct the conjunction and disjunction of the set of simple statements,
Indicate the truth value of each.
(1)(The first prime number is 2.
(8 is not a multiple of 2.
(2) (3 x 5 - 14 + 1
moat fire engines are rea
6 is not a multiple of 3
(3) ALL eagles are reptiles
The United States of America is & dictatora
2. Make a truth table for the conjunction AABAC
3. Make
truth table for the disjunction pv gut.
Fahip
D. 簡單機械及物性學
.問答計算題題解
『解(設 LM為棒長=20
N 為棒與液面
之炎点
G & M N Z
中
3為此棒單位体積之重 駟為液体單位体積之量 又為浸入液内之棒長
A-为棒之橫截面積
T為紬縄之張力
#2-A2RW
排去上液体重= A.x手00
#7# Fy
= 0
TTAX W za Aw-----(( (1)
·(AX $N) AC
(2
AB LG
AC LG
x2-4ax+3a2=0
a. A.
☆ x=3a
AB
za-2x
必小热棒長故x=a 出液面之捧長為金棒之半
代入り得
TTA a w = 2a Aw
T == A aw = £#£. 3A w= 2
2解設此圓球之体積為V-
則此圓球之重量為 VP 此圆球在水中之失量為
d = vp - Vx1
此圓球在液体之相對密度
d=
dvp - dvd vp-dvp apa = ap ap
p dp + d'
Vpl - up
ap-dp
-dp = dpi
3.解設水銀密度=13: 61
V為氯泡在池底時之体積 衣為池深
亦即静
)中失重為
p為池底之压力强度=(+30)吋水
銀柱鳥
必為池面之压力强度=30吋水银柱高 作為氣泡在池面之体積=10V 依波義耳定律得:
(13.6 +30 V = 30 X 10V,
d = 13.6 X (300 ~ 3.0)
13.6 × 270,=3672 (25)
··306 (R) 深為306呎
4.解設氣压計之截面積為A平方厘米
端空氣之压力為p
至米水銀柱高時管
1厘米水銀柱高
=(85-76) A±** =9A * * * 大氣压力為他時,氧压斟水银柱高75
上場室氣圧力為れ客積為
依波義耳定律得
1X 9 A
= 10A IS #
p=0.9(厘米水银柱高)
= 75+ 1 = 75 +0.9
75.9 米水銀柱高)
二正確三大氣圧力為75.9重米水銀柱嵩