1969-01-23 — Page 23

有教備 華 黃三第張六第 日六初月二十年申戊醫璽 WAH NG TAT

丸日僑華 四期星

1969

文中學會考試題預習

„Aasociative law and Denorgan's lawy

Commaltative law 1

中中會考 數學科

日三廿月一年九六九一层公年八十五國民劑中

(接第六張第二頁)

(†=) 我們常用的同系函數艺本間伍公式,有下面幾種:- 工逆數關係(

·喬仲雙

現代數學科

Intions:

(PZ)

MODERN MATHEMATICS (12)

The necessary and; sufficient condition for

13x - 18 19 x 6. Let p be: "3x Ȥ 18" and q ba

6.Then, por p=1atore)

(5) A is a subset of 8 if and only fif*A A BU■"0.

Let p be "A fa a subset of Band a be "A (B")

—P.

Flyg or p = qorq = P (c) Jack goes to swim when it is summer. Whenever

19 summer, Helen stays at home.

Let p be "Jack goes to swim, q be" it is summer," and 'r' be "Helen stays at home. Then

(d) "It is not true that he is brightiena diligent

and "he is not bright or he is not diligent." Let p be "he is bright," and qbe "he is dilig

Then

B) A

FFFETT

FFFFTE

FTTFT

TETETT.

FT FT FTF T F.T

Hence it is proved,

FFT

TT

FTTT

TE

TF F

Hance it is proved,

3. (a) Il n is a multiple of 6, then it is a multipla 3. If n is not a multiple of 3, then it is not multiple of 6.

Let p be "n is a multiple of 6," and a be "n is a multiple of 3. Then the statements are P and non. but pa=N Therefore the 2 statements are equivalent.

ir a 1s even, 17 19 divisable by 2. n 13 not even or n is divisible by 2, 2

The symbolic forms of the two statements ar

pg and vp vg. But p→ 9 Hence the 2 statements are equivalent

"If James playa footba II, then Feter plays piano, Peter plays piano, then Lily sings. Therefore, t

Jame a plays football, Lily singe.

The symbolic form of the statements is

(p => QA8 — r })→→→→→ (p->r), which is a tautology. Hence the conclusion is valid.

Relations of composite propositions (continued). (b) Properties of the equivalence of propositions.

and the Era Melism between equality of sets and equivalence, of propositions:

In algebra, we have the equality of numbers; in geometry, we have the congruence of polygons and the similarity of polygons; and in logic, we have the equivalence of propositions. All these are called equivalence relations. Every type of equivalence relatim obeys the same laws, namlj the reflexive low, symmetrical law, transitive law, and substitution law. Hence if P(p,q,

elp,q,r,...), than

(1) They are reflexive P{0.0.r...

P(p,q,r,...).

(2) They are symmetrical: Ir rip,g,r,..

ulp,o,r,...) then Q(p,a,r,.. = P(p,q, (3) They are transitive: If P(p,q,r,...) =

elp,q,r,...) and Qlp,q,r,...1= R(p,q,", (then P(p.q.,...) = £{p,q,t,...). One can be replaced by the other.

Since, the structure or soolean algebra applies to propositions as well as to sets, the parallelism of the properties of both must be a natural consequence, Each law of sets concerning intersection, union and complementation is applicable to the algebra of propositions. We can obtained a table of the laws of algebra of propositions from the table of the algebraic laws of sets by just replacing the symbol of union by symbol of disjunction, symbol of Intersection by symbol of conjunction, symbor or complementation by symbol of negation, symbols of universe and the empty set by T and respectively, and the symbol of equality by symbol of equivalence. Thus for propositions we have the Idempotent laws,

(a) PVP=P. PAP= }

Associative laws

(b) (PVQ}VR=PV(QVB) (PAQ) AR=PA (QAR)

Commutative laws

(c) PV QQVP [PAGE QAP

Distributive laws

(d) PV(QAR)= (PVQ)A ~PA(QVR)=(PÅ Q}V,

Identity law? ✨

(e) PV PET.

PVT=T Complement

PVR (PAR)

Demorgan's laws

PARER PMEN

NIPVQ) =~PANO ~(PAREMP

Employing these laws we can simplify

propositions or prove an equivalence of

propositions just as we deal the same with sets. Examples:

(1) Simplify: (PAQ)V(PA@()

„Solution: (PA Q)V(P^^ q).

SE PACQU~~ Q).......Distributive law

ERNAL.......Complement law

**Identity law

[(2) Simplify: (PVQVNB)A (MPA~QAR) (Solution: (PVQU~R) A (~PAMQARI

=[~ (PVQ)AR] ^ [(PVQJVMB]

= [~(PVQ)^B^(PVG)] V

[M(PVQ)AR AMR, Distributive low.)

{~ (BVQ)^(PVQ)) A_R]V

[~{RV=Q) A (RAR)]..Commutative law and

Associative law

"[(FAR]V [7(PVQ)]. Complement law)

PVP P

Identity law,

Idempotent law,

It means that the proposition is Falsen (3) Prove:/PVQVMR V (PAQAR) is true,

~ PVM G V N RV (PAQAR)

= N(PAQAR) V (PAGAR)... Demorgan's law

*

COS O ARCO

tano coto

工商數間係:

Tan) — sinė,

COL 8

COLO

Complement law.

I

cae o

tano+

(莊生图解)

Johnson's Diagram

其間係可用在生图解,以助記憶

(在六角形名頂中第一行為弦

(

(4) Prove: PV (~FAQ) = PVO

Proof.

PV(~PAGE(PV~P) ^ (PV Q)

Implication

.....Distributive law

TA (PVQ)...Complement BW PVQ Identity law.

There is a second important relatim between composite propositions. This is called implication. P(p,q,r,.....) implies Q(p.q.r,...? if one of the following conditions is satisfied, (a) P(p,q,r...

Q(p,q,r,...) is a tautology

(b) P(p,q,r...) 1~ Q(p,q,r,...) is a contrad -

iction

(c) P(p,q,rumAPTER.../ 30 a tautology,

Notice that if the third condition is satisfied we say P(p,q,r,...)' implies Q and conversely, denoted by

FAQ or Pen

In all cases, except when condition (c) ia satisfied, the implication relation is

reflexive, antisymmetrical, and transitive. That

(a) PP, (b) If PQ then Q does not imply P

(c) If P⇒ QAQ⇒R, then P⇒⇒ R.

In order to verary that one proposition Implies another, we have to follow some rules. They are known as the rule of modus paneus (or law of detachment), rule of substitutim, and transitive law of implication

(a) Rule of modus, paneus:

Proof :s

TTTTTT

TRIPE T

FEETT TE

FFET ET

Since the conditional is a tantólogy,

PA (P-Q) implies Q

ibile” of substitution

If two propositions are equivalent,

either may be substituted for the other Without altering the truth of the

(e) Transitive law of implication:

(P→ QAQ→→R) = (P > R.) Proof:

TTTTTTTTTTT.

TITP

TFFERTT T

TFF.F

T FP

ET T

FTTTTTT I

FTTTF FT

F

FTET FTTT

FTT

FTFTFTFTPT F

Since the conditional is a tautology;) the implication is verified.

(d). Worked azamgi86, İrove the validity or îna

following.

(1) If Jona dusen't like candy, then apples are) fattening. If apples are fattening, then Jane will not eat then, Jane will eat Appian or she will become 111, Jane is in fact 111. Therefore; John Likes candy. Solution:

Let P be Jonn likes candy, a te apples are fattening, r be Jane eats apples, and s be Jane becomes 111. Then the sentences can be, symbolized as below.

The validity of the propositim is tested as)

A{q— ~r}}^{rVe}A_S (~p—> ~r)^(rVs)^ s⇒ ... Associative, law and transitive law of implication.

Rule of substitution

{~ p-> B}A_8 →

essociative law and} transitive law of Implication.

''SA (~£6→→→ pl..................... Commutative Law and

rule of substitution

The last camót imply p, because [s ^ {ws →→→ p)] →→→ p is not a tantólogy

(2) Is I go to camp, then Benry goes to visit grandma and Jim goes to summer school. Henry does not go to visit grandmother. Therefore,

I do not go to camp. Solution.

Let p be I go to camp, q be Henry goes "to' visit grandma, and r be Jin goes to summer school. Then the sentences above can be symbolized as follows.

p(q Ar) A ~ (=>> ~

The validity of the proposition is testõu as below.

Rule of substitution.

Rule of substitution' and distributive law.

Rule of substitution and associative Law Rule of substitution and complement law.

豚) 以下依次笃地函數,割函数,又左方為正函數 (正弦正切,正割),右方為鲜函數。

(2)在对角线上士函數,有運數闔

(3)各边上, 以一函數除其鄰近函数可得文商數即 為第三個函數(不論順鐘向或反鐘向)例如 ← C«CO (7)^*) (4)頂奥在下方之三角形中(图中配有陰影者) 其上方 两数之平方和,等於下方正數之平方

第十二次預習題

(1)証明下刺坦学式

(a) (1+ sind + cos 8 ) === 2 ( 1+ king) (19 001 ()

(b) (I+sica tan 8) ( 1 + tan ß sin3()

·(1+ acß tam 8) (1+ tana sin ()

(c) (1+ tanx)2+ (1-tank)a

(d).

(1+ ctx) + (1-cotx)2

COS - CON 20

sinX+ sin tj.

cold + col's + cos"! + cos "(2+5+V}

= 2+2 001 (x+8) cos(B+9) cor(Y+α) [$k$Y]

(2)一索繁统二輪 其半径各為1呎.3呎 两輪中心相

距5呎 求此索之長

(3) - 园半径6时 由园外一奌至此国作两切线着两切: 线成 55°角求两切线典其丽吏苏孤钙成之面積 (4) - El (1) a) (13) 1) < § 1. 15 8:41 (142) 3*

求上 至2°C間在斜高上之距離(园錐体積= (5)~塊三角形板各边為5吋,6吋.7吋在水平枱

上沿最長边轉 30°求其子对之頂真(2)典枱面之驱

(6)蒋经軌跡之長度。

(6)設如右图:

△ABC為正三角形“AD=气 AB

試証:AP 约爲o园内接正 七边形的一边

(用數值証明)

Identity law and rula

of substitution.

Wping does implrep, because

(~pang) wvp is a tautology,

(3) Prove that (^^ B→→→

Proof

T TTT

TFF TEF

IT T FTT F

TETT

Since the biconditioni the implication de valid.

FTT

tentology

Prove that (pe->q} = [(p_A_q} V_{~PANG)

Proof

TTTAT PEN TPP F

FFT FTABLE

FETT

Since the truth Eables of pensq and (p ^ q) V (~paq) are identical, they are] equivalent.

Nork for the week

I. Test the validity of each of the following argumente : 1. It is not trus that if prime numbers are odd, then

groups have identities.If groups do not "have identities, then there are nongommutative rings. Also if there are non factorable expressions, there must be non-commutative rings. Therefore there are no nonfactorable expressions.

If rich people are unhappy, then money 18 SVIL. If medical care is expensive, then money is not evil. As a matter of fact medical care is? expensive and rich people are unhappy. Thereror medical care is both expensive and inexpensive, Price rise only when wages rise. If employee fringe benefits are not increased, wages rise, and conversely. Since prices have risen, it follows that employee fringe benefits have not been increased.

II. Test the validity of each of the following:

1. ~(AAB)~Y~3

2, ~(AVB) 4~A ANË

3. JABAN B) GID MA

AVB

h