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Unformatted text preview: itacl‘ﬂ l C MW 5 {78 {0‘59 1.6.12’b. Proof By Contradiction. [The original statement is: Vx,y e R s.t. x ¢ 0 and x is ratn'l: y being irratn'l ——> xy is irratn'l.]
Let x,y e' R be such that x ¢ 0 and x is rational. Suppose, by hypothesis, that y is irrational. For the sake of contradiction, suppose that xy is not irrational. So xy is rational. Hence there exist a,b e Z such that b ¢ 0 and xy = a/b. Since x is rational, there exist c,d e Z such that d ¢ 0 and x = c/d.
Since x ¢ 0, we may divide xy by x to produce y. This gives us y = (a/b)/(c/d) = (ad/bc). Since x ¢ 0, we have c ¢ 0. Since b¢0 and c¢0, wehave bc¢0. Note that ad,bc e Z. I Hence y is rational. But this contradicts a hypothesis. Therefore our assumption that xy is not irrational is impossible.
Therefore xy must be irrational. I As noted in lecture: c. Vx,y E R s.t. x¢0 and x is ratn'l: y is irratn'l <——> xy is irratn'l. (1. Direct Proof of Converse. [The reformulated statement in Part (a) was:
Vx,y e R such that x ¢ 0 and x is ratn'l: xy being rat'n'l —> y is rat'n'l.
So we are being asked here to prove: , Vx,y e R such that x ¢ 0 and x is ratn'l: y is rat‘n'l —> xy is rat'n'l.] Let x,y e R be such that x ¢ 0 and x is rational. Suppose, by hypothesis, that y is rational. Since x is rational, there exist a,b e Z . such that b ¢ 0 and x = a/b.
Since y is rational, there exist c,d e Z such that d ¢ 0 and x = c/d.
Note that xy = (a/b)(c/d) = (ab/cd); here cd ¢ 0 since c ¢ 0 and d ¢ 0.
Note that ab,cd e Z. ' 56 Y}! irrational *7 5/22. We give a proof by contradiction. Suppose that we don’t get a pair of blue socks or a pair of black socks. \ Then we drew at most one of each color. This accounts for only two socks. But we are drawing three socks Therefore our supposition that we did not get a pair of blue socks or a pair of blank socks is incorrect and
our proof is complete. ‘ , (24. We give a proof by contradiction. If there were at most two days falling in the same month then we could have at most 2  12 = 24 days, since there are 12 months. Since we have chosen 25 days, at least three of
them must fall in the same month. b9 32. We give % 44. We want propositional functions ,5 2M and that (iii) implies, (i). That will sufiisc:
Then :r/2 = p/(Zq), and t ,th ’ébl 1w t 0 implies (a), that W) implies (“'0’ th q a 0 ’ 'Wl ' d
= Where P and q are Damages 2 = where P an q
For the ﬁrst; suppose that m  ti/gqers With 2‘1 ¢ 0' For the second, suppose m)? mind tli'iéqis rational, Since
alsincepandZQﬂIe1Il 3 f1=<6p)/q_.1=(6pq q _ s
’ = 2 , SO ‘3 d axe mteger
0. Then :1: ( Pl/‘I h last suppose that 3.3 _ 1 = p/q where :0 an ‘1' rs With
With q % 0" For t e a 31 since 13+ 4 and 3‘1 are mtege , . q + 1)/3 == (p+ q)/(3q), and this is ration direct proofs that ( ration
areintegers with q ¢
6? _ q and q are integers
with qyéO. Then :5: (p/ 3a?” 4vl 6. The number 1 has this property, since the only positive integer not exceeding 1 is 1 itself, and therefore the sum is 1. This is a constructive proof. 138. a)16——>8—+4~+2—+1 d. 36. a) Since 12 = 1, this statement is false; :3 = 1 is a counterexample. So is :r = 0 (these are the only two counterexamples) .
b) There are two counterexamples: x = x/i and z = —ﬁ. c) There is one counterexample: :z: = O. P and Q that are sometimes, but not always, true (so that the second aking one true and the other false. For mean that a: is an even number (a multiple of 2) and (2(2) to mean that a: is example, we can take FCC) to = 9 shows that V$(P(z) H Q(x)) is false. 3. multiple of 3. Then an example like a: = 4 or :1: 1.3.50. The added instruction on the Part TH of Section 1.6 posted document
said that students who believe that the two expressions are not always logically
equivalent must: Specify some particular domain of discourse D and some
particular propositions P and Q such that the two expressions evaluate to different truth values. General Cnterexmp: Initially take the domain of discourse to be D := {x1, x2}.
This means that the two given expressions become (P(X1)AP(X2))V(Q(X1)AQ(X2))
and (P(X1)VQ(X1))A(P(X1)VQ(X2)) respectively. Let p, q, r, s repectively denote P(x1), P(x2), Q(xl), Q(xz). Then the two expressions are (pAq)v(r/\s) and (pvr)/\(qu). To show that two logical expressions are not logically equivalent, there is no
need to compute the entire truth table (which would have 16 rows in this case).
One only needs to find the same row in each table which yields different T/F
values. After some thought, we see that one combination of four T/F values
which works here is: p = T, q = F, r = F, s = T, since this yields (pAq)v(r/\s) =
’ FvF = F but (pvr)/\(qu) = TAT = T. Hence the two given expressions cannot be logically equivalent. Speciﬁc Counterexample: Take x1 = 1 and x2 = 2. (So D = {1,2}.) Let P(x) be "x S 1" and Q(x) be "x _>. 2". Then P(x) and Q(x) take on the pattern of T/F values needed above for an rn  b) We can rewrite y2 — :c < 100 as y2 < 100 + as. Since s .  quares can never be ne ative ' '
is, say, —200. This :1: provides a counterexample. g ’ no suCh y eXlSts If x c) This is not true, since sixth powers are both squares and cubes. 'Irivi _ __ al counterexam les would ' 1
m—y—Oandm=y=l,butwecanalsotakesomethingﬁke33:27 and p mcude y=9,since 272:35=93_ 1.6.12’e. , (Re)Prove or Disprove each direction: [The correctedbyemajl version is as in Part (c), but without the "nonzero":
Vx,y e R s.t. x is ratn'l: y is irratn'l <—> xy is irratn'l.]. l (i) Counterexample to: _ l _ [Vx,y e R s.t. x is ratn'l: y is irratn'l —> xy is irratn'l.] ‘ Consider x = 0 and y = 7t. Then x is rational and y is irrational. However, xy = O is rational and is thus not irrational. (ii) Proof. [Note that the contrapositive of
Vx,ye R s.t. x is ratn‘l: xy is irratn'l —> y is irratn'l is
Vx,ye R s.t. x is ratn'l: y is ratn'l —>_ xy is ratn'1.] C The direct proof of Part ((1) can be reused here after we note that the hypothesis
of x ¢ 0 there was never used in that proof. I ' ﬂ 34. No. This line of reasoning shows that if V2232 — =2 :12, then we must have :1: = l or :1: = —1. These are
therefore the only possible solutions, but we have no guarantee that they are solutions, since not all of our
steps were reversible (in particular, squaring both sides). Therefore we must substitute these values back into
the original equation to determine whether they do indeed satisfy it. ﬁ' 38. We must ﬁnd a number that cannot be written as the sum of the squares of three integers. We claim that 7
is such a number (in fact, it is the smallest such number). The only squares that can be used to contribute
to the sum are 0 1 and 4 We cannot use two 4’s, because their sum exceeds 7. Therefore we can use at a 7  most one 4 which means that we must get 3 using just 0’s and 1’s. Clearly three 1’s are required for this,
5 I 
ber of squares used, to four. Thus 7 cannot be written as the sum of three squares. bringing the total num It .9... 221 W W “(t ‘i 1.6.96. True. . Proof. Let 110 e Z.
The hypothesis is that n0 is even and n0 is odd.
According to Theorem 1.6.2 of the lecture notes, this hypothesis is False. Hence this theorem's implication is vacuously true. I 1.6.983. Each of these statements would have been called "propositional
functions" in Section 1.3. We'll see in Section 2.3 that one should never refer to
a function without at least implicitly indicating what the domain of the function
is. (Unfortunately the concept of "domain" is not introduced in Section 1.6 until
p. 34, and then only in the context of universally quantiﬁed statements.) The domain for each of these propositional functions is R. b. The propositional function of Statement ( 1) is explicitly existentially
quantified within all of R, and implicitly this existential quantiﬁcation within all
of R is (correctly) carried forward to the propositional function of Statement 2.
The only particular value for which these two propositional functions hold is x0 = 2 (= c). The error occurs in Step (21/2): This step is attempting to guarantee that
Statement (3) also holds for at least one value within R, by manipulating
Statement (2). For this approach to work, that "at least one value" would also
have to be x0 = 2 in Statement (3). But this would mean that within this
instantiated—tox0=2 environment, Step (21/2)'would be dividing by 0. There is
no law stating "equals divided by 0 are equal". So it is not valid to claim
"There exists x e R such that x + 1 = x + 2", which is what Statement (3) is implicitly claiming. I ...
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