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graded hw 4 - Math 300 Fall 2011 Homework 4 Due Thursday 13...

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Unformatted text preview: Math 300, Fall 2011 Homework 4 Due Thursday 13 October, 2011 ((9% (1) Prove that p ==> q and -:q ==> up are logically equivalent but that 03‘ W“ p => q and q ==> p are not logically equivalent. ‘ S l (2) Prove that -i(p A q) and (p /\ fiq) V pp are logically equivalent. (3) Prove that p (=1) q has the same truth table as (p => q) /\ (q => p). » r1 e 1n sym 0 1c an ua e an prove t e s a emen a per ect square IS 1V1 e 4W‘t' bl'lgg d htt t‘If f 'd"dd by 4, the remainder is either 0 or 1’. ~" (5) Prove by contradiction that for integers a, b, c, if a2 + b2 = c2 then a is even or b is even. «6) Prove by contradiction that for real numbers a and b, if their product is irrational, then a and b can’t both be rational. “(7) Write in symbolic language, determine if true or false and negate ‘Foriavery positive real number a: there is a positive real number y less than a: such that for all positive real numbers 2, the difference between 2 and the product of y with z is non positive’ . 1 “WWW P10111113}: p =>01 and 'lqfi 11) e111 mmm 11311)111511211113 3011 111.131 1373M}, and 69>? @111 11’): “$135111 quwakzm 01— ”IP- 101 @13:[email protected]’*7PQ1:P T T “T \O T 1: T F F 1/ / F T F '1' T O p —.— \ T "T .1-11'1’“ T 'Wmmmsw 9P0; Q1111 1q=51p m_1g211111;11‘%9+m Cormwcm CQVHeE CNN W131 Mm mmmuw Eqvwawm , HWQN 11111111111111.1313qu 1) 22% am 05>? £1211 1101131151. so mem 9.311 191 c11awnt11m M11. mum \0C3\§a1\1 equawflh 2 mew 111011 11111 “111) Aqfl 81111 (p A1111 1/ 11? 31121631131111 111111va1c1n1 P ‘1’“1éP M11 1P 191 LPMOD V 11> . T F F F F T F T F 1’ T F T T '1‘ F ‘T F F T T T T “11151 11111133111101 'IWAOQ 3111 cp A1011 V 112 111.1 1611111131.,831112 Conausm. 16.11119: mum 111111 1m $111 1111111111111 1111111111111. 3 11101103111011 1211111 111211 1:.) i=1 01, 111151111 3311121111111 41311111 as wacf) :1 @155?) P 11 P @171 (Pm/1 (0115137): T +TT>T TF_F _P FT 1: F 1F? 'r '1 1.1111111111141111 Par p my 1111 C9 am A 1991)}. 111. 11.121111131129111 10111.1 .11 . Can 191 mm W 1'11»), @111 Wigwam wvaWH 5;: 1111110311151 11 :1 13:11:11 $112313; 15 1111111111}, 3.531111 WNW '13 13W 9 .111 \. / . . 1 5111110111 18143119031 W'X Qi‘LW 99:31 311117151. 1.111. 111.111 1<111,111e21.01=1,1_v111=111" H mm 30112111: 111111 1.1111 11111111111111 13 0. 11011111 1111111112121 1611 L1 11111011 A1120 1111121 1141111, 51121 A1‘= (n+1? +11?~1111+1-+1113+511+1 411% e; 41n12¢+2><1= 9112 11111111 Z114 ,Owd ‘1 1n {- . 31.093603 3008:0100 Imm 0000 saw: me? 0.st 0mm an, H . , '0 New 00200 wacvkawxowxmaw. w '0 1s cod ,_ 0m ’x 00$ 0 mmamwmt W . 00mm 4.000 @0800 @3300 \n’rmp‘s ,wmmw mm?- ampqsmflkmm 90000me m0 mm 0,00 , '0 0"“0" =0 Mun 0x3 waWMSQW\ , ‘Wmma 000000 0= rum b=uu 000% ai+b1=c q” L1Kfl3‘rLUn)‘ = £00th w +4}?wa ' LKK“ 301100 +2 :E 3000/ O , 0 W000 W momwa 00006000020400 3 v00 00M d- i 7 MW \¥\3WWM\¥ m0 0 000 000 CM 0 0. \qu wwwrbmxmw nomvflcfiaaodd 0000,0000? 0000 02 0 @600000000 \0 , t9 Wfilml mam WWW W££0f 0m 000000 a 0000 W 100! 000 ’O 13 \[email protected]’\ ”MM. 0 0mm Q30“: 09,0039: mm “A 8070062 0.0061 09%. amb—le 0,110,001": (000% 11’; z? . . fimxmwel +00 “0350 000m 0 00000 030mm ’03; a m mamm. mama 0*b’ ”34-11:? a «000 000661 0» £0! await , (avbeT) 3. mmm: “£0 WNW“ 000va x mm \s 0. 00M \W‘! ms. M30 0 80:0 00,» mam mm «0000mm 4; )0 <3me W 1 000m 006m ®4wmmksmnmmm“ 30me mm (001700133000110330Hefibo3L0«2M0<® , '10s 00mm mm: 01000000 @0000: 00m $5000 0000 \s (0000 \o 1 00000000 \Mogxsmmsxmwm 3000 000mm 0 00000005 W “W QEW‘GX‘M—ng‘OXCfifi‘fimiaWOOX’t‘ Lima \/0 ...
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