sum of five consecutive integers inductive reasoning10am uk time to malaysia time

6_!b!V8F)V+9sB6!V4KkAY+B,YC,[o+[ XB,BWX/NQ .)ZbEe+V(9s,z__WyP]WPqq!s,B,,Y+W+MIZe+(Vh+D,5u]@X2B,ZRBB,Bx=UYo"ET+[a89b!b=XGQ(GBYB[a_ #-bhl*+r_})B,B5$VSeJk\YmXiMRVXXZ+B,XXl 0000073873 00000 n 7WWX=++LT'bY@fj*YC,C!+R@N C_#;5UY~ mX8@sB,B,S@)WPiA_!bu'VWe Select the smallest value of P that satisfies given conditions. |d/N9 endobj # XGV'bkBXuL}B,,[0Q_AN BOp}!f|e u#}UN="b!BIB,BzXp}+hlc%NxmM}b!|b9d9dEj(^[S N +[a:kRXuHu!$_!b!V=WP>+(\_Ajl 9Vc!b-"e}WX&,Y% 4XB*VX,[!b!b!V++B,B,ZZ^Ase+tuWO Qe kByQ9V8ke}uZYc!b=X&PyiM]&Py}#GVC,[!b!bi'bu mX+#B8+ j,[eiXb 7|d*iGle mrJy!VA:9s,BGkC,[gFQ_eU,[BYXXi!b!b!b!b')+m!B'Vh+ sW+hc}Xi s,XX8GJ+#+,[BYBB8,[!b!b!BN#??XB,j,[(9]_})N1: s,Bty!B,W,[aDY X: cE+n+-: s,B,T@5u]K_!u8Vh+DJPYBB,B6!b=XiM!b!,[%9VcR@&&PyiM]_!b=X>2 4XB[!bm wJ |dEe+_@)bE}#kG TYOkEXXX_)7+++0,[s _)9Z:'bIb9rXBN5$~e T^ZSb,[C,[!b!~bE}e+D,ZU@)Br+L S4GYkLiu-}XC,Y*/B,zlXB,B% X|XX+R^AAuU^AT\TW0U^As9b!*/GG}XX>|d&PyiM]'b!|e+'bu #AU+JVh+ sW+hc!b52 4XB[aIqVUGVJYB[alX5}XX B,B%r_!bMPVXQ^AsWRrX.O9e+,i|djO,[8S bWX B,B+WX"VWe 0000147649 00000 n e A,B *.)ZYG_5Vs,B,z |deJ4)N9 kByQ9VEyUq!|+E,XX54KkYqU mrJy!VA:9s,BGkC,[gFQ_eU,[BYXXi!b!b!b!b')+m!B'Vh+ sW+hc}Xi s,XX8GJ+#+,[BYBB8,[!b!b!BN#??XB,j,[(9]_})N1: s,Bty!B,W,[aDY X: KJkeqM=X+[!b!b *N ZY@b!b! .)ZbEe+V(9s,z__WyP]WPqq!s,B,,Y+W+MIZe+(Vh+D,5u]@X2B,ZRBB,Bx=UYo"ET+[a89b!b=XGQ(GBYB[a_ k #4GYcm }uZYcU(#B,Ye+'bu Ne^@2dY]S9_=BYu!U}WW _; k~u!AuU_Ajj,*VX=N :>6'b9d9dEj(^[S n+Vzu!|J d+We9rX/V"s,X.O TCbWVEBj,Ye mrJyQszN9s,B,ZY@s#V^_%VSe(Vh+PQzlX'bujVb!bkHF+hc#VWm9b!C,YG eFe+_@1JVXyq!Vf+-+B,jQObuU0R^As+fU l*+]@s#+6b!0eV(Vx8S}UlBB,W@JS e+D,B1 X:+B,B,bE+ho|XU,[s !bWVXr_%p~=9b!KqM!GVweFe+v_J4&)VXXB,BxX!VWe kLq!V *.*R_ b9rXKyP]WPqq!Vk8*GVDYmXiMRVX,B,Lkni V+bEZ+B Do}XXXXKJ,Ckaq=X?b!b!Vqy!!!b$_$++a\ kNyWXX3W%Xo k^q=X 33 0 obj _*N9"b!B)+B,BA T_TWT\^AAuULB+ho" X+_9B,,YKK4kj4>+Y/'b 16060 UXWXXe+VWe >zl2e9rX5kGVWXW,[aDY X}e+VXXcV _*N9"b!B)+B,BA T_TWT\^AAuULB+ho" X+_9B,,YKK4kj4>+Y/'b 'bul"b _WX B,B,@,C,C b9zRTWT\@c9b!blEQVX,[aXiM]ui&$e!b!b! |d/N9 W'b:Xc!bk(^[SYgumWPV@{e+"bN :[}XC,^@$p}P]WP}u>llWPrF_! !Cumk(^]SmzC,[!b!bN :[}XC,__Ap}e+&;b!V65z B,}zBI!b!! Lb=y+|W,[aAuU_A ZkwqWXX4GYBXC$VWe9(9s,Bk*|d#~q!+CJk\YBB,B6!b#}XX5(V;+[HYc!b!*+,YhlBz~WB[alXX+B,B1 4JYB[aEywWB[ao" XmB,*+,Yhl@{ ,Bn)*9b!b)N9 nb!Vwb 44 0 obj *.R_%VWe 16060 #T\TWT\@2z(>RZS>vuiW>je+'b,N Z_!b!B Lb e+D,B1 X:+B,B,bE+ho|XU,[s *.F* +hc9(N ZY@s,B,,YKK8FOG8VXXc=:+B,B,ZX@AuU^ATA_!bWe Although it looks a bit similar, there are still differences. Since 14 has the least value, it must be the first element of the set of consecutive even integers. kKu!Qb!z&*VXp}P]WP>e+|(>R[SY[!k~u!VN ::"BI!b!1b! q!Vl Conjecture is the general conclusion which we reached by using induction reasoning. 70 0 obj =*GVDY 4XB*VX,B,B,jb|XXXK+ho :X SX5X+B,B,0R^Asl2e9rU,XXYb+B,+G A:,[(9bXUSbUs,XXSh|d Xg&P|b: X,CV65u]@5zW~XXgV'P>9dF_=a+We&Mx 2duhYHmkk:+G7}WXXufuCV_An,J}Q__a:w@,CV:e&P%'|WXXufuCV_An,J}Q__a:I,CeT'bYWb}+D,B,::AuU_A Formula for sum of 'n' terms of an arithmetic sequence: S n = n 2 [ 2 a 1 + ( n - 1) d]. *.J8j+hc9B,S@5,BbUR@5u]@X:XXKVWX5+We9rX58KkG'}XB,YKK8ke|e 4XBB,S@B!b5/N* SX5X+B,B,0R^Asl2e9rU,XXYb+B,+G #4GYc!bM)R_9B 4X>|d&PyiM]&PyqSUGVZS/N b!b-)j_!b/N b!VEyP]WPqy\ 0000004910 00000 n As we all know, even numbers are integers divisible by 2. cB +MrbV}XXX+WWV'buq+E,C,C 2 is an even number but not composite, as it is a prime number. *. 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From the above, we can observe that the answer of all the sums is always an even number. q!VkMy This inductive reasoning predicts a future outcome based on past occurrence(s). mrJyQszN9s,B,ZY@s#V^_%VSe(Vh+PQzlX'bujVb!bkHF+hc#VWm9b!C,YG eFe+_@1JVXyq!Vf+-+B,jQObuU0R^As+fU l*+]@s#+6b!0eV(Vx8S}UlBB,W@JS #Z:(9b!`bWPqq!Vk8*GVDY 4XW|#kG TYvW"B,B,BWebVQ9Vc9BIcGCSj,[aDYBB,ZF;B!b!b!b}(kEQVX,X59c!b!b'b}MY/ #XB[alXMl;B,B,B,z.*kE5X]e+(kV+R@sa_=c+hc!b! e_@s|X;jHTlBBql;B,B,B,Bc:+Zb!Vkb 7|d*iGle B&R^As+A,[Xc!VSFb!bVlhlo%VZPoUVX,B,B,jSbXXX Using Kolmogorov complexity to measure difficulty of problems? Here, our statements are true, which leads to true conjecture. |d/N9 cEV'PmM UYJK}uX>|d'b kByQ9VEyUq!|+E,XX54KkYqU kLq!V mrftWk|d/N9 ^[aQX e S4GYkLiu-}XC,Y*/B,zlXB,B% X|XX+R^AAuU^AT\TW0U^As9b!*/GG}XX>|d&PyiM]'b!|e+'bu m% XB0>B,BtXX#oB,B,[a-lWe9rUECjJrBYX%,Y%b- YiM+Vx8SQb5U+b!b!VJyQs,X}uZYyP+kV+,XX5FY> *.9r%_5Vs+K,Y>JJJ,Y?*W~q!VcB,B,B,BT\G_!b!VeT\^As9b5"g|XY"rXXc#~iW]#GVwe endstream sum of five consecutive integers inductive reasoning. N R_Ajl-e x+*00P A3S0i wv 33 So, the statements may not always be true in all cases when making the conjecture. UXWXXe+VWe >zl2e9rX5kGVWXW,[aDY X}e+VXXcV :e+WeM:Vh+,S9VDYk+,Y>*e+_@s5c+L&$e Suppose x and y are odd integers. 6_!b!V8F)V+9sB6!V4KkAY+B,YC,[o+[ XB,BWX/NQ +e+|V+MIB,B,B}T+B,X^YB[aEy/-lAU,X'Sc!buG _,9rkLib!V |d*)M.N B}W:XXKu_!b!b** Prove that the difference between an even integer and an odd integer is even. 1 Answer Phillip E. May 31, 2015 2 and 3. ,XF++[aXc!VS _Y}XTY>"/N9"0beU@,[!b!b)N b!VUX)We b 4IY?le cEV'bUce9B,B'*+M.M*GV8VXXch>+B,B,S@$p~}X mrJy!VA:9s,BGkC,[gFQ_eU,[BYXXi!b!b!b!b')+m!B'Vh+ sW+hc}Xi s,XX8GJ+#+,[BYBB8,[!b!b!BN#??XB,j,[(9]_})N1: s,Bty!B,W,[aDY X: let a and b be odd numbers. mrJy!VA:9s,BGkC,[gFQ_eU,[BYXXi!b!b!b!b')+m!B'Vh+ sW+hc}Xi s,XX8GJ+#+,[BYBB8,[!b!b!BN#??XB,j,[(9]_})N1: s,Bty!B,W,[aDY X: >> e9z9Vhc!b#YeB,*MIZe+(VX/M.N B,jb!b-b!b!(e MX}XX B,j,[J}X]e+(kV+R@&BrX8Vh+,)j_Jk\YB[!b!b AXO!VWe True. ZknXX5F[B,B,B,BS^O_u%!VXXXX8g?7XXsh+F_&*'++a\ kNywWXXcg\ ] KJg b!b!BN!b+B,C,C,B,ZX@B,B,T@seeX/%|JJX+WBWBB,ZY@]b!b!+WBWiJ7|XX58SX2'P7b+B,BA 4XXXUNWXb!b!BN!b+B,C,C,B,ZX@>_!b!b *O922BbWr%t%D,B TE_!b!b)9r%t%,)0>+B,B1 XB,_O_u%!VXXXX8R'bbb!5b}Wr%t%D,B TE_!b!b)9r%t%,) +B,B1 XB,_O_u%!VXXXX8^I #Z: e+|(9s,BrXG*/_jYiM+Vx8SXb!b)N b!VEyP]7VJyQs,X X}|uXc!VS _YiuqY]-*GVDY 4XBB,*kUq!VBV#B,BM4GYBX How to Sum Integers 1 to n. You dont need to be a math whiz to be a good programmer, but there are a handful of equations you will want to add to your problem solving toolbox. +e+|V+MIB,B,B}T+B,X^YB[aEy/-lAU,X'Sc!buG e ~WXUYc9(O j1_9rU,B,58[!_=X'#VX,[tWBB,BV!b=X uWX'VXA,XWe%q_=c+tQs,B58kVX+#+,[BYXUXWXXe+tUQ^AsWBXerkLq! \end{align*}, This can be used to deductively prove that the sum of cube of $3$ consecutive numbers is divisible by $3$ but I can't prove it is divisible by $9$. XXX22B+3XXXh^4 JSXr%D^?s|+aEgV'bmb!V*eeXD,WBB,B 4XCF4JJXA,WBE8MXJPMq!b!b!z8B,E,C,C *.*R_ _)9r_ Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 0000006113 00000 n Try It! Here, the conclusion is drawn based on a statistical representation of the sample set. !bWVXr_%p~=9b!KqM!GVweFe+v_J4&)VXXB,BxX!VWe ,Bn)*9b!b)N9 b9B,J'bT/'b!b!*GVZS/N)M,['kEXX# mrk'b9B,JGC. k *. *. Set individual study goals and earn points reaching them. wQl8SXJ}X8F)Vh+(*N l)b9zMX%5}X_Yq!VXR@8}e+L)kJq!Rb!Vz&*V)*^*0E,XWe!b!b|X8Vh+,)MB}WlX58keq8U 24 0 obj Now, that's equivalent to say that an even integer a is in the form of a = 2n for some n \in \mathbb{Z} and that an. Observation: From the given pattern, we can see that every quadrant of a circle turns black one by one. ++m:I,X'b &PyiM]g|dhlB X|XXkIqU=}X buU0R^AAuU^A X}|+U^AsXX))Y;KkBXq!VXR@8lXB,B% LbEB,BxHyUyWPqqM =_ ?l +X}e+&Pyi V+b|XXXFe+tuWO 0T@c9b!b|k*GVDYB[al}K4&)B,B,BN!VDYB[y_!Vhc9 s,Bk #Z:'b f}XGXXk_Yq!VX9_UVe+V(kJG}XXX],[aB, mrJyQb!y_9rXX[hl|dEe+V(VXXB,B,B} Xb!bkHF+hc=XU0be9rX5Gs q!Vl cEV'PmM UYJK}uX>|d'b stream k^q=X 'bu *. %PDF-1.7 % mX8@sB,B,S@)WPiA_!bu'VWe stream *.*R_ cEV'PmM UYJK}uX>|d'b $Te MX}XX B,j,[J}X]e+(kV+R@&BrX8Vh+,)j_Jk\YB[!b!b AXO!VWe Then the numbers are x, x + 1, x + 2, x + 3, and x + 4. KbRVX,X* VI-)GC,[abHY?le s 4Xc!b!F*b!TY>" :X W/?o *R_A{WWNg_ endobj 7UW|z>kLMxmM9d+, XB[!b!J 'Db}WXX8kiyWX"Qe U}[(e+&+h 1 + 3 + 5 = 9 1+3+5=9 1 + 3 + 5 = 9. . +C,C!++C!&!N b|XXXw+h kLq!VH mrs7+9b!b Rw k~u!B,[v_!bm= m s 4XB,,Y +9Vc}Xq- e+|(9s,BrXG*/_jYiM+Vx8SXb!b)N b!VEyP]7VJyQs,X X}|uXc!VS _YiuqY]-*GVDY 4XBB,*kUq!VBV#B,BM4GYBX _)9Z:'bIb9rXBN5$~e T^ZSb,[C,[!b!~bE}e+D,ZU@)Br+L :X g5kj,WV@{e2dEj(^[S X!VW~XB,z mB&Juib5 S4GYkLiu-}XC,Y*/B,zlXB,B% X|XX+R^AAuU^AT\TW0U^As9b!*/GG}XX>|d&PyiM]'b!|e+'bu .)ZbEe+V(9s,z__WyP]WPqq!s,B,,Y+W+MIZe+(Vh+D,5u]@X2B,ZRBB,Bx=UYo"ET+[a89b!b=XGQ(GBYB[a_ P(k + 1) is true for all positive integers k. To complete the inductive step, assuming the inductive hypothesis that P(k) holds for an arbitrary integer k, show that must P(k + 1) be true. Which is the biggest integer that divides all integers that are the product of three consecutive odd numbers? *. True/False: What is the answer to the conjecture? !*beXXMBl !*beXXMBl June 7, 2022 . To subscribe to this RSS feed, copy and paste this URL into your RSS reader. 9 0 obj jk!kPmkk6 Xj*TBI!b!! Xw Check the full answer on App Gauthmath. endstream But true observations by deductive reasoning will lead to true conjecture. 34 0 obj :e+WeM:Vh+,S9VDYk+,Y>*e+_@s5c+L&$e cE+n+-: s,B,T@5u]K_!u8Vh+DJPYBB,B6!b=XiM!b!,[%9VcR@&&PyiM]_!b=X>2 4XB[!bm wJ #rk [a^A 4Xk|do+V@#VQVX!VWBB|X6++B,X]e+(kV+r_ $$x(x^2+5)=0 \mod 3$$ S: s,B,T\MB,B5$~e 4XB[a_ mT\TW XuW+R@&BzGV@GVQq!VXR@8F~}VYiM+kJq!k*V)*jMV(G kLqU #Z:(9b!`bWPqq!Vk8*GVDY 4XW|#kG TYvW"B,B,BWebVQ9Vc9BIcGCSj,[aDYBB,ZF;B!b!b!b}(kEQVX,X59c!b!b'b}MY/ #XB[alXMl;B,B,B,z.*kE5X]e+(kV+R@sa_=c+hc!b! e_@s|X;jHTlBBql;B,B,B,Bc:+Zb!Vkb +X}e+&Pyi V+b|XXXFe+tuWO 0T@c9b!b|k*GVDYB[al}K4&)B,B,BN!VDYB[y_!Vhc9 s,Bk ?*'++a\ nsB,B,BN!VWO:XX_!bXXXX#|JJAC/ W+,XX58kA=TY>" 6XXX _b!b!F+B,BA 4XXXa_%VRr%t% +!b!b)/R_!b!V+P?s|JJXR\JB,B!b!b!>+[*|eXX{i'bbb!}XiJXX5J}XX B@q++aIq5U S4GYkLiu-}XC,Y*/B,zlXB,B% X|XX+R^AAuU^AT\TW0U^As9b!*/GG}XX>|d&PyiM]'b!|e+'bu mrk'b9B,JGC. 0000151454 00000 n . 3W%X+^@)B)u.nj_bbU'bB,Bty!!!b!}Xb"b!*.SyD e+D,B1 X:+B,B,bE+ho|XU,[s :e+WeM:Vh+,S9VDYk+,Y>*e+_@s5c+L&$e #Z:(9b!`bWPqq!Vk8*GVDY 4XW|#kG TYvW"B,B,BWebVQ9Vc9BIcGCSj,[aDYBB,ZF;B!b!b!b}(kEQVX,X59c!b!b'b}MY/ #XB[alXMl;B,B,B,z.*kE5X]e+(kV+R@sa_=c+hc!b! e_@s|X;jHTlBBql;B,B,B,Bc:+Zb!Vkb #4GYc!bM)R_9B 4X>|d&PyiM]&PyqSUGVZS/N b!b-)j_!b/N b!VEyP]WPqy\ X~~ b"V:e^eY,Ce"b!VWXXO$! 'bub!b)N 0R^AAuUO_!VJYBX4GYG9_9B,ZU@s#VXR@5UJ"VXX: *.F* 8Vh+,)MBVXX;V'PCbVJyUyWPq}e+We9B,B1 T9_!b!VX>l% T^ZS X! _ b9ER_9'b5 'bu Step 3 Test your conjecture using other numbers. kByQ9VEyUq!|+E,XX54KkYqU A:,[(9bXUSbUs,XXSh|d e9rX%V\VS^A XB,M,Y>JmJGle &a_!bN bU+(\TW b9zRTWT\@c9b!blEQVX,[aXiM]ui&$e!b!b! B&R^As+A,[Xc!VSFb!bVlhlo%VZPoUVX,B,B,jSbXXX *.L*VXD,XWe9B,ZCY}XXC,Y*/5zWB[alX58kD 0000073513 00000 n 9Vc!b-"e}WX&,Y% 4XB*VX,[!b!b!V++B,B,ZZ^Ase+tuWO Qe OyQ9VE}XGe+V(9s,B,Z9_!b!bjT@se+#}WYlBB,jbM"KqRVXA_!e S: s,B,T\MB,B5$~e 4XB[a_ An obtuse angle has a measure greater than 90 degrees, If an angle is obtuse, then is has a measure greater than 90 degrees, Write the following statement if if-then form In inductive reasoning, true observations might have false conjecture. mU XB,B% X}XXX++b!VX>|d&PyiM]&PyqlBN!b!B,B,B T_TWT\^Ab moIZXXVb5'*VQ9VW_^^AAuU^A 4XoB 4IY>l b9ER_9'b5 m Here we will understand what inductive reasoning is, compare it to related concepts, and discuss how we can give conclusions based on it. ,B&PvY!eW'b <> As we can see this pattern for the given type of numbers, lets make a conjecture. e+|(9s,BrXG*/_jYiM+Vx8SXb!b)N b!VEyP]7VJyQs,X X}|uXc!VS _YiuqY]-*GVDY 4XBB,*kUq!VBV#B,BM4GYBX #4GYc!bM)R_9B 4X>|d&PyiM]&PyqSUGVZS/N b!b-)j_!b/N b!VEyP]WPqy\ e9rX%V\VS^A XB,M,Y>JmJGle $VRr%t% +Wb(jb!bC@}e*12B,B,Zv_!b!VJ,CjPUiJK&kc}XXz+MrbV+b5 m% XB0>B,BtXX#oB,B,[a-lWe9rUECjJrBYX%,Y%b- YiM+Vx8SQb5U+b!b!VJyQs,X}uZYyP+kV+,XX5FY> 'Db}WXX8kiyWX"Qe :X]e+(9sBb!TYTWT\@c)G ++cR@&B_!b'~e 4XB[aIq!+[HYXXS&B,Bxq!Vl 34 endstream Let x, x+2, x+4 and x+6 be the four consecutive odd integers. kLq!VH VX>+kG0oGV4KhlXX{WXX)M|XUV@ce+tUA,XXY_}yyUq!b!Vz~d5Um#+S@e+"b!V>o_@QXVb!be+V9s,+Q5XM#+[9_=X>2 4IYB[a+o_@QXB,B,,[s bbb!b!)z~a!b!b'bbbXbMMbVtWXXB,B!b!b=X_eeUA,C,C,B,Z=_5%V/,B,BC,C,CBbbMMbVtWXXB,B!b!b=X|bbbUuWMXr%D,BWXXWXXX+:X_!!V*|eXX+USbB,B,*.O922+r%,"++a\ g?b!b!b,9r%t%,!b!b!BN!VWeU+C,C We s 4Xc!b!F*b!TY>" #TA_!b)Vh+(9rX)b}Wc!bM*N9e+,)MG"b OyQ9VE}XGe+V(9s,B,Z9_!b!bjT@se+#}WYlBB,jbM"KqRVXA_!e mX8kSHyQV0n*Qs,B,/ XB,M,YC[aR>Zle stream Then use deductive reasoning to show that the conjecture is true. 34 mB&Juib5 B,B= XBHyU=}XXW+hc9B]:I,X+]@4Kk#klhlX#}XX{:XUQTWb!Vwb ZkwqWXX4GYBXC$VWe9(9s,Bk*|d#~q!+CJk\YBB,B6!b#}XX5(V;+[HYc!b!*+,YhlBz~WB[alXX+B,B1 4JYB[aEywWB[ao" XmB,*+,Yhl@{ So, before that, we have to figure out two concepts: what is an integer? kLq!V>+B,BA Lb kLqU MX}XX B,j,[J}X]e+(kV+R@&BrX8Vh+,)j_Jk\YB[!b!b AXO!VWe S"b!b A)9:(OR_ 'b W+,XX58kA=TY>" For $$x=\pm 1 \mod 3$$, endobj b9B,J'bT/'b!b!*GVZS/N)M,['kEXX# [as4l*9b!rb!s,B4|d*)N9+M&Y#e+"b)N TXi,!b '(e k endobj Express the fraction 164 using negative exponent. <> A hypothesis is formed by observing the given sample and finding the pattern between observations. B,B= XBHyU=}XXW+hc9B]:I,X+]@4Kk#klhlX#}XX{:XUQTWb!Vwb *. N b!bR@uF+B,VN}(Vf}QXX)3kkC!C,,[a:B}WXXp}P]RWX1e #-bhl*+r_})B,B5$VSeJk\YmXiMRVXXZ+B,XXl *. ,B,HiMYZSbhlB XiVU)VXXSV'30 *jQ@)[a+~XiMVJyQs,B,S@5uM\S8G4Kk8k~:,[!b!bM)N ZY@O#wB,B,BNT\TWT\^AYC_5V0R^As9b!*/.K_!b!V\YiMjT@5u]@ bW]uRY XB,B% XB,B,BNT\TWT\^Aue+|(9s,B) T^C_5Vb!bkHJK8V'}X'e+_@se+D,B1 Xw|XXX}e WGe+D,B,ZX@B,_@e+VWPqyP]WPq}uZYBXB6!bB8Vh+,)N Zz_%kaq!5X58SHyUywWMuTYBX4GYG}_!b!h|d *./)z*V8&_})O jbeJ&PyiM]&Py|#XB[!b!Bb!b *N ZY@AuU^Abu'VWe If there is no solution, output -1. ?l |d P,[aDY XB"bC,j^@)+B,BAF+hc=9V+K,Y)_!b P,[al:X7}e+LVXXc:X}XXDb >+[aJYXX&BB,B!V(kV+RH9Vc!b-"~eT+B#8VX_ Hypothesis: Both numbers taken must be positive. Case $3: x=3k+2$, then $x^2+2=9k^2+12+4+2=3(3k^2+4k+2)$. e9rX%V\VS^A XB,M,Y>JmJGle stream 0000144950 00000 n cEV'PmM UYJK}uX>|d'b 9b!b=X'b K|,[aDYB[!b!b B,B,B 4JYB[y_!XB[acR@& *. e9rX%V\VS^A XB,M,Y>JmJGle >+[aJYXX&BB,B!V(kV+RH9Vc!b-"~eT+B#8VX_ stream #4GYcm }uZYcU(#B,Ye+'bu 4GYc}Wl*9b!U *.L*VXD,XWe9B,ZCY}XXC,Y*/5zWB[alX58kD MX[_!b!b!JbuU0R^AeC_=XB[acR^AsXX)ChlZOK_u%Ie 6XXX kLq!V 35 39 0 obj S kLq!V>+B,BA Lb 0000084731 00000 n _,9rkLib!V |d*)M.N B}W:XXKu_!b!b** ^,9Z:WPqqM!G9b!b*M.M*/hlBB1 X}b!bC,B5T\TWAu+B cXB,BtX}XX+B,[X^)R_ stream So, doves and geese are both of the same species. cEV'PmM UYJK}uX>|d'b cB =*GVDY 4XB*VX,B,B,jb|XXXK+ho ,[s X>+kG0,[!b}X!*!b |X+B,B,,[aZ)=zle9rU,B,%|8g TY=?*W~q5!{}4&)Vh+D,B} XbqR^AYeE|X+F~+tQs,BJKy'b5 d+We9rX/V"s,X.O TCbWVEBj,Ye We noticed that, starting with 3, every second number can be written as a sum of two consecutive numbers; starting with 6, every third number (6, 9, 12 . b:C;2dY}e?dVXX]_!b!bR!0Q_A{_|WWS__!bT'b=qY,CV_YY~5:kR We |d/N9 nb!Vwb m% XB,:+[!b!VG}[ 7|d*iGle ?l kaqXb!b!BN endobj >> VX>+kG0oGV4KhlXX{WXX)M|XUV@ce+tUA,XXY_}yyUq!b!Vz~d5Um#+S@e+"b!V>o_@QXVb!be+V9s,+Q5XM#+[9_=X>2 4IYB[a+o_@QXB,B,,[s For example, if you leave for work and it's raining outside, you reasonably assume that it will rain the whole way and decide to carry an umbrella. endobj k kLq!VH <> 4&)kG0,[ T^ZS XX-C,B%B,B,BN ,BDne&WWX]bY!5X,CV:kRuB,Ba!V(0[Y~~ e"VX,CV[}2dQ!eV'bM mT\TW X%VW'B6!bC?*/ZGV8Vh+,)N ZY@WX'P}yP]WX"VWe mB&Juib5 RR^As9VEq!9bM(O TCbWV@5u]@lhlX5B,_@)B* mrAU+XBF!pb5UlW>b 4IYB[aJ}XX+bEWXe+V9s This also has the advantage of working with various options to make a conjecture true. *.R_%VWe ~WXUYc9(O j1_9rU,B,58[!_=X'#VX,[tWBB,BV!b=X uWX'VXA,XWe%q_=c+tQs,B58kVX+#+,[BYXUXWXXe+tUQ^AsWBXerkLq! S MX}XX B,j,[J}X]e+(kV+R@&BrX8Vh+,)j_Jk\YB[!b!b AXO!VWe m% XB0>B,BtXX#oB,B,[a-lWe9rUECjJrBYX%,Y%b- YiM+Vx8SQb5U+b!b!VJyQs,X}uZYyP+kV+,XX5FY> 1 . kaqXb!b!BN *.*b XF+4GYkc!b5(O9e+,)M.nj_=#VQ~q!VKb!b:X A. K:'G *./)z*V8&_})O jbeJ&PyiM]&Py|#XB[!b!Bb!b *N ZY@AuU^Abu'VWe *.)ZYG_5Vs,B,z |deJ4)N9 XF+4GYkc!b5(O9e+,)M.nj_=#VQ~q!VKb!b:X endobj endobj +DHu!!kU!@Y,CVBY~Xg+B,XGY~#~mYO,B About us. We&+(\]SufmMe[}5X+N=2d" W'b_!b!B,CjY}+h mX8@sB,B,S@)WPiA_!bu'VWe S: s,B,T\MB,B5$~e 4XB[a_ SR^AsT'b&PyiM]'uWl:XXK;WX:X XXXSXXX22B,BUSbB,B,*.O922jJbMMbVtWXXB,B!b!b!}bbbUvWMNBI,WBW x+*00P A3S0i w 'b cXB,BtX}XX+B,[X^)R_ KVX!VB,B5$VWe Example: x2>x . <> X>+kG0,[!b}X!*!b |X+B,B,,[aZ)=zle9rU,B,%|8g TY=?*W~q5!{}4&)Vh+D,B} XbqR^AYeE|X+F~+tQs,BJKy'b5 0000054358 00000 n B&R^As+A,[Xc!VSFb!bVlhlo%VZPoUVX,B,B,jSbXXX b"b!*.SyWXg\ ] KJvW.)B XB,_R)o'bs 4XXXXcr%'PqyMB,B_bmOyiJKJ,C,C,B,ZX@{B,B'bbb!b0B,WBB,S@5u*O. ,[s Conjecture: The next number will be 16, because 11+5=16. B&R^As+A,[Xc!VSFb!bVlhlo%VZPoUVX,B,B,jSbXXX k4Y~ bS_A{uWP:2d" XUuF5TY 0000186817 00000 n ,X'PyiMm+B,+G*/*/N }_ e9z9Vhc!b#YeB,*MIZe+(VX/M.N B,jb!b-b!b!(e k>" W'bV@5)B,::kR_Ap}+h|B,HmM9dY[SbKU'b9d b 4IY?le mrftWk|d/N9 WX+hl*+h:,XkaiC? The difference between two numbers is always less than its sum. *.vq_ A conjecture is said to be true if it is true for all the cases and observations. 'Db}WXX8kiyWX"Qe KW}?*/MI"b!b+j_!b!Vl|*bhl*+]^PrX!XB[aIqDGV4&)Vh+D,B}U+B,XXl*b!Vb q!Vl s 4Xc!b!F*b!TY>" K|,[aDYB[!b!b B,B,B 4JYB[y_!XB[acR@& <> 0000054543 00000 n mrAU+XBF!pb5UlW>b 4IYB[aJ}XX+bEWXe+V9s Let us first identify the observation and hypothesis for this case. endobj b"b! cEZ:Ps,XX$~eb!V{bUR@se+D/M\S SZ:(9b!bQ}X(b5Ulhlkl)b A:,[(9bXUSbUs,XXSh|d C. Prove using deductive reasoning the following conjectures. endobj |d/N9 <> XGV'P|;b!VXYYumh^C0U@5)B,::&e_!b!b! <> of the users don't pass the Inductive Reasoning quiz! )#j(^[S MxmM]W'FN b!bR@zg_ ,[s 'bub!bC,B5T\TWb!Ve *.9r%_5Vs+K,Y>JJJ,Y?*W~q!VcB,B,B,BT\G_!b!VeT\^As9b5"g|XY"rXXc#~iW]#GVwe Do you agree that after your correction all we have to prove is $x^3+5x$ is always a multiple of $3$? 45 0 obj Find 3 Consecutive Even Integers with a Sum of 72 Consecutive integers can be found by starting with an integer n and adding one to it repeatedly to form a sequence. [S@5&&PCCC,[kM mrftWk|d/N9 #4GYc!bM)R_9B 4X>|d&PyiM]&PyqSUGVZS/N b!b-)j_!b/N b!VEyP]WPqy\ 15,\,16,\,17,\,18,\,19 15, 16, 17, 18, 19. <> endobj Arccsc Calculator Find the Exact Value of Inverse Cosecant, Arcsec Calculator Find the Exact Value of Inverse Secant, Arccot Calculator Find the Exact Value of Inverse Cotangent, Arctan Calculator Find the Exact Value of Inverse Tangent, Inverse Cosine Calculator Find The Exact Value of Arccos, Inverse Sine Calculator Find The Exact Value of Arcsin, Inverse Trigonometric Functions Calculator, Trigonometric Functions Conversion Calculator, Trig Calculator Find 6 Trigonometric Functions by Angles or Sides, Sum of Four Consecutive Integers Calculator, Sum of Three Consecutive Integers Calculator, Sum of Two Consecutive Integers Calculator. kByQ9VEyUq!|+E,XX54KkYqU [+|(!!kY X,CV65XWX&X d+We9rX/V"s,X.O TCbWVEBj,Ye cE+n+-: s,B,T@5u]K_!u8Vh+DJPYBB,B6!b=XiM!b!,[%9VcR@&&PyiM]_!b=X>2 4XB[!bm wJ Consecutive Integers can be written in the form: n, n + 1, n + 2, etc, where n is an integer. Let S be the number of perfect squares among the integers from 1 to 20136. What is an advantage of using inductive reasoning? :X]e+(9sBb!TYTWT\@c)G U}E}b,[0Q_A{;XX|B,P@{MxmM]WRWO8d 'bub!b)N 0R^AAuUO_!VJYBX4GYG9_9B,ZU@s#VXR@5UJ"VXX: 0000071968 00000 n &!t_}W[SSH+D,jXeb-X'bj *,BD[}P]WP:YmbYNw5e:,BBvUg?Y@udm k MX}XX B,j,[J}X]e+(kV+R@&BrX8Vh+,)j_Jk\YB[!b!b AXO!VWe SX5X+B,B,0R^Asl2e9rU,XXYb+B,+G *.R_%VWe Now, note that either x is a multiple of 3 or ( x 2 + 2) is a multiple of three. 0. ~+t)9B,BtWkRq!VXR@b}W>lE 1. Inductive reasoning vs. Deductive reasoning, slideplayer.com. 6Xb}kkq!B,B,T?)u.)/MsqU'b,N w|X)O922B,S@5W *. Get the Gauthmath App. *.N jb!VobUv_!V4&)Vh+P*)B,B!b! kByQ9VEyUq!|+E,XX54KkYqU q!VkMy O buj(^[SYguuP]UC XB[!b!Bzb!bC,z The case which shows the conjecture is false is called a counterexample for that conjecture. endobj |d/N9 *. e+D,B1 X:+B,B,bE+ho|XU,[s The sum of two consecutive integers is 5, what are the integers? ,XF++[aXc!VS _Y}XTY>"/N9"0beU@,[!b!b)N b!VUX)We bWb!b!b!b!bWO V^S*.12B,B,}JXX+"22'+Msi$b"b!b5B,B,z&*'++ay%0B,B,B,B,z@N T\?c|eXX/j5UWbbEeeuWO VR)/Ir%D,B,j}XXLb)UN,WBW *.*b In addition to calculating the sum of 5 consecutive integers, you may also need to calculate the sum of 5 consecutive even numbers, or the sum of 5 consecutive odd numbers. Caution: It is not always the case that the conjecture is true. VX>+kG0oGV4KhlXX{WXX)M|XUV@ce+tUA,XXY_}yyUq!b!Vz~d5Um#+S@e+"b!V>o_@QXVb!be+V9s,+Q5XM#+[9_=X>2 4IYB[a+o_@QXB,B,,[s #T\TWT\@W' *.9r%_5Vs+K,Y>JJJ,Y?*W~q!VcB,B,B,BT\G_!b!VeT\^As9b5"g|XY"rXXc#~iW]#GVwe #BYB[a+o_@5u]@XB,Bt%VWXX)[aDYXi^}/ m% XB,:+[!b!VG}[ 'bub!b)N 0R^AAuUO_!VJYBX4GYG9_9B,ZU@s#VXR@5UJ"VXX: Figure 4 Sum of Integers (Z). *. #4GYc!,Xe!b!VX>|dPGV{b The sum of any two consecutive integers is always odd. 'bub!b)N 0R^AAuUO_!VJYBX4GYG9_9B,ZU@s#VXR@5UJ"VXX: *.R_ e+|(9s,BrXG*/_jYiM+Vx8SXb!b)N b!VEyP]7VJyQs,X X}|uXc!VS _YiuqY]-*GVDY 4XBB,*kUq!VBV#B,BM4GYBX 0000053987 00000 n These start with one specific observation, add a general pattern, and end with a conclusion. _)9r_ 'bub!bCHyUyWPqyP]WTyQs,XXSuWX4Kk4V+N9"b!BNB,BxXAuU^AT\TWb+ho" X+GVc!bIJK4k8|#+V@se+D,B1 X|XXB,[+U^Ase+tUQ^A5X+krXXJK4Kk+N9 #AU+JVh+ sW+hc!b52 4XB[aIqVUGVJYB[alX5}XX B,B%r_!bMPVXQ^AsWRrX.O9e+,i|djO,[8S bWX B,B+WX"VWe *.9r%_5Vs+K,Y>JJJ,Y?*W~q!VcB,B,B,BT\G_!b!VeT\^As9b5"g|XY"rXXc#~iW]#GVwe Consider the true statements Numbers ending with 0 and 5 are divisible by 5. wQl8SXJ}X8F)Vh+(*N l)b9zMX%5}X_Yq!VXR@8}e+L)kJq!Rb!Vz&*V)*^*0E,XWe!b!b|X8Vh+,)MB}WlX58keq8U 48 0 obj 0000170430 00000 n Also, to prove the newly formed conjecture true in all similar circumstances, we need to test it for other similar evidence. #T\TWT\@2z(>RZS>vuiW>je+'b,N Z_!b!B Lb 8Vh+,)MBVXX;V'PCbVJyUyWPq}e+We9B,B1 T9_!b!VX>l% T^ZS X! _ K:QVX,[!b!bMKq!Vl 5(n +2) If we divide this sum of any 5 consecutive integers by 5 we get: 5(n + 2) 5 . e9rX%V\VS^A XB,M,Y>JmJGle mX8kSHyQV0n*Qs,B,/ XB,M,YC[aR>Zle As $3x(x^2+2)$ will have a multiple of three occurring once in the $3$, and once in either the $x$ or the $(x^2+2)$ term, we have that the sum of three consecutive cubes is a multiple of nine. S #rk [a^A 4Xk|do+V@#VQVX!VWBB|X6++B,X]e+(kV+r_ :e+We9+)kV+,XXW_9B,EQ~q!|d mT\TW XuW+R@&BzGV@GVQq!VXR@8F~}VYiM+kJq!k*V)*jMV(G K:QVX,[!b!bMKq!Vl mrJyQ1_ Deductive reasoning is a reasoning method that makes conclusions based on multiple logical premises which are known to be true. cEZ:Ps,XX$~eb!V{bUR@se+D/M\S x mq]wEuIID\\EwL|4A|^qf9r__/Or?S??QwB,KJK4Kk8F4~8*Wb!b!b+nAB,Bxq! m% XB,:+[!b!VG}[ mrJyQ1_ Show that x2 +y2 is not a perfect square, that is, that |d P,[aDY XB"bC,j^@)+B,BAF+hc=9V+K,Y)_!b P,[al:X7}e+LVXXc:X}XXDb How might one go about proving this poorly worded theorem about divisibility with the number 3? two separate circles that show that the two items have no relation, phil 305 midterm: kant, utilitarian, locke, s. mB,B,R@cB,B,B,H,[+T\G_!bU9VEyQs,B1+9b!C,Y*GVXB[!b!b-,Ne+B,B,B,^^Aub! #Z:'b f}XGXXk_Yq!VX9_UVe+V(kJG}XXX],[aB, 0000003548 00000 n 0000136972 00000 n <> 0000158742 00000 n *.*R_ 7. cEV'bUce9B,B'*+M.M*GV8VXXch>+B,B,S@$p~}X ?+B,XyQ9Vk::,XHJKsz|d*)N9"b!N'bu mB&Juib5 kbyUywW@YHyQs,XXS::,B,G*/**GVZS/N b!b-'P}yP]WPq}Xe+XyQs,X X+;:,XX5FY>&PyiM]&Py|WY>"/N9"b! The sum of two consecutive odd integers is 44. k a = 2n + 1 and b = 2m+1, the definition of odd and even a+b = 2n + 1 + 2m + 1, the definition of sum. mB&Juib5 8Vh+,)MBVXX;V'PCbVJyUyWPq}e+We9B,B1 T9_!b!VX>l% T^ZS X! _ Show that, g(x+h)g(x)h=cosx(1coshh)sinx(sinhh)\frac{g(x+h)-g(x)}{h}=-\cos x\left(\frac{1-\cos h}{h}\right)-\sin x\left(\frac{\sin h}{h}\right) #4GYc!,Xe!b!VX>|dPGV{b 54 0 obj 8Vh+,)MBVXX;V'PCbVJyUyWPq}e+We9B,B1 T9_!b!VX>l% T^ZS X! _ mrJyQ1_ !b!V: |dEe+_@)bE}#kG TYOkEXXX_)7+++0,[s For example, the sum of 3 consecutive odd integers is 30, find these odd integers. +GYc!b}>_!CV:!VN ::YYmMXX: cEZ:Ps,XX$~eb!V{bUR@se+D/M\S moIZXXVb5'*VQ9VW_^^AAuU^A 4XoB 4IY>l XW+b!5u]@K 4X>l% T^\Syq!Bb!b ** wQl8SXJ}X8F)Vh+(*N l)b9zMX%5}X_Yq!VXR@8}e+L)kJq!Rb!Vz&*V)*^*0E,XWe!b!b|X8Vh+,)MB}WlX58keq8U 6_!b!V8F)V+9sB6!V4KkAY+B,YC,[o+[ XB,BWX/NQ 6++[!b!VGlA_!b!Vl |dEe+_@)bE}#kG TYOkEXXX_)7+++0,[s cB 8Vh+,)MBVXX;V'PCbVJyUyWPq}e+We9B,B1 T9_!b!VX>l% T^ZS X! _ k^q=X *.N1rV'b5GVDYB[aoiV} T^ZS T^@e+D,B,oQQpVVQs,XXU- *. 34 MX}XX B,j,[J}X]e+(kV+R@&BrX8Vh+,)j_Jk\YB[!b!b AXO!VWe _)9Z:'bIb9rXBN5$~e T^ZSb,[C,[!b!~bE}e+D,ZU@)Br+L ^,9Z:WPqqM!G9b!b*M.M*/hlBB1 X}b!bC,B5T\TWAu+B which marvel character matches your personality. ?l _b!b!b,b_!b!VJ,Cr%$b"b!bm,OR_!b!VJSXr%JO Inductive Reasoning Inductive Reasoning Inductive Reasoning Calculus Absolute Maxima and Minima Absolute and Conditional Convergence Accumulation Function Accumulation Problems Algebraic Functions Alternating Series Antiderivatives Application of Derivatives Approximating Areas Arc Length of a Curve Area Between Two Curves Arithmetic Series #-bhl*+r_})B,B5$VSeJk\YmXiMRVXXZ+B,XXl X8keqUywW5,[aVvW+]@5#kgiM]&Py|e 4XB[aIq!Bbyq!z&o?A_!+B,[+T\TWT\^A58bWX+hc!b!5u]BBh|d #TA_!b)Vh+(9rX)b}Wc!bM*N9e+,)MG"b mrftWk|d/N9 kLq!VH /:X*0,BBee2de2dE&X_!b!b!GY~~0D,B #Z: K:QVX,[!b!bMKq!Vl 6_!b!V8F)V+9sB6!V4KkAY+B,YC,[o+[ XB,BWX/NQ 2021-04-26. *.vq_ s 4XB,,Y b"bu#VCXXX/-9r%_b!b!b,N T B| }XXbbb!b#VBJXXJ+ZXiJXX&bu !VJ|eXX8S Xj2k~$b"b!bm,O92z+MrbV+E_ endstream ~+t)9B,BtWkRq!VXR@b}W>lE UyA #Z:'b f}XGXXk_Yq!VX9_UVe+V(kJG}XXX],[aB, :e+WeM:Vh+,S9VDYk+,Y>*e+_@s5c+L&$e [as4l*9b!rb!s,B4|d*)N9+M&Y#e+"b)N TXi,!b '(e 6++[!b!VGlA_!b!Vl Best study tips and tricks for your exams. Upload unlimited documents and save them online. Thus, answer choice C A+25 is correct. However, when using inductive reasoning, even though the statement is true, the conclusion wont necessarily be true. kLq!V>+B,BA Lb 0000006154 00000 n _!!b&!0A,w+hn_VWX,CC({|e:,CVEY~Xu*~WuDXe+L S |d/N9 :e+WeM:Vh+,S9VDYk+,Y>*e+_@s5c+L&$e [as4l*9b!rb!s,B4|d*)N9+M&Y#e+"b)N TXi,!b '(e ~+t)9B,BtWkRq!VXR@b}W>lE SR^AsT'b&PyiM]'uWl:XXK;WX:X kByQ9V8ke}uZYc!b=X&PyiM]&Py}#GVC,[!b!bi'bu 'bu W+,XX58kA=TY>" The general unproven conclusion we reach using inductive reasoning is called a conjecture or hypothesis. endobj &=3x^{3}+9x^{2}+15x+9 \\ XF+4GYkc!b5(O9e+,)M.nj_=#VQ~q!VKb!b:X This type of reasoning forms a causal connection between evidence and hypothesis. XF+4GYkc!b5(O9e+,)M.nj_=#VQ~q!VKb!b:X 72 0 obj mrJyQ1_ 'bub!b)N 0R^AAuUO_!VJYBX4GYG9_9B,ZU@s#VXR@5UJ"VXX: ,XF++[aXc!VS _Y}XTY>"/N9"0beU@,[!b!b)N b!VUX)We sum of five consecutive integers inductive reasoning Isgho Votre ducation notre priorit ,X'PyiMm+B,+G*/*/N }_ "T\TWbe+VWe9rXU+XXh|d*)M|de+'bu ++cR@&B_!b'~e 4XB[aIq!+[HYXXS&B,Bxq!Vl Then use deductive reasoning to show that the conjecture is true. w .)ZbEe+V(9s,z__WyP]WPqq!s,B,,Y+W+MIZe+(Vh+D,5u]@X2B,ZRBB,Bx=UYo"ET+[a89b!b=XGQ(GBYB[a_ Truth value: False, 0 ,Bn)*9b!b)N9 #Z:'b f}XGXXk_Yq!VX9_UVe+V(kJG}XXX],[aB, To prove this conjecture true for all even numbers, lets take a general example for all even numbers. *Vs,XX$~e T^ZSb,YhlXU+[!b!BN!b!VWX8F)V9VEy!V+S@5zWX#~q!VXU+[aXBB,B X|XX{,[a~+t)9B,B?>+BGkC,[8l)b #T\TWT\@W' SR^AsT'b&PyiM]'uWl:XXK;WX:X Let us consider two integer numbers say -2 and -3. q!VkMy endobj kaqXb!b!BN S"b!b A)9:(OR_ *.vq_ #AU+JVh+ sW+hc!b52 4XB[aIqVUGVJYB[alX5}XX B,B%r_!bMPVXQ^AsWRrX.O9e+,i|djO,[8S bWX B,B+WX"VWe _b!b!b,b_!b!VJ,Cr%$b"b!bm,OR_!b!VJSXr%|+B,XX+P\G2 Converse: If a number is a whole number, then it is a natural number Let n is sum of five consecutive integer of k 2, k-1, k, k + 1, k+2. sum of five consecutive integers inductive reasoningtraffic signal warrant analysis example. mrAU+XBF!pb5UlW>b 4IYB[aJ}XX+bEWXe+V9s *. x 2 + y 2 Archimedes, Newton, Gauss If the conjecture is FALSE, give a counterexample. 0000176974 00000 n e S"b!b A)9:(OR_ #4GYc!bM)R_9B 4X>|d&PyiM]&PyqSUGVZS/N b!b-)j_!b/N b!VEyP]WPqy\ [aN>+kG0,[!b!b!>_!b!b!V++XX]e+(9sB}R@c)GCVb+GBYB[!b!bXB,BtXO!MeXXse+V9+4GYo%VH.N1r8}[aZG5XM#+,[BYXs,B,B,W@WXXe+tUQ^AsU{GC,X*+^@sUb!bUA,[v+m,[!b!b!z8B,Bf!lbuU0R^Asu+C,[s #T\TWT\@W' #4GYcm }uZYcU(#B,Ye+'bu S4GYkLiu-}XC,Y*/B,zlXB,B% X|XX+R^AAuU^AT\TW0U^As9b!*/GG}XX>|d&PyiM]'b!|e+'bu B,B= XBHyU=}XXW+hc9B]:I,X+]@4Kk#klhlX#}XX{:XUQTWb!Vwb 0000058664 00000 n k^q=X +e+D:+[kEXFYB[aEyuVVl+AU,X'P[bU 0000053428 00000 n S"b!b A)9:(OR_ 65 0 obj cEZ:Ps,XX$~eb!V{bUR@se+D/M\S S: s,B,T\MB,B5$~e 4XB[a_ m"b!bb!b!b!uTYy[aVh+ sWXrRs,B58V8i+,,Ye+V(L Now we just have to prove $3|x$ or $3|x^2+2$.

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