Question Number 61738 by rajesh4661kumar@gamil.com last updated on 07/Jun/19
Answered by MJS last updated on 07/Jun/19
![strictly logical A B A⇒B false false true false true true true false false true true true [(1+2=6)∧(2+3=8)∧(3+4=9)]⇒8+9=(7/(13))π (or whatever you want) but of course the question isn′t meant this way](https://www.tinkutara.com/question/Q61741.png)
$$\mathrm{strictly}\:\mathrm{logical} \\ $$$$\:\:\:\:{A}\:\:\:\:\:\:\:\:\:\:{B}\:\:\:\:\:\:\:\:\:\:{A}\Rightarrow{B} \\ $$$${false}\:\:\:{false}\:\:\:\:\:\:\:\:{true} \\ $$$${false}\:\:\:\:{true}\:\:\:\:\:\:\:\:\:{true} \\ $$$$\:{true}\:\:\:\:{false}\:\:\:\:\:\:\:{false} \\ $$$$\:{true}\:\:\:\:\:{true}\:\:\:\:\:\:\:\:\:{true} \\ $$$$\left[\left(\mathrm{1}+\mathrm{2}=\mathrm{6}\right)\wedge\left(\mathrm{2}+\mathrm{3}=\mathrm{8}\right)\wedge\left(\mathrm{3}+\mathrm{4}=\mathrm{9}\right)\right]\Rightarrow\mathrm{8}+\mathrm{9}=\frac{\mathrm{7}}{\mathrm{13}}\pi\:\left(\mathrm{or}\:\mathrm{whatever}\:\mathrm{you}\:\mathrm{want}\right) \\ $$$$ \\ $$$$\mathrm{but}\:\mathrm{of}\:\mathrm{course}\:\mathrm{the}\:\mathrm{question}\:\mathrm{isn}'\mathrm{t}\:\mathrm{meant}\:\mathrm{this}\:\mathrm{way} \\ $$
Answered by Rasheed.Sindhi last updated on 08/Jun/19
$$\:\:\:\:\underset{\smile} {\overset{\frown} {\:\langle\:\:\mathrm{1}\:+\:\mathrm{2}\:=\:\mathrm{6}\:\:\rangle\:}}\: \\ $$$$\:\:\:\:\:\:\lceil\frac{\mathrm{12}}{\mathrm{2}}\rceil=\lceil\mathrm{6}\rceil=\mathrm{6} \\ $$$$\:\:\:\:\underset{\smile} {\overset{\frown} {\:\langle\:\:\mathrm{2}\:+\:\mathrm{3}\:=\:\mathrm{8}\:\rangle\:}} \\ $$$$\:\:\:\:\:\:\lceil\frac{\mathrm{23}}{\mathrm{3}}\rceil=\lceil\mathrm{7}.\overline {\mathrm{6}}\rceil=\mathrm{8} \\ $$$$\:\:\:\:\underset{\smile} {\overset{\frown} {\:\langle\:\:\mathrm{3}\:+\:\mathrm{4}\:=\:\mathrm{9}\:\rangle\:}} \\ $$$$\:\:\:\:\:\:\lceil\frac{\mathrm{34}}{\mathrm{4}}\rceil=\lceil\mathrm{8}.\mathrm{5}\rceil=\mathrm{9} \\ $$$$\:\:\:\:\underset{\smile} {\overset{\frown} {\:\langle\:\:\mathrm{8}\:+\:\mathrm{9}\:=\:?\:\rangle\:}} \\ $$$$\:\:\:\:\lceil\frac{\mathrm{89}}{\mathrm{9}}\rceil=\lceil\mathrm{9}.\overline {\mathrm{8}}\rceil=\mathrm{10} \\ $$$$ \\ $$$$\:\: \\ $$$$\boldsymbol{\mathrm{t}}\oplus\left(\boldsymbol{\mathrm{t}}+\mathrm{1}\right)=\lceil\frac{\mathrm{10t}+\left(\mathrm{t}+\mathrm{1}\right)}{\mathrm{t}+\mathrm{1}}\rceil \\ $$$$\left[\:\:\rceil\mathrm{is}\:\mathrm{ceiling}\:\mathrm{function}.\right. \\ $$