Question Number 31711 by gunawan last updated on 13/Mar/18 $$\mathrm{Given}\:{p}\:\mathrm{is}\:\mathrm{primes}\:\mathrm{and}\:\mathrm{A}=\left\{−\frac{{m}}{{n}}−{p}\frac{{n}}{{m}}\:\mid\:{m}\:,\:{n}\:\in\mathbb{N}\right\} \\ $$$$\mathrm{find}\:\mathrm{sup}\:\mathrm{A} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 31669 by gunawan last updated on 12/Mar/18 $$\mathrm{A}=\left\{\frac{{m}}{{n}}+\frac{\mathrm{8}{n}}{{m}}\::\:{m},\:{n}\:\in\:\mathrm{N}\right\}\:\mathrm{N}=\:\mathrm{natural}\:\mathrm{numbers} \\ $$$$\mathrm{supremum}\:? \\ $$$$\mathrm{infimum}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 31237 by pieroo last updated on 04/Mar/18 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{A}−\mathrm{B}\:=\:\mathrm{B}'\:\cap\:\mathrm{A}. \\ $$ Commented by Tinkutara last updated on 04/Mar/18 You can also see on internet https://www.wyzant.com/resources/answers/272263/prove_a_b_a_intersection_b Commented by pieroo last updated…
Question Number 161623 by amin96 last updated on 20/Dec/21 $$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}+\mathrm{1}} }{\boldsymbol{{n}}\left(\boldsymbol{{n}}+\mathrm{2}\right)}=? \\ $$ Answered by TheSupreme last updated on 20/Dec/21 $$\frac{{A}}{{n}}+\frac{{B}}{{n}+\mathrm{2}}=\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}\left({n}+\mathrm{2}\right)} \\…
Question Number 161439 by amin96 last updated on 17/Dec/21 Commented by bobhans last updated on 18/Dec/21 $$\left(\mathrm{Q2}\right)\:\mathrm{y}=\:\frac{\mathrm{x}−\psi}{\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{6}}\:\Rightarrow\mathrm{yx}^{\mathrm{2}} −\mathrm{5yx}+\mathrm{6y}=\mathrm{x}−\psi \\ $$$$\:\:\mathrm{yx}^{\mathrm{2}} −\left(\mathrm{5y}+\mathrm{1}\right)\mathrm{x}+\left(\mathrm{6y}+\psi\right)=\mathrm{0} \\ $$$$\:\bigtriangleup\geqslant\mathrm{0}\:\Rightarrow\mathrm{25y}^{\mathrm{2}} +\mathrm{10y}+\mathrm{1}−\mathrm{4y}\left(\mathrm{6y}+\psi\right)\geqslant\mathrm{0}…
Question Number 29014 by A.Ayo last updated on 03/Feb/18 $${Prove}\:\:{that}\:{A}\cup{A}^{{c}} ={A} \\ $$ Commented by Tinkutara last updated on 03/Feb/18 $${A}\cup{A}'={U} \\ $$ Commented by…
Question Number 28806 by pieroo last updated on 30/Jan/18 $$\mathrm{If}\:\mathrm{n}\left(\mathrm{A}\right)=\mathrm{15}\:\mathrm{and}\:\mathrm{n}\left(\mathrm{B}\right)=\mathrm{25},\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{and}\:\mathrm{least}\:\mathrm{values}\:\mathrm{of}\:\mathrm{n}\left(\mathrm{AuB}\right)? \\ $$$$\left(\mathrm{b}\right)\:\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{and}\:\mathrm{least}\:\mathrm{value}\:\mathrm{of}\:\mathrm{n}\left(\mathrm{AnB}\right)? \\ $$$$\left(\mathrm{c}\right)\:\mathrm{Draw}\:\mathrm{Venn}\:\mathrm{diagrams}\:\mathrm{to}\:\mathrm{illustrate}\:\mathrm{the}\:\mathrm{four}\: \\ $$$$\:\:\:\:\:\:\:\mathrm{situations}\:\mathrm{in}\:\left(\mathrm{a}\right)\:\mathrm{and}\:\left(\mathrm{b}\right)\:\mathrm{above} \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 93950 by oustmuchiya@gmail.com last updated on 16/May/20 $${Let}\:\ast\:{be}\:{the}\:{binary}\:{operation}\:{on}\:\mathrm{N} \\ $$$${given}\:{by}\:\mathrm{a}\ast\mathrm{b}=\mathrm{L}.\mathrm{C}.\mathrm{M}.\:{of}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{\mathrm{b}}.\:\mathrm{F}{ind} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{5}\ast\mathrm{7}\:,\:\:\mathrm{20}\ast\mathrm{16}\:\:\:\left(\mathrm{ii}\right)\:\mathrm{is}\:\ast\:{communitative}? \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{is}\:\ast\:{associative}? \\ $$$$\left(\mathrm{iv}\right)\mathrm{F}{ind}\:{the}\:{identity}\:{of}\:\ast\:{in}\:\boldsymbol{\mathrm{N}} \\ $$$$\left(\mathrm{v}\right)\:\mathrm{which}\:\mathrm{elements}\:\mathrm{of}\:\boldsymbol{\mathrm{N}}\:{are}\:{invertible} \\ $$$$\:\:\:\:\:\:\:{for}\:{the}\:{operation}\:\ast? \\ $$ Answered…
Question Number 93933 by oustmuchiya@gmail.com last updated on 16/May/20 $${Let}\:\ast'\:{be}\:{the}\:{binary}\:{operation}\:{on}\:{the}\:{set}\:\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\}\:{defined}\:{by}\:\mathrm{a}\ast'\mathrm{b}=\mathrm{H}.\mathrm{C}.\mathrm{F}{of}\:{a}\:{and}\:{b}.\: \\ $$$${Is}\:{the}\:{operation}\:\ast'\:{same}\:{as}\:{the}\:{operation}\:\ast\:{defined}\:{above}?\:{justify}\:{your}\:{answer}. \\ $$ Commented by Rasheed.Sindhi last updated on 16/May/20 $${Operation}\:\ast'\:{is}\:{defined}\:{in}\:{the} \\ $$$${question}.{What}\:{do}\:{you}\:{mean}\:{by} \\…
Question Number 93567 by oustmuchiya@gmail.com last updated on 13/May/20 $${Given}\:{that}\:{A}=\left\{\mathrm{0},\mathrm{1},\mathrm{3},\mathrm{5}\right\}\:{B}=\left\{\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{7}\right\}\:{and}\:{C}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8}\right\}\:{prove}\:{that} \\ $$$$\left(\mathrm{i}\right)\:\left(\mathrm{A}\cap\mathrm{B}\right)\cap\mathrm{C}\:=\:\mathrm{A}\cap\left(\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\left(\mathrm{ii}\right)\:\left(\mathrm{A}\cup\mathrm{B}\right)\cup\mathrm{C}\:=\:\mathrm{A}\cup\left(\mathrm{B}\cup\mathrm{C}\right) \\ $$$$\left(\mathrm{iii}\right)\:\left(\mathrm{A}\cup\mathrm{B}\right)\cap\mathrm{C}\:=\:\left(\mathrm{A}\cup\mathrm{C}\right)\cup\left(\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\left(\mathrm{iv}\right)\:\left(\mathrm{A}\cap\mathrm{C}\right)\cup\mathrm{B}\:=\:\left(\mathrm{A}\cup\mathrm{B}\right)\cap\left(\mathrm{C}\cup\mathrm{B}\right) \\ $$ Commented by Ritu Jain last…