# what-is-a-and-b-such-that-a-b-a-b-and-a-b-a-b-

Question Number 76531 by benjo 1/2 santuyy last updated on 28/Dec/19
$${what}\:{is}\:{a}\:{and}\:{b}\:{such}\:{that}\:{a}+{b}\:=\:{a}×{b}\: \\$$$${and}\:{a}+{b}\:={a}/{b}\:? \\$$
Commented by MJS last updated on 28/Dec/19
$$\left(\mathrm{1}\right)\:{a}+{b}={ab}\:\Rightarrow\:{a}=\frac{{b}}{{b}−\mathrm{1}}\:\Rightarrow\:{b}\neq\mathrm{1} \\$$$$\left(\mathrm{2}\right)\:\frac{{b}}{{b}−\mathrm{1}}+{b}=\frac{\mathrm{1}}{{b}−\mathrm{1}} \\$$$$\:\:\:\:\:\:\:\frac{{b}^{\mathrm{2}} }{{b}−\mathrm{1}}=\frac{\mathrm{1}}{{b}−\mathrm{1}} \\$$$$\:\:\:\:\:\:\:{b}^{\mathrm{2}} =\mathrm{1}\wedge{b}\neq\mathrm{1}\:\Rightarrow\:{b}=−\mathrm{1}\:\Rightarrow\:{a}=\frac{\mathrm{1}}{\mathrm{2}} \\$$
Commented by john santu last updated on 28/Dec/19
$${testing}\:{b}\:=\:\mathrm{1}\:{to}\:{equation}\:\rightarrow\:{a}\:+\:\mathrm{1}\:={a}×\mathrm{1}\:{that}\:{rigt},\:{for}\:{any}\: \\$$$${value}\:{of}\:{a} \\$$
Commented by benjo 1/2 santuyy last updated on 28/Dec/19
$${but}\:{a}\:\neq\:\mathrm{0} \\$$
Commented by MJS last updated on 28/Dec/19
$${b}=\mathrm{1}\:??? \\$$$${a}+\mathrm{1}={a} \\$$$$\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{a} \\$$
Answered by john santu last updated on 28/Dec/19
$$\rightarrow\:{a}×{b}=\frac{{a}}{{b}}\:\left[{b}\neq\mathrm{0}\right] \\$$$${ab}^{\mathrm{2}} −{a}=\mathrm{0}\:\rightarrow\:{a}=\mathrm{0}\:\vee\:{b}=\pm\mathrm{1}\: \\$$
Commented by MJS last updated on 28/Dec/19
$$\mathrm{you}\:\mathrm{forgot}\:\mathrm{to}\:\mathrm{check} \\$$$${a}+{b}=\frac{{a}}{{b}}\:\Rightarrow\:{b}\neq\mathrm{0} \\$$$${a}+{b}={ab} \\$$$$\:\:\:\:\:{a}=\mathrm{0}\:\Rightarrow\:{b}=\mathrm{0} \\$$$$\:\:\:\:\:{b}=−\mathrm{1}\:\Rightarrow\:{a}−\mathrm{1}=−{a}\:\Rightarrow\:{a}=\frac{\mathrm{1}}{\mathrm{2}} \\$$$$\:\:\:\:\:{b}=\mathrm{1}\:\Rightarrow\:{a}+\mathrm{1}={a}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{a} \\$$$${a}+{b}=\frac{{a}}{{b}} \\$$$$\:\:\:\:\:{a}=\mathrm{0}\:\Rightarrow\:{b}=\frac{\mathrm{0}}{{b}}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{b} \\$$$$\:\:\:\:\:{b}=−\mathrm{1}\:\Rightarrow\:{a}−\mathrm{1}=−{a}\:\Rightarrow\:{a}=\frac{\mathrm{1}}{\mathrm{2}} \\$$$$\:\:\:\:\:{b}=\mathrm{1}\:\Rightarrow\:{a}+\mathrm{1}={a}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{a} \\$$$$\Rightarrow \\$$$${a}=\frac{\mathrm{1}}{\mathrm{2}}\wedge{b}=−\mathrm{1} \\$$
Commented by john santu last updated on 28/Dec/19
$${oo}\:{yes}\:{i}\:{agree} \\$$