Question-221359

Question Number 221359 by Engr_Jidda last updated on 31/May/25

Answered by Rasheed.Sindhi last updated on 31/May/25

x∗y=2x+2y−((xy)/5) let e is an identity element of ∗ x∗e=e∗x=x ⇒2x+2e−((xe)/5)=x e(((10−x)/5))=−x e=((5x)/(x−10)) e is not constant let x^(−1) is inverse element w.r.t. ∗ x∗x^(−1) =e ⇒2x+2x^(−1) −((xx^(−1) )/5)=((5x)/(x−10)) x^(−1) (((10−x)/5))=((5x)/(x−10))−2x =((5x−2x^2 +20x)/(x−10)) x^(−1) =((25x−2x^2 )/(x−10))×(5/(10−x)) =((2x^2 −25x)/(10−x))×(5/(10−x)) x^(−1) =((5(2x^2 −25x))/((10−x)^2 ))

$${x}\ast{y}=\mathrm{2}{x}+\mathrm{2}{y}−\frac{{xy}}{\mathrm{5}} \\ $$$${let}\:{e}\:{is}\:{an}\:{identity}\:{element}\:{of}\:\ast \\ $$$${x}\ast{e}={e}\ast{x}={x} \\ $$$$\Rightarrow\mathrm{2}{x}+\mathrm{2}{e}−\frac{{xe}}{\mathrm{5}}={x} \\ $$$${e}\left(\frac{\mathrm{10}−{x}}{\mathrm{5}}\right)=−{x} \\ $$$${e}=\frac{\mathrm{5}{x}}{{x}−\mathrm{10}} \\ $$$${e}\:{is}\:{not}\:{constant} \\ $$$$\: \\ $$$${let}\:{x}^{−\mathrm{1}} \:{is}\:{inverse}\:{element}\:{w}.{r}.{t}.\:\ast \\ $$$${x}\ast{x}^{−\mathrm{1}} ={e} \\ $$$$\Rightarrow\mathrm{2}{x}+\mathrm{2}{x}^{−\mathrm{1}} −\frac{{xx}^{−\mathrm{1}} }{\mathrm{5}}=\frac{\mathrm{5}{x}}{{x}−\mathrm{10}} \\ $$$${x}^{−\mathrm{1}} \left(\frac{\mathrm{10}−{x}}{\mathrm{5}}\right)=\frac{\mathrm{5}{x}}{{x}−\mathrm{10}}−\mathrm{2}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{5}{x}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{20}{x}}{{x}−\mathrm{10}} \\ $$$${x}^{−\mathrm{1}} =\frac{\mathrm{25}{x}−\mathrm{2}{x}^{\mathrm{2}} }{{x}−\mathrm{10}}×\frac{\mathrm{5}}{\mathrm{10}−{x}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{25}{x}}{\mathrm{10}−{x}}×\frac{\mathrm{5}}{\mathrm{10}−{x}} \\ $$$$\:\:\:\:{x}^{−\mathrm{1}} =\frac{\mathrm{5}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{25}{x}\right)}{\left(\mathrm{10}−{x}\right)^{\mathrm{2}} } \\ $$

Commented by Rasheed.Sindhi last updated on 31/May/25

Verification: Let x=5 e=((5x)/(x−10))=((25)/(−5))=−5 5^(−1) =((5(50−125))/(25))=((5(−75))/(25))=−15 5∗5^(−1) =−5 5∗(−15)=−5 10−30−((−75)/5)=−5 −20+15=−5

$${Verification}: \\ $$$${Let}\:\:{x}=\mathrm{5} \\ $$$${e}=\frac{\mathrm{5}{x}}{{x}−\mathrm{10}}=\frac{\mathrm{25}}{−\mathrm{5}}=−\mathrm{5} \\ $$$$\mathrm{5}^{−\mathrm{1}} =\frac{\mathrm{5}\left(\mathrm{50}−\mathrm{125}\right)}{\mathrm{25}}=\frac{\mathrm{5}\left(−\mathrm{75}\right)}{\mathrm{25}}=−\mathrm{15} \\ $$$$\mathrm{5}\ast\mathrm{5}^{−\mathrm{1}} =−\mathrm{5} \\ $$$$\mathrm{5}\ast\left(−\mathrm{15}\right)=−\mathrm{5} \\ $$$$\mathrm{10}−\mathrm{30}−\frac{−\mathrm{75}}{\mathrm{5}}=−\mathrm{5} \\ $$$$−\mathrm{20}+\mathrm{15}=−\mathrm{5} \\ $$

Commented by Engr_Jidda last updated on 31/May/25

$${thanks} \\ $$

Answered by mehdee7396 last updated on 31/May/25

$$\nexists{e}\in\mathbb{R};\:{x}\ast{e}={x}\Rightarrow\nexists{y}\in\mathbb{R};{x}\ast{y}={e}\:\:;{e}\:\:{is}\:\:{cons} \\ $$$$ \\ $$

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