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Question Number 62372 by Rasheed.Sindhi last updated on 20/Jun/19

Solve for x , y  3x>2y ∧ 2x<3y   where x,y∈N

$${Solve}\:{for}\:{x}\:,\:{y} \\ $$ $$\mathrm{3}{x}>\mathrm{2}{y}\:\wedge\:\mathrm{2}{x}<\mathrm{3}{y}\: \\ $$ $${where}\:{x},{y}\in\mathbb{N} \\ $$

Answered by MJS last updated on 20/Jun/19

(2/3)y<x<(3/2)y ∧ y>0

$$\frac{\mathrm{2}}{\mathrm{3}}{y}<{x}<\frac{\mathrm{3}}{\mathrm{2}}{y}\:\wedge\:{y}>\mathrm{0} \\ $$

Commented byRasheed.Sindhi last updated on 20/Jun/19

Thanks Sir!  Guide me please, what′s wrong  in the following:       3x>2y ∧ 2x<3y (Given)       ⇒3x>2y  ∧  3y >2x      ⇒3x×3y >2y×2x      ⇒9xy>4xy      ⇒9xy−4xy>0      ⇒5xy>0      ⇒xy>0       ⇒ x>0 ∧ y>0           ???

$$\mathcal{T}{hanks}\:{Sir}! \\ $$ $$\mathcal{G}{uide}\:{me}\:{please},\:{what}'{s}\:{wrong} \\ $$ $${in}\:{the}\:{following}: \\ $$ $$\:\:\:\:\:\mathrm{3}{x}>\mathrm{2}{y}\:\wedge\:\mathrm{2}{x}<\mathrm{3}{y}\:\left({Given}\right) \\ $$ $$\:\:\:\:\:\Rightarrow\mathrm{3}{x}>\mathrm{2}{y}\:\:\wedge\:\:\mathrm{3}{y}\:>\mathrm{2}{x} \\ $$ $$\:\:\:\:\Rightarrow\mathrm{3}{x}×\mathrm{3}{y}\:>\mathrm{2}{y}×\mathrm{2}{x} \\ $$ $$\:\:\:\:\Rightarrow\mathrm{9}{xy}>\mathrm{4}{xy} \\ $$ $$\:\:\:\:\Rightarrow\mathrm{9}{xy}−\mathrm{4}{xy}>\mathrm{0} \\ $$ $$\:\:\:\:\Rightarrow\mathrm{5}{xy}>\mathrm{0} \\ $$ $$\:\:\:\:\Rightarrow{xy}>\mathrm{0} \\ $$ $$\:\:\:\:\:\Rightarrow\:{x}>\mathrm{0}\:\wedge\:{y}>\mathrm{0}\: \\ $$ $$\:\:\:\:\:\:\:\:??? \\ $$

Commented byMJS last updated on 20/Jun/19

xy>0 ⇏ x>0∧y>0 !!!    3x>2y ⇒ x>(2/3)y  3y>2x ⇒ x<(3/2)y  ⇒ (2/3)y<x<(3/2)y  1. obviously x≠0 ∧ y≠0  2. x<0 ∧ y<0  (2/3)>(x/y)>(3/2)  ⇒ (x/y)∈]−∞; (2/3)[∩](3/2); +∞[={}  ⇒ no solution  3. x<0 ∧ y>0  (2/3)<(x/y)<(3/2) (with (x/y)<0)  ⇒ no solution  4. x>0 ∧ y<0  (2/3)>(x/y)>(3/2)  ⇒ (x/y)∈]−∞; (2/3)[∩](3/2); +∞[={}  ⇒ no solution  5. x>0 ∧ y>0  (2/3)<(x/y)<(3/2) ⇔ (2/3)y<x<(3/2)y ⇔ (2/3)x<y<(3/2)x

$${xy}>\mathrm{0}\:\nRightarrow\:{x}>\mathrm{0}\wedge{y}>\mathrm{0}\:!!! \\ $$ $$ \\ $$ $$\mathrm{3}{x}>\mathrm{2}{y}\:\Rightarrow\:{x}>\frac{\mathrm{2}}{\mathrm{3}}{y} \\ $$ $$\mathrm{3}{y}>\mathrm{2}{x}\:\Rightarrow\:{x}<\frac{\mathrm{3}}{\mathrm{2}}{y} \\ $$ $$\Rightarrow\:\frac{\mathrm{2}}{\mathrm{3}}{y}<{x}<\frac{\mathrm{3}}{\mathrm{2}}{y} \\ $$ $$\mathrm{1}.\:\mathrm{obviously}\:{x}\neq\mathrm{0}\:\wedge\:{y}\neq\mathrm{0} \\ $$ $$\mathrm{2}.\:{x}<\mathrm{0}\:\wedge\:{y}<\mathrm{0} \\ $$ $$\frac{\mathrm{2}}{\mathrm{3}}>\frac{{x}}{{y}}>\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $$\left.\Rightarrow\:\frac{{x}}{{y}}\in\right]−\infty;\:\frac{\mathrm{2}}{\mathrm{3}}\left[\cap\right]\frac{\mathrm{3}}{\mathrm{2}};\:+\infty\left[=\left\{\right\}\right. \\ $$ $$\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$ $$\mathrm{3}.\:{x}<\mathrm{0}\:\wedge\:{y}>\mathrm{0} \\ $$ $$\frac{\mathrm{2}}{\mathrm{3}}<\frac{{x}}{{y}}<\frac{\mathrm{3}}{\mathrm{2}}\:\left(\mathrm{with}\:\frac{{x}}{{y}}<\mathrm{0}\right) \\ $$ $$\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$ $$\mathrm{4}.\:{x}>\mathrm{0}\:\wedge\:{y}<\mathrm{0} \\ $$ $$\frac{\mathrm{2}}{\mathrm{3}}>\frac{{x}}{{y}}>\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $$\left.\Rightarrow\:\frac{{x}}{{y}}\in\right]−\infty;\:\frac{\mathrm{2}}{\mathrm{3}}\left[\cap\right]\frac{\mathrm{3}}{\mathrm{2}};\:+\infty\left[=\left\{\right\}\right. \\ $$ $$\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$ $$\mathrm{5}.\:{x}>\mathrm{0}\:\wedge\:{y}>\mathrm{0} \\ $$ $$\frac{\mathrm{2}}{\mathrm{3}}<\frac{{x}}{{y}}<\frac{\mathrm{3}}{\mathrm{2}}\:\Leftrightarrow\:\frac{\mathrm{2}}{\mathrm{3}}{y}<{x}<\frac{\mathrm{3}}{\mathrm{2}}{y}\:\Leftrightarrow\:\frac{\mathrm{2}}{\mathrm{3}}{x}<{y}<\frac{\mathrm{3}}{\mathrm{2}}{x} \\ $$

Commented byRasheed.Sindhi last updated on 20/Jun/19

 Learnt something new from you!  Lot of thanks Sir!

$$\:{Learnt}\:{something}\:{new}\:{from}\:{you}! \\ $$ $${Lot}\:{of}\:{thanks}\:\mathcal{S}{ir}! \\ $$

Commented byMJS last updated on 20/Jun/19

you′re welcome

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$

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