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Question Number 98259 by Hassen_Timol last updated on 12/Jun/20

Commented by Hassen_Timol last updated on 12/Jun/20

$$\mathrm{Please}$$

Commented by PRITHWISH SEN 2 last updated on 12/Jun/20

$$\mathrm{ab}+\mathrm{a}<\mathrm{ab}+\mathrm{b}\Rightarrow \mathrm{a}(\mathrm{b}+1)<\mathrm{b}(\mathrm{a}+1)$$$$\frac{\mathrm{a}}{\mathrm{b}}<\frac{\mathrm{a}+1}{\mathrm{b}+1}$$$$\mathrm{similarly}$$$$\mathrm{ab}-\mathrm{b}<\mathrm{ab}-\mathrm{a}\Rightarrow \mathrm{b}(\mathrm{a}-1)<\mathrm{a}(\mathrm{b}-1)\Rightarrow \frac{\mathrm{a}-1}{\mathrm{b}-1}<\frac{\mathrm{a}}{\mathrm{b}}$$$$\frac{\mathrm{a}-1}{\mathrm{b}-1}<\frac{\mathrm{a}}{\mathrm{b}}<\frac{\mathrm{a}+1}{\mathrm{b}+1}$$

Commented by Hassen_Timol last updated on 12/Jun/20

$$\mathbf{Thank}\mathbf{you}\mathbf{very}\mathbf{much}:-)$$

Commented by PRITHWISH SEN 2 last updated on 12/Jun/20

$$\mathrm{welcome}$$

Answered by mr W last updated on 12/Jun/20

$$anotherway:$$$$sayb=x$$$$a=b-m=x-m$$$$f\left(x\right)=\frac{a}{b}=\frac{x-m}{x}=1-\frac{m}{x}$$$$f{}^{\prime }\left(x\right)=\frac{m}{x^2}>0$$$$\Rightarrow f\left(x\right)isstrictlyincreasing.$$$$\Rightarrow f(x-1)<f\left(x\right)<f(x+1)$$$$\Rightarrow 1-\frac{m}{x-1}<1-\frac{m}{x}<1-\frac{m}{x+1}$$$$\Rightarrow \frac{x-m-1}{x-1}<\frac{x-m}{x}<\frac{x-m+1}{x+1}$$$$\Rightarrow \frac{a-1}{b-1}<\frac{a}{b}<\frac{a+1}{b+1}$$

Commented by PRITHWISH SEN 2 last updated on 12/Jun/20

$$\mathrm{wow}!$$

Answered by mathmax by abdo last updated on 12/Jun/20

$$\mathrm{let}\frac{\mathrm{a}}{\mathrm{b}}=\mathrm{t}\Rightarrow 0<\mathrm{t}<1\mathrm{so}$$$$\frac{\mathrm{a}}{\mathrm{b}}-\frac{\mathrm{a}-1}{\mathrm{b}-1}=\mathrm{t}-\frac{\mathrm{bt}-1}{\mathrm{b}-1}=\frac{\mathrm{bt}-\mathrm{t}-\mathrm{bt}+1}{\mathrm{b}-1}=\frac{1-\mathrm{t}}{\mathrm{b}-1}>0$$$$\frac{\mathrm{a}+1}{\mathrm{b}+1}-\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{bt}+1}{\mathrm{b}+1}-\mathrm{t}=\frac{\mathrm{bt}+1-\mathrm{bt}-\mathrm{t}}{\mathrm{b}+1}=\frac{1-\mathrm{t}}{\mathrm{b}+1}>0\Rightarrow \frac{\mathrm{a}-1}{\mathrm{b}-1}<\frac{\mathrm{a}}{\mathrm{b}}<\frac{\mathrm{a}+1}{\mathrm{b}+1}$$

Commented by PRITHWISH SEN 2 last updated on 13/Jun/20

$$\mathrm{nice}$$

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