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Question Number 117603 by Ar Brandon last updated on 12/Oct/20

$$\mathrm{Let}\mathrm{f}:\mathbb{R}\to \mathbb{R}\mathrm{be}\mathrm{a}\mathrm{function}\mathrm{satisfying}\mathrm{the}\mathrm{following}:$$$$\left(\mathrm{a}\right)f(-x)=-f\left(x\right)$$$$\left(\mathrm{b}\right)f(x+1)=f\left(x\right)+1$$$$\left(\mathrm{c}\right)f\left(\frac{1}{x}\right)=\frac{f\left(x\right)}{x^2}\mathrm{for}\mathrm{all}x\ne 0$$$$\mathrm{Show}\mathrm{that}$$$$\left(\mathrm{i}\right)f\left(x\right)=x\mathrm{for}\mathrm{all}x,\mathrm{y}\in \mathbb{R}$$$$\left(\mathrm{ii}\right)f(x+y)=f\left(x\right)+f\left(\mathrm{y}\right)\mathrm{for}\mathrm{all}x,\mathrm{y}\in \mathbb{R}$$$$\left(\mathrm{iii}\right)f\left(xy\right)=f\left(x\right)f\left(\mathrm{y}\right)\mathrm{for}\mathrm{all}x,\mathrm{y}\in \mathbb{R}$$$$\left(\mathrm{iv}\right)f\left(\frac{x}{\mathrm{y}}\right)=\frac{f\left(x\right)}{f\left(\mathrm{y}\right)}\mathrm{for}\mathrm{all}x,\mathrm{y}\in \mathbb{R}\mathrm{with}\mathrm{y}\ne 0$$

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