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Question

# If f : Q → Q, g : Q → Q are two functions defined by f(x) = 2 x and g(x) = x + 2, show that f and g are bijective maps. Verify that (gof)−1 = f−1 og −1.

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Solution

## Injectivity of f: Let x and y be two elements of domain (Q), such that f(x) = f(y) $⇒$2x = 2y $⇒$x = y So, f is one-one. Surjectivity of f: Let y be in the co-domain (Q), such that f(x) = y. $⇒2x=y\phantom{\rule{0ex}{0ex}}⇒x=\frac{y}{2}\in Q\left(\text{domain}\right)$ $⇒$ f is onto. So, f is a bijection and, hence, it is invertible. Finding f -1: $\text{Let}{f}^{-1}\left(x\right)=y...\left(1\right)\phantom{\rule{0ex}{0ex}}⇒x=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒x=2y\phantom{\rule{0ex}{0ex}}⇒y=\frac{x}{2}\phantom{\rule{0ex}{0ex}}\text{So,}{f}^{-1}\left(x\right)=\frac{x}{2}\left(\text{from}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}$ Injectivity of g: Let x and y be two elements of domain (Q), such that g(x) = g(y) $⇒$x + 2 = y + 2 $⇒$x = y So, g is one-one. Surjectivity of g: Let y be in the co domain (Q), such that g(x) = y. $⇒x+2=y\phantom{\rule{0ex}{0ex}}⇒x=2-y\in Q\left(\text{domain}\right)$ $⇒$ g is onto. So, g is a bijection and, hence, it is invertible. Finding g -1: $\text{Let}{g}^{-1}\left(x\right)=y...\left(2\right)\phantom{\rule{0ex}{0ex}}⇒x=g\left(y\right)\phantom{\rule{0ex}{0ex}}⇒x=y+2\phantom{\rule{0ex}{0ex}}⇒y=x-2\phantom{\rule{0ex}{0ex}}\text{So,}{g}^{-1}\left(x\right)=x-2\left(\text{From}\left(2\right)\right)\phantom{\rule{0ex}{0ex}}$ Verification of (gof)−1 = f−1 og −1: $f\left(x\right)=2x;g\left(x\right)=x+2\phantom{\rule{0ex}{0ex}}\mathrm{and}{f}^{-1}\left(x\right)=\frac{x}{2};{g}^{-1}\left(x\right)=x-2\phantom{\rule{0ex}{0ex}}\mathrm{Now},\left({f}^{-1}o{g}^{-1}\right)\left(x\right)={f}^{-1}\left({g}^{-1}\left(x\right)\right)\phantom{\rule{0ex}{0ex}}⇒\left({f}^{-1}o{g}^{-1}\right)\left(x\right)={f}^{-1}\left(x-2\right)\phantom{\rule{0ex}{0ex}}⇒\left({f}^{-1}o{g}^{-1}\right)\left(x\right)=\frac{x-2}{2}...\left(3\right)\phantom{\rule{0ex}{0ex}}\left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\phantom{\rule{0ex}{0ex}}=g\left(2x\right)\phantom{\rule{0ex}{0ex}}=2x+2\phantom{\rule{0ex}{0ex}}\text{Let}{\left(gof\right)}^{-1}\left(x\right)\text{=y ....}\left(4\right)\phantom{\rule{0ex}{0ex}}x=\left(gof\right)\left(y\right)\phantom{\rule{0ex}{0ex}}⇒x=2y+2\phantom{\rule{0ex}{0ex}}⇒2y=x-2\phantom{\rule{0ex}{0ex}}⇒y=\frac{x-2}{2}\phantom{\rule{0ex}{0ex}}⇒{\left(gof\right)}^{-1}\left(x\right)=\frac{x-2}{2}\text{[from}\left(4\right)...\left(5\right)\right]\phantom{\rule{0ex}{0ex}}\text{From}\left(3\right)\text{and}\left(5\right)\text{,}\phantom{\rule{0ex}{0ex}}{\left(gof\right)}^{-1}={f}^{-1}o{g}^{-1}$

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