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माना `f:ZtoZ` जबकि
`f(n)={{:((n-1),,"यदि", "न विषम है"),((n+1),,"यदि", "न सम है" ):}`
दर्शाइए की f व्यूत्क्रणीय है व् `f^(-1)` ज्ञात कीजिय

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(i) एकेकी है `f^(-1)`
माना `f(n_1)=f(n_2)`
Case (i) यदि `n_1` सम है ---
तब `f(n_1)=f(n_2)rArrn_1-1=n_2+1`
`rArr" "n_1-n_2=2`
yadi `n_1` visham hai, `n_2` सम है तब `n_1-n_2ne2`
`f(n_1)nef(n_2)` jab `n_2nen_2` [जब `n_1` सम व् `n_2` विषम होगा ]
Case (ii) जब `n_1` व् `n_2` दोनों विषम है `f(n_1)=f(n_2)hArrn_1-1=n_2-1`
`hArr n_1=n_2`
अतः सभी cases से f एकेकी है
(ii) f आच्छादक है - माना `n in Z`
यदि n विषम है तब `(n-1)` सम है
व् `f(n-1)=(n-1)+1=n`
तथा यदि n सम है, तब `(n+1)` विषम है
तब `f(n+1)=(n+1)-1=n`
`rArr" f"` प्रत्येक `y inZ` अपना पूर्व प्रतिबम्ब Z में रखता है
`rArrf` आच्छादक है
`f^(-1)` ज्ञात करना - स्पष्तः
`f^(-1)(n)={{:((n-1),,"यदि", "न विषम है"),((n+1),,"यदि", "न सम है"):}`

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