Let a function f : N → N be defined by

$f\left(n\right)=[\begin{array}{ccc}\hfill 2n\hfill & \hfill ,\hfill & \hfill n=2,4,6,8....\hfill \\ \hfill n-1\hfill & \hfill ,\hfill & \hfill n=3,7,11,15,....\hfill \\ \hfill \frac{n+1}{2}\hfill & \hfill ,\hfill & \hfill n=1,5,9,13,.....\hfill \end{array}$

then, f is

<p>&nbsp;<math><mrow><mi>f</mi><mrow><mo> (</mo><mrow><mi>x</mi></mrow><mo>)</mo></mrow><mo>=</mo><mrow><mo> [</mo><mrow><mtable><mtr columnalign="center"><mtd columnalign="center"><mn>2</mn><mi>n</mi></mtd><mtd columnalign="center"><mo></mo></mtd><mtd columnalign="center"><mi>n</mi><mo>=</mo><mn>2</mn><mo>, </mo><mn>4</mn><mo>, </mo><mn>6</mn><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo></mtd></mtr><mtr columnalign="center"><mtd columnalign="center"><mi>n</mi><mo>−</mo><mn>1</mn></mtd><mtd columnalign="center"><mo></mo></mtd><mtd columnalign="center"><mi>n</mi><mo>=</mo><mn>3</mn><mo>, </mo><mn>7</mn><mo>, </mo><mn>1</mn><mn>1</mn><mo>, </mo><mn>1</mn><mn>5</mn><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo></mtd></mtr><mtr columnalign="center"><mtd columnalign="center"><mfrac><mrow><mi>n</mi><mo>+</mo><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mtd><mtd columnalign="center"><mo></mo></mtd><mtd columnalign="center"><mi>n</mi><mo>=</mo><mn>1</mn><mo>, </mo><mn>5</mn><mo>, </mo><mn>9</mn><mo>, </mo><mn>1</mn><mn>3</mn><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo></mtd></mtr></mtable></mrow></mrow></mrow></math></p><p>for n = 2, 4, 6 ……</p><p>f (n) = 4, 8, 12, ….4 (n) form</p><p>for n = 3, 7, 11, 15, ….</p><p>f (n) = 1, 3, 5, 7, …. (4n + 1) or, (4n + 3) from</p><p><math><mrow><mo>∴</mo></mrow></math>&nbsp;f is one and onto.</p>

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