Show that the function

Question:

Show that the function $f: N \rightarrow Z$, defined by

$\mathrm{f}(\mathrm{n})=\left\{\begin{array}{l}\frac{1}{2}(\mathrm{n}-1), \text { when } \mathrm{n} \text { is odd } \\ -\frac{1}{2} \mathrm{n}, \text { when } \mathrm{n} \text { is even }\end{array}\right.$

is both one - one and onto.

 

 

Solution:

$f(n)=\left\{\begin{array}{l}\frac{1}{2}(n-1), \text { when } n \text { is odd } \\ -\frac{1}{2} n, \text { when } n \text { is even }\end{array}\right.$

$f(1)=0$

$f(2)=-1$

$f(3)=1$

$f(4)=-2$

$f(5)=2$

$f(6)=-3$

Since at no different values of $x$ we get same value of $y \therefore: f(n)$ is one -one

And range of $f(n)=Z=Z$ (codomain)

$\therefore$ the function $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{Z}$, defined by

$f(n)=\left\{\begin{array}{l}\frac{1}{2}(n-1), \text { when } n \text { is odd } \\ -\frac{1}{2} n, \text { when } n \text { is even }\end{array}\right.$

is both one - one and onto.

 

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