Question

Is this function a surjection, 1 to 1 or a bijection, or none? Show each property. Z f: W (-1)" given by f(n) = where{x} is the greatest integer which is less than or equal to x.

Answer 1

: W Z
is defined by , $f(n)=(-1)^{n}\left [ \frac{n+1}{2} \right ]$ .

Surjection :  Let   .

mZ

Case 1 : If $m\geq 0$ then if we choose   then ,

n = 2m

$f(n)=f(2m)$

$\Rightarrow f(n)=(-1)^{2m}\left [ \frac{2m+1}{2} \right ]$

$\Rightarrow f(n)=\left [ \frac{2m+1}{2} \right ]$

$\Rightarrow f(n)=\left [ m+\frac{1}{2} \right ]$

► f(n) = m

Case 2 : If $m\leq 0$    then if we choose  $n=2(m+1)$ then ,

$f(n)=f(2(m+1))$

$\Rightarrow f(n)=(-1)^{2(m-1)}\left [ \frac{2m+2+1}{2} \right ]$

$\Rightarrow f(n)=\left [ \frac{2(m+1)-1}{2} \right ]$

$\Rightarrow f(n)=\left [ m+1-\frac{1}{2} \right ]$

► f(n) = m

So for all  
mZ
there exist  $n\in w$ such that  $f(n)=w$ .

Hence   is surjective .

One-to-one : Let $f(m)=f(n)$

$\Rightarrow (-1)^{n}\left [ \frac{n+1}{2} \right ]=(-1)^{m}\left [ \frac{m+1}{2} \right ]$

So either both n and m are even or both are odd

If both are even then , n= 2k and m=2p say then ,

$\Rightarrow (-1)^{2k}\left [ \frac{2k+1}{2} \right ]=(-1)^{2p}\left [ \frac{2p+1}{2} \right ]$

$\Rightarrow \left [k+ \frac{1}{2} \right ]=\left [p+ \frac{1}{2} \right ]$

$\Rightarrow k=p$

$\Rightarrow 2k=2p$

$\Rightarrow m=n$

Also if both m and n are odd the n = 2k+1 and m =2p +1

$\Rightarrow (-1)^{2k+1}\left [ \frac{2k+1+1}{2} \right ]=(-1)^{2p+1}\left [ \frac{2p+1+1}{2} \right ]$

$\Rightarrow -\left [ \frac{2k+1+1}{2} \right ]=-\left [ \frac{2p+1+1}{2} \right ]$

$\Rightarrow \left [ \frac{2k+1+1}{2} \right ]=\left [ \frac{2p+1+1}{2} \right ]$

$\Rightarrow \left [k+1 \right ]=\left [ p+1 \right ]$

$\Rightarrow k+1=p+1$

$\Rightarrow k=p$

$\Rightarrow 2k+1=2p+1$

$\Rightarrow m=n$

So if  $f(m)=f(n)$ then  $m=n$ . So   is one-to-one .

As   is both  one-to-one and  surjective so it is an bijection .

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