Question # Show that the signum function ( f: R rightarrow R, ) defined by

# Show that the signum function ( f: R rightarrow R, ) defined by

[

f(x)=left{begin{array}{l}

1, text { if } x>0

0, text { if } x=0

-1, text { if } x<0

end{array}right.

]

is neither one-one nor onto.

Solution

Clearly, ( f(2)=1 ) and ( f(3)=1 ) Thus, ( f(2)=f(3) ) while ( 2 eq 3 ) ( therefore f ) is not one-one. ( operatorname{Range}(f)={1,0,-1} subset R )

So, ( f ) is into.

Hence, ( f ) is neither one-one nor onto.