Show that the signum function f:R→R...
Question

# 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<0end{array}right.]is neither one-one nor onto.

11th - 12th Class
Maths
Solution
65
4.0 (1 ratings)

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.