Show that the Signum Function f : R → R , given by is neither one-one nor onto.

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Solution

The given function f:R→R is defined by,

f(x)={ 1, if x>0 0, if x=0 −1, if x<0

Assume, x 1 =5 and x 2 =10.

f( 5 )=1 f( 10 )=1

Therefore, f( x 1 )=f( x 2 ), but x 1 ≠ x 2 .

Thus, f is not one-one.

For, −2∈R, there does not exist any x in R such that, f( x )=−2 since, f( x ) takes only 3 values { −1,0,1 }.

Therefore, f is not onto.

Thus, the signum function, f:R→R , defined by f(x)={ 1, if x>0 0, if x=0 −1, if x<0 is neither one-one nor onto.

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