Question

show that signum function f : R→R, given by:
f(x) = 1, if x>0
0,if x=0
-1,if x<0

is neither one-one nor onto.

Answer 1

y = sgn(x )
signum function is in actually |x|/x
y = |x|/x
when , x > 0 then, y = 1
when, x = 0 then, y = 0
when, x< 0 then, y = -1
hence, domain of y = sgn(x) is all real number ,
range of y = sgn(x) is { -1 , 0, 1 }

now,
we know,
function is an one - one function when,
function is strictly increasing or decreasing , e.g dy/dx > 0 or < 0
but y = sgn(x) doesn't differentiate at x = 0 , also function is neither increasing nor decreasing , so , y = sgn(x) is not one-one function .

onto function :- co-domain = range
here , co- domain € R
but range € { -1 , 0, 1}
so, co-domain ≠ range
hence, y = sgn(x) is not onto function

Answer 2

Question= that the Signum Function f: R → R, given by

f(x)=⎧⎩⎨⎪⎪⎪⎪⎪⎪10−1for x>0for x=0 is neither one−one nor ontofor x<0

Solution:

Check for one to one function:

For example:

f(0) = 0

f(-1) = -1

f(1) = 1

f(2) = 1

f(3) = 1

Since, for the different elements say f(1), f(2) and f(3), it shows the same image, then the function is not one to one function.

Check for Onto Function:

For the function,f: R →R

f(x)=⎧⎩⎨⎪⎪10−1for x>0for x=0for x<0

In this case, the value of f(x) is defined only if x is 1, 0, -1

For any other real numbers(for example y = 2, y = 100) there is no corresponding element x.

Thus, the function “f” is not onto function.

Hence, the given function “f” is neither one-one nor onto.