Question

Let f: R→ R be given by f(x) = x2. Prove that the function is not injective.

Answer 1

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.