Functions f R R g R R are both one to one on the set of re

Functions f: R -> R, g: R -> R are both one to one on the set of real numbers R. Is the function f + g also one to one? Prove your answer.

Solution

f+g is not a one to one function.

We can disprove that f+g is a one to one function by taking a counter example.

Assume that f(x) = x and g(x) = -x. Now, (f+g)(x) = f(x) + g(x) = x-x = 0.

Hence, f+g gives the value 0 for all the values of x. Hence, f+g is not one to one.