True or false: If 𝑓 and 𝑔 are
both one-to-one functions, then 𝑓 plus 𝑔 must be a one-to-one function.
In order to prove that this
statement is true, we must prove it is true for all possible functions in this
case. However, in order to prove a
statement is false, we can do so via a counterexample. We begin by recalling that a
function is one-to-one if each element of the range of the function corresponds to
exactly one element of the domain. In mapping diagrams, this means
that each element of the range has exactly one arrow pointing to it.
Let’s consider the functions 𝑓 and
𝑔 represented by the mapping diagrams shown. We see that both functions 𝑓 and
𝑔 are injective or one-to-one because each element of the range has exactly one
arrow from the domain pointed at it. When we add 𝑓 and 𝑔 as required,
we obtain the following diagram. Noting that all three lines in the
range give the same number, we can redraw the mapping diagram. Since the range element 10 has
three arrows pointing to it, this range element corresponds to more than one of the
domain elements. We can therefore conclude the
function 𝑓 plus 𝑔 is not injective. Using a counterexample, we have
shown that the statement “if 𝑓 and 𝑔 are both one-to-one functions, then 𝑓 plus
𝑔 must be a one-to-one function” is false.