Q5 of 107 Page 29

Show that the function f : R → R given by f (x) = x3 is injective.

Let f : R → R given by f (x) = x3.

Suppose f(x) = f(y), where x, y ϵ R.


⇒ x3 = y3 …(1)


Now, we need to show that x = y.


Suppose x ≠ y, their cubes will also not be equal.


⇒ x3 ≠ y3


However, this will be contraction to (1).


Thus, x = y


Therefore, f is injective.


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