Answer to Question #290710 in Discrete Mathematics for Yaku

Question

a) Consider the following functional relation, f, defined as:

f : R \rightarrow R, f(x) = x2

Determine whether or not f is a bijection. If it is, prove it. If it is not, show why it is not.

b) Consider the set

F = {y | y = ax3 + b},

a, b being constants such that a \ne 0 and x \in R.
Is F equivalent to R? If so, prove it. If not, explain in details why it is not the case.

Answer 1

a.)

$f:\R \to \R, f(x)=x^2$

This function is not a bijection since it is not one-to-one. We can see this as follows:

$f(-2)=f(2)=4$, but $-2 \neq 2$. Thus $f$ is not injective. Hence it is not a bijection.

b.)

$F=\{y|y=ax^3+b\}$

Yes, $F$ is equivalent to $\R$ since will can always find a bijection between the two set.

$f:F\to \R$