Pre-Algebra Examples

(- 2, 5)

$(- 2, 5)$ ,

(- 1, 0)

$(- 1, 0)$ ,

(1, - 2)

$(1, - 2)$

Step 1

Since for each value of

x

$x$ there is one and only one value of

y

$y$ , the given relation

(- 2, 5), (- 1, 0), (1, - 2)

$(- 2, 5), (- 1, 0), (1, - 2)$ is a function.

The relation is a function.

Step 2

Since the relation is a function and for each value of

y

$y$ there is one and only one value of

x

$x$ , the given relation

(- 2, 5), (- 1, 0), (1, - 2)

$(- 2, 5), (- 1, 0), (1, - 2)$ is a one-to-one function.

The relation is a one-to-one function.

Step 3

Every point in the range is the value of

y

$y$ for at least one point

x

$x$ in the domain, so this is a surjective function.

Surjective function

Step 4

Since

(- 2, 5), (- 1, 0), (1, - 2)

$(- 2, 5), (- 1, 0), (1, - 2)$ is injective (one to one) and surjective, then it is bijective function.

Bijective function