Real Function of Two Variables/Examples/Root of Minus (x^2 + y^2 + 1)

Examples of Real Functions of Two Variables

Let $z$ denote the function defined as:

$z = \sqrt {-\paren {x^2 + y^2 + 1} }$

The domain of $z$ is:

$\Dom z = \O$

That is, there are no points of the Cartesian plane for which $z$ is defined.

Proof

The domain of $z$ is given implicitly and conventionally.

What is meant is:

$z: S \to \R$ is the function defined on the largest possible subset $S$ of $\R^2$ such that:

$\forall \tuple {x, y} \in S: \map z {x, y} = \sqrt {-\paren {x^2 + y^2 + 1} }$

From Domain of Real Square Root Function, in order for the real square root function to be defined, its argument must be non-negative.

Hence for $z$ to be defined, it is necessary for:

\(\ds -\paren {x^2 + y^2 + 1}\)

\(\ge\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds x^2 + y^2\)

\(\le\)

\(\ds -1\)

But:

$\forall x, y \in \R: x^2 + y^2 \ge 0$

Hence the result.

$\blacksquare$

Sources

• 1963: Morris Tenenbaum and Harry Pollard: Ordinary Differential Equations ... (previous) ... (next): Chapter $1$: Basic Concepts: Lesson $2 \text C$: Function of Two Independent Variables: Example $2.62: 5$