Real Function of Two Variables/Examples/Root of Minus (x^2 + y^2 + 1)
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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 $2.62: 5$