ThmDex – An index of mathematical definitions, results, and conjectures.
Reflection invariance of Euclidean volume function
Formulation 0
Let $\mathbb{R}^N$ be a D816: Euclidean real Cartesian product.
Let $\mathsf{Vol}$ be the D1738: Euclidean real volume function on $\mathbb{R}^N$.
Let $I \subseteq \mathbb{R}^N$ be D3036: Real N-interval.
Then \begin{equation} \mathsf{Vol} (- I) = \mathsf{Vol} (I) \end{equation}
Proofs
Proof 0
Let $\mathbb{R}^N$ be a D816: Euclidean real Cartesian product.
Let $\mathsf{Vol}$ be the D1738: Euclidean real volume function on $\mathbb{R}^N$.
Let $I \subseteq \mathbb{R}^N$ be D3036: Real N-interval.
By definition, $I$ is a Cartesian product of some basic real intervals $I_1, \dots, I_N$. Let $a_1, b_1, \dots, a_N, b_N$ be the endpoints of $I_1, \dots, I_N$, respectively. Result R2973: The four classes of real intervals under reflection states that now $-b_1, - a_1, \dots, -b_N, -a_N$ are the endpoints of the reflected intervals $- I_1, \dots, - I_N$, respectively. Applying result R2969: Euclidean real Cartesian product of scaled sets now yields \begin{equation} \begin{split} \mathsf{Vol}(- I) & = \mathsf{Vol} \Big( - \prod_{n = 1}^N I_n \Big) \\ & = \mathsf{Vol} \Big( \prod_{n = 1}^N -I_n \Big) \\ & = \prod_{n = 1}^N |(-b_n) - (-a_n)| \\ & = \prod_{n = 1}^N |b_n - a_n| \\ & = \mathsf{Vol} \Big( \prod_{n = 1}^N I_n \Big) \\ & = \mathsf{Vol}(I) \end{split} \end{equation} $\square$