Subtraction of Subring is Subtraction of Ring
Jump to navigation
Jump to search
Theorem
Let $\struct {R, +, \circ}$ be an ring.
For each $x, y \in R$ let $x - y$ denote the subtraction of $x$ and $y$ in $R$.
Let $\struct {S, + {\restriction_S}, \circ {\restriction_S}}$ be a subring of $R$.
For each $x, y \in S$ let $x \sim y$ denote the subtraction of $x$ and $y$ in $S$.
Then:
- $\forall x, y \in S: x \sim y = x - y$
Proof
Let $x, y \in S$.
Let $-x$ denote the ring negative of $x$ in $R$.
Let $\mathbin \sim x$ denote the ring negative of $x$ in $S$.
Then:
\(\ds x \sim y\) | \(=\) | \(\ds x \mathbin {+ {\restriction_S} } \paren {\mathbin \sim y}\) | Definition of Ring Subtraction | |||||||||||
\(\ds \) | \(=\) | \(\ds x + \paren {\mathbin \sim y}\) | Definition of Addition on Subring | |||||||||||
\(\ds \) | \(=\) | \(\ds x + \paren {-y}\) | Negative of Subring is Negative of Ring | |||||||||||
\(\ds \) | \(=\) | \(\ds x - y\) | Definition of Ring Subtraction |
$\blacksquare$