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Appendix F. Ruler Explanations

Here we'll provide explanations for a few theorems that deal with the geometry of rulers. Unfortunately, these explanations are fairly complicated: working through each will not necessarily lead to a greater understanding.
Theorem.
Consider a ruler and a vector.
\[ \begin{align} r &: \Ruler_{(x, y)} & v &: \Vector_{(x, y)} \end{align} \]
If the vector \( v \) travels from a marking \( m_1 \) to a marking \( m_2 \), then the measurement is equal to the difference of the markings.
\[ \begin{align} \langle{r}\mathbin{|}{v}\rangle &= m_2 - m_1 \end{align} \]
Why?
Suppose the ruler \( r = a \mult \bar{\standard}_{x} + b \mult \bar{\standard}_{y} \) is centered at the point \( p = (x_p, y_p) \), and suppose the vector \( v : \Vector_{(x, y)} \) travels from \( p_1 = (x_1, y_1) \) to \( p_2 = (x_2, y_2) \). On the one hand, we can start by rewriting the difference of markings.
\[ \begin{align} m_2 - m_1 &= \bigl(a \mult (x_2 - x_p) + b \mult (y_2 - y_p)\bigr) - \bigl(a \mult (x_1 - x_p) + b \mult (y_1 - y_p)\bigr) \\ &= a \mult (x_2 - x_1) + b \mult (y_2 - y_1) \end{align} \]
On the other, we can rewrite the measurement.
\[ \begin{align} \langle{r}\mathbin{|}{v}\rangle &= \langle{r}\mathbin{|}{\position p_2 - \position p_1}\rangle \\ &= \bigl\langle{a \mult \bar{\standard}_{x} + b \mult \bar{\standard}_{y}}\mathbin{\big|}{(x_2 - x_1) \mult \vec{\standard}_{x} + (y_2 - y_1) \mult \vec{\standard}_{y}}\bigr\rangle \\ &= a \mult (x_2 - x_1) + b \mult (y_2 - y_1) \end{align} \]
We've found the same result!
Theorem.
Suppose \( r : \Ruler_{(x, y)} \) is a ruler.
  • The markings of \( r \) are perpendicular to the transpose vector, \( \transpose r \).
  • The transpose, \( \transpose r \), points in the direction of increasing markings.
Why?
Let's write our ruler \( r \) and its transpose, \( v = \transpose r \), in their standard forms
\[ \begin{align} r &= a \mult \bar{\standard}_{x} + b \mult \bar{\standard}_{y}, \\ v &= a \mult \vec{\standard}_{x} + b \mult \vec{\standard}_{y}. \end{align} \]
Looking to geometry, we see that the vector
\[ \begin{align} w &= b \mult \vec{\standard}_{x} - a \mult \vec{\standard}_{y} \end{align} \]
is perpendicular to \( v \).
We'll have verified our first statement if we can show that \( w \) is parallel to the markings of \( r \). In other words, it suffices to check that measuring \( w \) with \( r \) gives zero.
\[ \begin{align} \langle{r}\mathbin{|}{w}\rangle &= \bigl\langle{a \mult \bar{\standard}_{x} + b \mult \bar{\standard}_{y}}\mathbin{\big|}{b \mult \vec{\standard}_{x} - a \mult \vec{\standard}_{y}}\bigr\rangle \\ &= a b \mult \bigl\langle{\bar{\standard}_{x}}\mathbin{\big|}{\vec{\standard}_{x}}\bigr\rangle - a^2 \mult \bigl\langle{\bar{\standard}_{x}}\mathbin{\big|}{\vec{\standard}_{y}}\bigr\rangle + b^2 \mult \bigl\langle{\bar{\standard}_{y}}\mathbin{\big|}{\vec{\standard}_{x}}\bigr\rangle - b a \mult \bigl\langle{\bar{\standard}_{y}}\mathbin{\big|}{\vec{\standard}_{y}}\bigr\rangle \\ &= a b - b a \\ &= 0. \end{align} \]
To explain the second statement, we can use \( r \) to measure its transpose vector, \( v \).
\[ \begin{align} \langle{r}\mathbin{|}{v}\rangle &= \bigl\langle{a \mult \bar{\standard}_{x} + b \mult \bar{\standard}_{y}}\mathbin{\big|}{a \mult \vec{\standard}_{x} + b \mult \vec{\standard}_{y}}\bigr\rangle \\ &= a^2 \mult \bigl\langle{\bar{\standard}_{x}}\mathbin{\big|}{\vec{\standard}_{x}}\bigr\rangle + a b \mult \bigl\langle{\bar{\standard}_{x}}\mathbin{\big|}{\vec{\standard}_{y}}\bigr\rangle + b a \mult \bigl\langle{\bar{\standard}_{y}}\mathbin{\big|}{\vec{\standard}_{x}}\bigr\rangle + b^2 \mult \bigl\langle{\bar{\standard}_{y}}\mathbin{\big|}{\vec{\standard}_{y}}\bigr\rangle \\ &= a^2 + b^2 \end{align} \]
This measurement is positive and so the transpose must point from smaller to larger markings.
Theorem.
A ruler's magnitude is equal to the measurement the ruler gives for the radius of the unit circle.
Why?
Let's consider a ruler
\[ \begin{align} r &: \Ruler_{(x, y)} \end{align} \]
and its transpose vector
\[ \begin{align} v &: \Vector_{(x, y)} \\ v &= \transpose r. \end{align} \]
The transpose is perpendicular to the markings of \( r \). Dividing \( v \) by its magnitude
\[ \begin{align} v^{\prime} &= \frac{v}{\magnitude v} \end{align} \]
we find a unit vector pointing in the same direction as \( v \). To measure the unit circle with our ruler \( r \), we can measure the unit vector \( v^{\prime} \).
\[ \begin{align} \langle{r}\mathbin{|}{v^{\prime}}\rangle &= \biggl\langle{r}\mathbin{\bigg|}{\frac{v}{\magnitude v}}\biggr\rangle = \frac{\langle{r}\mathbin{|}{v}\rangle}{\magnitude v} = \frac{\langle{r}\mathbin{|}{\transpose r}\rangle}{\magnitude \transpose r} = \frac{(\magnitude r)^2}{\magnitude r} = \magnitude r \end{align} \]
We've found the magnitude of \( r \).
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