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This page is not of significant importance, but thinking about it I have no formal definition of a line, how it pertains to linear maps ect, even though it's "obvious" in my head, Remember:

We must be able to account for lines in "weird" spaces, like hyperbolic lines, (arguably) lines in polar coordinates, like $r(\theta)\eq 1$ giving a circle.

Alec (talk) 13:00, 14 December 2018 (UTC)

$\newcommand{\v}[1]{\mathbf{#1} }$

See also: line (terminology) for an overview of line vs (line) segment vs ray

Definition

Parametrised line

Let $\v a\in\mathbb{R}^n$ be given (for some $n\neq 0$ ) and let $\v b\in\big(\mathbb{R}^n-\{\v 0\}\big)$ be given also^{[Note 1]}

Define \ell:\mathbb{R}\rightarrow\mathbb{R}^n^{Warning:We may be able to relax these}^{[Note 2]}, we will use t\in\mathbb{R} for the parameter to the function \ell
- Where $\ell$ is given by: $\ell:t\mapsto \v a +t\cdot\v b$

If we take a point $\v c\in\mathbb{R}^n$ in addition to $\v a$ (taken above) if we define:

$\v b:\eq \v c-\v a$ then

$\ell(t)$ is at $\v a$ for $t\eq 0$ and moves at a uniform speed towards $\ell(t)\eq \v c$ when $t\eq 1$ , that is to say:

\ell\big\vert_{[0,1]\subseteq\mathbb{R} }:[0,1]\rightarrow\mathbb{R}^n ^{(Notation: restriction)} yields the line segment from \v a to \v b
- Furthermore, this degenerates correctly in the case that $\v a\eq \v c$ (which would give $\v b\eq \v 0$

Restricting $t$ to $\mathbb{R}_{\ge 0}$ yields the ray based at $\v a$ through $\v c$ (provided we may call the case of $\ell$ given by $\v c\eq \v a$ a "ray" at all, as it's a point, same as the segment and line in this case)

Not restricting $t$ at all, so $t\in\mathbb{R}$ , yields the line through $\v a$ and $\v c$ , provided again we count the degenerate case $\v a\eq \v c$ a line.

Notes

Jump up ↑
This is written verbosely as an introduction to the style of proofs. We seek to show that $\ell(t)$ actually moves as $t$ changes without using calculus
We want \ell(t) to actually move with t, notice that:
- $\ell(t+\delta t)-\ell(t)\eq \big(\v a+(t+\delta t)\cdot\v b\big)-\big(\v a+t\cdot\v b\big)$ (for $\Vert\cdot\Vert$ denoting a norm and $\vert\cdot\vert$ denoting an absolute value object, but specifically here just the absolute value as we're dealing with $\mathbb{R}$ )
  $\eq \v a-\v a+t\cdot\v b-t\cdot \v b+\delta t\cdot \v b$
  
  $\eq \delta t\cdot \v b$
Thus:
- $\big\Vert \ell(t+\delta t)-\ell(t)\big\Vert \eq \Vert \delta t\cdot \v b\Vert \eq \vert \delta t\vert \cdot\Vert \v b\Vert$
So by the properties of a norm (using any norm for which we may write \Vert\v b\Vert) we see:
- \big(\vert\delta t\vert\cdot\Vert \v b\Vert\eq 0\in\mathbb{R}\big) $\iff$ \big((\vert\delta t\vert\eq 0) $\vee$ (\Vert\v b\Vert\eq 0)\big)
  - Notice \delta t\neq 0\in\mathbb{R} $\implies$ \vert\delta t\vert\neq 0 (for \vert\cdot\vert being the absolute value of course)
    - So to have $\big((\vert\delta t\vert\eq 0)$ $\vee$ $(\Vert\v b\Vert\eq 0)\big)$ holding (being true) means that $\Vert\v b\Vert\eq 0$ must be true (as the only way "A or B" can be true if A is known to be false is for B to be true)
Thus given \delta t\neq 0 we see that:
- \big(\big\Vert \ell(t+\delta t)-\ell(t)\big\Vert\eq 0\big)\iff\big(\big\Vert\v b\Vert \eq 0\big) which is equivalent to:
  - \big(\big\Vert \ell(t+\delta t)-\ell(t)\big\Vert\neq 0\big)\iff\big(\big\Vert\v b\Vert \neq 0\big)
    - Note that this $\iff$ is only true in the scope of $\delta t\neq 0$ being given.
Finally we observe again that as per the definition of a norm that \Vert\v b\Vert\eq 0\iff \v b\eq \v 0, combining this we have shown:
- $\forall \delta t\in\mathbb{R}\forall \v a,\v b\in\mathbb{R}^n\left[\delta t\neq 0\implies \Big(\big[\Big\Vert \ell(t+\delta t)-\ell(t)\big\Vert\neq 0\Big]\iff\big\Vert\v b\Vert \neq 0\iff \v b\neq \v 0\Big) \right]$
That is that provided \v b is non-zero, the line is "not a point", \ell(t) actually changes with t.
Jump up ↑ In fact we must be able to relax this to give lines in "weird" spaces

References

[NormAndZero-1] Jump up ↑
This is written verbosely as an introduction to the style of proofs. We seek to show that $\ell(t)$ actually moves as $t$ changes without using calculus
We want $\ell(t)$ to actually move with $t$ , notice that:
$\ell(t+\delta t)-\ell(t)\eq \big(\v a+(t+\delta t)\cdot\v b\big)-\big(\v a+t\cdot\v b\big)$ (for $\Vert\cdot\Vert$ denoting a norm and $\vert\cdot\vert$ denoting an absolute value object, but specifically here just the absolute value as we're dealing with $\mathbb{R}$ )
$\eq \v a-\v a+t\cdot\v b-t\cdot \v b+\delta t\cdot \v b$

$\eq \delta t\cdot \v b$
Thus:
$\big\Vert \ell(t+\delta t)-\ell(t)\big\Vert \eq \Vert \delta t\cdot \v b\Vert \eq \vert \delta t\vert \cdot\Vert \v b\Vert$
So by the properties of a norm (using any norm for which we may write $\Vert\v b\Vert$ ) we see:
$\big(\vert\delta t\vert\cdot\Vert \v b\Vert\eq 0\in\mathbb{R}\big)$ $\iff$ $\big((\vert\delta t\vert\eq 0)$ $\vee$ $(\Vert\v b\Vert\eq 0)\big)$
Notice $\delta t\neq 0\in\mathbb{R}$ $\implies$ $\vert\delta t\vert\neq 0$ (for $\vert\cdot\vert$ being the absolute value of course)
So to have $\big((\vert\delta t\vert\eq 0)$ $\vee$ $(\Vert\v b\Vert\eq 0)\big)$ holding (being true) means that $\Vert\v b\Vert\eq 0$ must be true (as the only way "A or B" can be true if A is known to be false is for B to be true)
Thus given $\delta t\neq 0$ we see that:
$\big(\big\Vert \ell(t+\delta t)-\ell(t)\big\Vert\eq 0\big)\iff\big(\big\Vert\v b\Vert \eq 0\big)$ which is equivalent to:
$\big(\big\Vert \ell(t+\delta t)-\ell(t)\big\Vert\neq 0\big)\iff\big(\big\Vert\v b\Vert \neq 0\big)$
Note that this $\iff$ is only true in the scope of $\delta t\neq 0$ being given.
Finally we observe again that as per the definition of a norm that $\Vert\v b\Vert\eq 0\iff \v b\eq \v 0$ , combining this we have shown:
$\forall \delta t\in\mathbb{R}\forall \v a,\v b\in\mathbb{R}^n\left[\delta t\neq 0\implies \Big(\big[\Big\Vert \ell(t+\delta t)-\ell(t)\big\Vert\neq 0\Big]\iff\big\Vert\v b\Vert \neq 0\iff \v b\neq \v 0\Big) \right]$
That is that provided $\v b$ is non-zero, the line is "not a point", $\ell(t)$ actually changes with $t$ .

[2] $\ell(t+\delta t)-\ell(t)\eq \big(\v a+(t+\delta t)\cdot\v b\big)-\big(\v a+t\cdot\v b\big)$ (for $\Vert\cdot\Vert$ denoting a norm and $\vert\cdot\vert$ denoting an absolute value object, but specifically here just the absolute value as we're dealing with $\mathbb{R}$ )
$\eq \v a-\v a+t\cdot\v b-t\cdot \v b+\delta t\cdot \v b$

$\eq \delta t\cdot \v b$

[3] $\big\Vert \ell(t+\delta t)-\ell(t)\big\Vert \eq \Vert \delta t\cdot \v b\Vert \eq \vert \delta t\vert \cdot\Vert \v b\Vert$

[4] $\big(\vert\delta t\vert\cdot\Vert \v b\Vert\eq 0\in\mathbb{R}\big)$ $\iff$ $\big((\vert\delta t\vert\eq 0)$ $\vee$ $(\Vert\v b\Vert\eq 0)\big)$
Notice $\delta t\neq 0\in\mathbb{R}$ $\implies$ $\vert\delta t\vert\neq 0$ (for $\vert\cdot\vert$ being the absolute value of course)
So to have $\big((\vert\delta t\vert\eq 0)$ $\vee$ $(\Vert\v b\Vert\eq 0)\big)$ holding (being true) means that $\Vert\v b\Vert\eq 0$ must be true (as the only way "A or B" can be true if A is known to be false is for B to be true)

[5] Notice $\delta t\neq 0\in\mathbb{R}$ $\implies$ $\vert\delta t\vert\neq 0$ (for $\vert\cdot\vert$ being the absolute value of course)
So to have $\big((\vert\delta t\vert\eq 0)$ $\vee$ $(\Vert\v b\Vert\eq 0)\big)$ holding (being true) means that $\Vert\v b\Vert\eq 0$ must be true (as the only way "A or B" can be true if A is known to be false is for B to be true)

[6] So to have $\big((\vert\delta t\vert\eq 0)$ $\vee$ $(\Vert\v b\Vert\eq 0)\big)$ holding (being true) means that $\Vert\v b\Vert\eq 0$ must be true (as the only way "A or B" can be true if A is known to be false is for B to be true)

[7] $\big(\big\Vert \ell(t+\delta t)-\ell(t)\big\Vert\eq 0\big)\iff\big(\big\Vert\v b\Vert \eq 0\big)$ which is equivalent to:
$\big(\big\Vert \ell(t+\delta t)-\ell(t)\big\Vert\neq 0\big)\iff\big(\big\Vert\v b\Vert \neq 0\big)$
Note that this $\iff$ is only true in the scope of $\delta t\neq 0$ being given.

[8] $\big(\big\Vert \ell(t+\delta t)-\ell(t)\big\Vert\neq 0\big)\iff\big(\big\Vert\v b\Vert \neq 0\big)$
Note that this $\iff$ is only true in the scope of $\delta t\neq 0$ being given.

[9] Note that this $\iff$ is only true in the scope of $\delta t\neq 0$ being given.

[10] $\forall \delta t\in\mathbb{R}\forall \v a,\v b\in\mathbb{R}^n\left[\delta t\neq 0\implies \Big(\big[\Big\Vert \ell(t+\delta t)-\ell(t)\big\Vert\neq 0\Big]\iff\big\Vert\v b\Vert \neq 0\iff \v b\neq \v 0\Big) \right]$

[2] Jump up ↑ In fact we must be able to relax this to give lines in "weird" spaces

[Note 1]

[Note 2]

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