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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 [ilmath]r(\theta)\eq 1[/ilmath] giving a circle.

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

[ilmath]\newcommand{\v}[1]{\mathbf{#1} } [/ilmath]

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

Definition

Parametrised line

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

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

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

[ilmath]\v b:\eq \v c-\v a[/ilmath] then

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

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

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

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

Notes

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

References

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

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

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

[ilmath]\eq \delta t\cdot \v b[/ilmath]

[3] [ilmath]\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[/ilmath]

[4] [ilmath]\big(\vert\delta t\vert\cdot\Vert \v b\Vert\eq 0\in\mathbb{R}\big)[/ilmath][ilmath]\iff[/ilmath][ilmath]\big((\vert\delta t\vert\eq 0)[/ilmath][ilmath]\vee[/ilmath][ilmath](\Vert\v b\Vert\eq 0)\big)[/ilmath]
Notice [ilmath]\delta t\neq 0\in\mathbb{R} [/ilmath][ilmath]\implies[/ilmath][ilmath]\vert\delta t\vert\neq 0[/ilmath] (for [ilmath]\vert\cdot\vert[/ilmath] being the absolute value of course)
So to have [ilmath]\big((\vert\delta t\vert\eq 0)[/ilmath][ilmath]\vee[/ilmath][ilmath](\Vert\v b\Vert\eq 0)\big)[/ilmath] holding (being true) means that [ilmath]\Vert\v b\Vert\eq 0[/ilmath] 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 [ilmath]\delta t\neq 0\in\mathbb{R} [/ilmath][ilmath]\implies[/ilmath][ilmath]\vert\delta t\vert\neq 0[/ilmath] (for [ilmath]\vert\cdot\vert[/ilmath] being the absolute value of course)
So to have [ilmath]\big((\vert\delta t\vert\eq 0)[/ilmath][ilmath]\vee[/ilmath][ilmath](\Vert\v b\Vert\eq 0)\big)[/ilmath] holding (being true) means that [ilmath]\Vert\v b\Vert\eq 0[/ilmath] 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 [ilmath]\big((\vert\delta t\vert\eq 0)[/ilmath][ilmath]\vee[/ilmath][ilmath](\Vert\v b\Vert\eq 0)\big)[/ilmath] holding (being true) means that [ilmath]\Vert\v b\Vert\eq 0[/ilmath] 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] [ilmath]\big(\big\Vert \ell(t+\delta t)-\ell(t)\big\Vert\eq 0\big)\iff\big(\big\Vert\v b\Vert \eq 0\big)[/ilmath] which is equivalent to:
[ilmath]\big(\big\Vert \ell(t+\delta t)-\ell(t)\big\Vert\neq 0\big)\iff\big(\big\Vert\v b\Vert \neq 0\big)[/ilmath]
Note that this [ilmath]\iff[/ilmath] is only true in the scope of [ilmath]\delta t\neq 0[/ilmath] being given.

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

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

[10] [ilmath]\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][/ilmath]

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

[Note 1]

[Note 2]

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References

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