Taking the square root of both sides.

[nothing]

Let unity be 1/1 as c.
Let A represent any identity displaced from c
whose constituency is s/t seeking to unify.

A = s/t
A → 1/1
A = 1/t, s/1
±√A = ±√1/t, s/±√1
√A= (+1/t, -1/t), (s/+1, s/-1)
A = √(+1/t, -1/t), √(s/+1, s/-1)

Valid?

 

[MarkFL]

Can you elaborate on:

"whose constituency is s/t seeking to unify" ?

 

[HallsofIvy]

That's all you didn't understand?

I am puzzled by "Let unity be 1/1 as c", and "Let A represent any identity displaced from c".

And I don't understand how "A= s/t" leads to "A= 1/t, s/1". Is that "." supposed to indicate a product?

 

[JeffM]

Let unity be 1/1 as c.
Let A represent any identity displaced from c
whose constituency is s/t seeking to unify.

A = s/t
A → 1/1
A = 1/t, s/1
±√A = ±√1/t, s/±√1
√A= (+1/t, -1/t), (s/+1, s/-1)
A = √(+1/t, -1/t), √(s/+1, s/-1)

Valid?

Because your language does not make much sense in English, we do not know what you are trying to ask.

[MATH]\text {Given: } A \equiv \dfrac{s}{t} \equiv 1. [/MATH]
[MATH]s = - \ 3 = t \implies A = \dfrac{s}{t}.[/MATH]
That is true, BUT

[MATH]1 \equiv A \ne \dfrac{-\ 3}{1} = -\ 3.[/MATH]
AND

[MATH]1 \equiv A \ne \dfrac{1}{-\ 3} = - \dfrac{1}{3}.[/MATH]
Your third line is true if and only if

[MATH]s = 1 = t \text { or } s = -\ 1 = t.[/MATH]
Moreover, you cannot take the square root of a negative number in the real number system.

It is true that [MATH]\pm \sqrt{A} = \pm 1[/MATH]
BUT IT IS FALSE that

[MATH](1)^2 = -\ 1.[/MATH]
IT IS ALSO FALSE that

[MATH](-\ 1)^2 = -\ 1.[/MATH]
Your fourth line is wrong for multiple reasons. Here is what you can say in terms of real numbers.

[MATH]s = 1 = t \text { and } A = \dfrac{s}{t} \implies \sqrt{A} = \dfrac{\sqrt{1}}{t} = \dfrac{s}{\sqrt{1}}. [/MATH]

 

[nothing]

Can you elaborate on:

"whose constituency is s/t seeking to unify" ?

To clarify: take space s, over time t, as simply a universal unit of motion(s), captured by A, in A = s/t,
1/1 being the speed of light as c, unity.

A is attempting to unify with c as 1/1, however has
a spacial displacement(s)/temporal displacement(s).

A = s/t
A → 1/1
A = 1/t, s/1
_________________________________________
last line captures each displacement(s) if/when setting the other to the desired 1.

i. A = s/t
ii. A → 1/1
iii. A = 1/t, s/1
(time and space displacement(s) from 1/1 as coordinates)
iv. ±√A = ±√1/t, s/±√1
(both roots of A taken: is taking the roots of the other side done equivalently?)
v. √A= (+1/t, -1/t), (s/+1, s/-1)
(± collapses on both sides, thus A can be either + or -, each with their respective spacial counter-part)
vi. A = √(+1/t, -1/t), √(s/+1, s/-1)

Please allow for the sake of argument that the root of a negative number(s) is a valid proposition.

If needed I can retry from the beginning.

That's all you didn't understand?

I am puzzled by "Let unity be 1/1 as c", and "Let A represent any identity displaced from c".

And I don't understand how "A= s/t" leads to "A= 1/t, s/1". Is that "." supposed to indicate a product?

Sorry, should have been clearer - hopefully clarified above. If not:

A is any motioning body, such as a person.
s is space, t is time, c is unity at the speed of light 1/1.
Assuming A ≠ 1 already, thus displaced, I am trying
to show how A returns to unity 1/1 if/when displaced.

Assume time and space have a multiplicative reciprocal relationship:
3d space and 1d time in the "real", and
1d space and 3d time in the "ethereal".

I want s/t to have their own respective reciprocal(s), hence
I'm hoping to confirm there is a quaternion in the OP.

 

[nothing]

Because your language does not make much sense in English, we do not know what you are trying to ask.

[MATH]\text {Given: } A \equiv \dfrac{s}{t} \equiv 1. [/MATH]
[MATH]s = - \ 3 = t \implies A = \dfrac{s}{t}.[/MATH]
That is true, BUT

[MATH]1 \equiv A \ne \dfrac{-\ 3}{1} = -\ 3.[/MATH]
AND

[MATH]1 \equiv A \ne \dfrac{1}{-\ 3} = - \dfrac{1}{3}.[/MATH]

Can you explain why? This interests me because I am working with the following universe in mind:

3d space / 1d time
(reciprocates into}
1d space / 3d time

I am wondering what the technical explanation is?

Your third line is true if and only if

[MATH]s = 1 = t \text { or } s = -\ 1 = t.[/MATH]

How would I capture A seeking 1/1 if not?

Moreover, you cannot take the square root of a negative number in the real number system.

It is permitted for my case - the root of a negative number needs to be unreal.

It is true that [MATH]\pm \sqrt{A} = \pm 1[/MATH]
BUT IT IS FALSE that

[MATH](1)^2 = -\ 1.[/MATH]
IT IS ALSO FALSE that

[MATH](-\ 1)^2 = -\ 1.[/MATH]
Your fourth line is wrong for multiple reasons. Here is what you can say in terms of real numbers.

[MATH]s = 1 = t \text { and } A = \dfrac{s}{t} \implies \sqrt{A} = \dfrac{\sqrt{1}}{t} = \dfrac{s}{\sqrt{1}}. [/MATH]

If the last permitted imaginary numbers, would:

√A = √1/t, s/√1
± A= (+1/t, -1/t), (s/+1, s/-1)

be coherent?

 

[JeffM]

Can you explain why? This interests me because I am working with the following universe in mind:

3d space / 1d time
(reciprocates into}
1d space / 3d time

I am wondering what the technical explanation is?

Your reciprocating of space and time is not a mathematical statement at all. It is making an assertion about the physical universe. Whether that statement can even begin to make physical sense or can be experimentally tested, I shall leave to physicists.

In any case, time and space must be measured in units so the mathematical translation will differ depending on what units are involved. That is, whatever the numeric formula is for translating cubic miles per hour into miles per cubic hour, it will differ from the numeric formula for translating cubic centimeters per second into centimeters per cubic second. (As I said, I do not know whether those mathematical formulas mean anything in terms of physics.)

How would I capture A seeking 1/1 if not?

I have no idea what you are trying to convey here.

If the last permitted imaginary numbers, would:

√A = √1/t, s/√1
± A= (+1/t, -1/t), (s/+1, s/-1)

be coherent?

No. If we are talking any kind of number and standard operations where

[MATH]A = 1 \text { and } A = \dfrac{s}{t}[/MATH]
Then all we can conclude in general is

[MATH]s = t \ne 0.[/MATH]
In general, it will be FALSE that

[MATH]A = \dfrac{1}{t} \text { or } A = \dfrac{s}{1}.[/MATH]
If you do not believe me try s = 3 = t. In that case, s/t = 1. But 1 does not equal 3 or 1/3.

Therefore it will also be FALSE in general that

[MATH]\pm \sqrt{A} = \pm \sqrt{\dfrac{1}{t}} \text { or } \pm \sqrt{A} = \pm \sqrt{\dfrac{s}{1}}.[/MATH]
And finally it will be FALSE in general that

[MATH]A = \dfrac{1}{t} \text { or } -\ \dfrac{1}{t} \text { or } \dfrac{s}{1} \text { or } -\ \dfrac{s}{1}.[/MATH]
The only time that the first false statement will be true is in the special case where s and t both equal 1.

[MATH]s = 1 = t, \text { and } A = \dfrac{s}{t}[/MATH]
DOES ENTAIL that

[MATH]A = \dfrac{1}{t} \text { and } A = \dfrac{s}{1}.[/MATH]
And that in turn DOES ENTAIL that

[MATH]\sqrt{A} = \sqrt{\dfrac{1}{t}} =\sqrt{\dfrac{s}{1}} \text { and } -\ \sqrt{A} = -\ \sqrt{\dfrac{1}{t}} = - \sqrt{\dfrac{s}{1}}[/MATH]
BUT EVEN IN THAT SPECIAL CASE, it will be FALSE that

[MATH]A = -\ \dfrac{1}{t} \text { or } -\ \dfrac{s}{1}.[/MATH]
I think you do not understand what the abbreviation [MATH]\pm[/MATH] means exactly and how careful you must be when using and thinking about that abbreviation.

 

[nothing]

Your reciprocating of space and time is not a mathematical statement at all. It is making an assertion about the physical universe. Whether that statement can even begin to make physical sense or can be experimentally tested, I shall leave to physicists.

How is a multiplicative reciprocal not a mathematical statement? For example:

"
'reciprocal' - any rationed identity(s) (as denoted by A)
whose variable multiplicative inverse(s),
if/when known, attained to and/or rationally operated on,
invariably reconciles to unity as 1/1. "

Let rA be the multiplicative reciprocal of A.
A x rA = 1/1
3/1 x 1/3 = 1/1

These two reciprocal regions are symmetrical:
3d space / 1d time as the 'real' sector,
1d space / 3d time as the 'ethereal' sector.

(3d space, 1d time) x (1d space, 3d time) = unity 1/1 denoted by c, the same taken as the speed of light.

This is the construct I am working with - I allowed perhaps the implicit relationship between space and time is important
to what the derivation ultimately means.

In any case, time and space must be measured in units so the mathematical translation will differ depending on what units are involved. That is, whatever the numeric formula is for translating cubic miles per hour into miles per cubic hour, it will differ from the numeric formula for translating cubic centimeters per second into centimeters per cubic second. (As I said, I do not know whether those mathematical formulas mean anything in terms of physics.)

The units are units of motion v:

v = s/t

For example, 3d space is (x, y, z) however 3d time would be like this:
t^1 = 24 hour day
t^2 = 1 solar year
t^3 = 25 920 great year
thus if v were being perceived from the 'ethereal sector' 1d space / 3d time,
v would only have one relevant spacial coordinate s, but three valid time coordinates
t1, t2 and t3 accordingly, because they all exist in/as a circle surrounding v.

I am trying to build an orientation system if/when placing any theoretical being into v.

I have no idea what you are trying to convey here.

In A = s/t, A is displaced from unity 1/1.
Therefor, A ≠ 1/1
however A would like to be united.
Therefor, if setting either s or t to 1,
we can find the displacement(s) of A in the other.

I need an expression which sets s=1 to find t and/or t=1 to find s
such to allow A to "know" which direction it is traveling: towards c, or away from c.

Another attempt to capture:

A = s/1, 1/t
√A = s/√1, √1/t
√A = (s/+1, s/-1), (1/+1, 1/-1)
A = √(s/+1, s/-1), √(t/+1, t/-1)

Maybe this is better? Is this a quaternion?

No. If we are talking any kind of number and standard operations where

[MATH]A = 1 \text { and } A = \dfrac{s}{t}[/MATH]
Then all we can conclude in general is

[MATH]s = t \ne 0.[/MATH]
In general, it will be FALSE that

[MATH]A = \dfrac{1}{t} \text { or } A = \dfrac{s}{1}.[/MATH]
If you do not believe me try s = 3 = t. In that case, s/t = 1. But 1 does not equal 3 or 1/3.

Therefore it will also be FALSE in general that

[MATH]\pm \sqrt{A} = \pm \sqrt{\dfrac{1}{t}} \text { or } \pm \sqrt{A} = \pm \sqrt{\dfrac{s}{1}}.[/MATH]
And finally it will be FALSE in general that

[MATH]A = \dfrac{1}{t} \text { or } -\ \dfrac{1}{t} \text { or } \dfrac{s}{1} \text { or } -\ \dfrac{s}{1}.[/MATH]
The only time that the first false statement will be true is in the special case where s and t both equal 1.

[MATH]s = 1 = t, \text { and } A = \dfrac{s}{t}[/MATH]
DOES ENTAIL that

[MATH]A = \dfrac{1}{t} \text { and } A = \dfrac{s}{1}.[/MATH]
And that in turn DOES ENTAIL that

[MATH]\sqrt{A} = \sqrt{\dfrac{1}{t}} =\sqrt{\dfrac{s}{1}} \text { and } -\ \sqrt{A} = -\ \sqrt{\dfrac{1}{t}} = - \sqrt{\dfrac{s}{1}}[/MATH]
BUT EVEN IN THAT SPECIAL CASE, it will be FALSE that

[MATH]A = -\ \dfrac{1}{t} \text { or } -\ \dfrac{s}{1}.[/MATH]
I think you do not understand what the abbreviation [MATH]\pm[/MATH] means exactly and how careful you must be when using and thinking about that abbreviation.

I did away with - thank you for the help.