Now, I have sat down and pondered this more, and you can more or less create any physical law using elementary algebra… But if there are mistakes here, please point them out… as I am sure you will do anyway if there is
Since
$$i^0$$
$$i^1=i$$
$$i^2=-1$$
$$i^3=-1.i=-i$$
And
$$i^4=-i.i=-i^2=-(-1)=1$$
Then we can integrate some variables, so long as we abide by the powers of
$$\sqrt{1}$$ and $$\sqrt{-1}$$
Then…
On a real right triangle, we have the Pythagorean Discipline $$a^2+b^2=c^2$$ where the solution is the $$c=\sqrt{a+b}$$ so $$c^2-b^2=a^2$$ or $$c^2-b^2=a^2$$, because $$a_{1}*b_{1}><b_{2}*a_{2}$$, so $$a$$ and $$b$$ are true, if there is no biased attack that $$b$$ cannot also be $$a$$.
These are called Logic Deductions, so $$\Inot (P\land\lnot P)$$ holds it as a pure logic that not only in a single vector, does one see it has no real directionality, even in three dimensions, vector + vector components cannot be biased either.
$$a^2-a(i)=b$$
So
$$a^2-a(i^0) \pm b$$
Then
$$i^0=-(-1)=-i^2=-1$$
Then it must hold true that the values of $$\sqrt{1}$$ and $$\sqrt{-1}$$ play exactly the same roles, when conditionally that each can never yield a positive value of when $$b>0$$. Here, we role on:
$$a^2-a(i^{0}_{(j)i}) \pm b >< (-1)=-i^2=\sqrt{-1}$$
Then we have yielded a negative value, from something which is positive to begin with, so we can then arbitrarily say the reverse is always the same. Again, there is no forward direction at these scales, if there is no complex-conjugate.
Now let the trace variables have
$$(j)i=-i^2$$ and $$i(j)=i^4$$
So then $${(j)i}x{(i(j)}=\sqrt{-1}$$
But as I thought about it some more, it came to realize that this is only true if $$i^2 \pm \sqrt{1}$$, hence them acting analogous to complex conjugated functions. So $$\psi \psi* =1$$ is one way to yield a real value in any vector function.
We can work out some of these real attributes, by working off the values of the known three dimensions we see in operation everyday we go to work, or read a book, or simply noticing the sides of a cube, that in what are called row vectors $$|V|=0,0,0$$, yields real values if we allow:
$$i^{2}=i*k^{2}$$
Then in a three-dimensional coordinated system, we can reduce the real parts of any connected vectors like so ~
$$a+b-i^2*k^{2^2}=0$$
So long as
$$a+b+c+k^{2^2}=0$$
Other than this being proof, if there are no mistakes, then you can also prove that a vector has two directionalities, (one negative and one positive), any power of $$i$$ can be either $$\sqrt{1}$$ or $$\sqrt{-1}$$ so long as the exponent is only ever a multiple of $$2$$, so we can have what we call ‘’odd powers,’’ since they are strictly given only as $$\sqrt{1}$$ or $$\sqrt{-1}$$ since $$-i=i^3$$.
Out of further analysis, going back to $$a^2-a(i^{0}_{(j)i}) \pm b >< (-1)=-i^2=\sqrt{-1}$$, we can allow (by geometrical methods and mathematical disciplines), that $$\sqrt{-1}=i^{n}$$, so long as it has a remainder of $$2$$, so as far as I have deducted,
$$i^{n}=i^{18}=i^{2}=\sqrt{-1}$$
So the $$i^0$$ in $$a^2-a(i^{0}_{(j)i}) \pm b >< (-1)=-i^2=\sqrt{-1}$$ is $$i^{40}$$ if there is absolutely no remainder at all, so $$i^0=(a^{2}-a(i^{40}))$$.
Let is know what you think, because if there are any errors, I would like to see why…Ta Ta 4 the now!
Since
$$i^0$$
$$i^1=i$$
$$i^2=-1$$
$$i^3=-1.i=-i$$
And
$$i^4=-i.i=-i^2=-(-1)=1$$
Then we can integrate some variables, so long as we abide by the powers of
$$\sqrt{1}$$ and $$\sqrt{-1}$$
Then…
On a real right triangle, we have the Pythagorean Discipline $$a^2+b^2=c^2$$ where the solution is the $$c=\sqrt{a+b}$$ so $$c^2-b^2=a^2$$ or $$c^2-b^2=a^2$$, because $$a_{1}*b_{1}><b_{2}*a_{2}$$, so $$a$$ and $$b$$ are true, if there is no biased attack that $$b$$ cannot also be $$a$$.
These are called Logic Deductions, so $$\Inot (P\land\lnot P)$$ holds it as a pure logic that not only in a single vector, does one see it has no real directionality, even in three dimensions, vector + vector components cannot be biased either.
$$a^2-a(i)=b$$
So
$$a^2-a(i^0) \pm b$$
Then
$$i^0=-(-1)=-i^2=-1$$
Then it must hold true that the values of $$\sqrt{1}$$ and $$\sqrt{-1}$$ play exactly the same roles, when conditionally that each can never yield a positive value of when $$b>0$$. Here, we role on:
$$a^2-a(i^{0}_{(j)i}) \pm b >< (-1)=-i^2=\sqrt{-1}$$
Then we have yielded a negative value, from something which is positive to begin with, so we can then arbitrarily say the reverse is always the same. Again, there is no forward direction at these scales, if there is no complex-conjugate.
Now let the trace variables have
$$(j)i=-i^2$$ and $$i(j)=i^4$$
So then $${(j)i}x{(i(j)}=\sqrt{-1}$$
But as I thought about it some more, it came to realize that this is only true if $$i^2 \pm \sqrt{1}$$, hence them acting analogous to complex conjugated functions. So $$\psi \psi* =1$$ is one way to yield a real value in any vector function.
We can work out some of these real attributes, by working off the values of the known three dimensions we see in operation everyday we go to work, or read a book, or simply noticing the sides of a cube, that in what are called row vectors $$|V|=0,0,0$$, yields real values if we allow:
$$i^{2}=i*k^{2}$$
Then in a three-dimensional coordinated system, we can reduce the real parts of any connected vectors like so ~
$$a+b-i^2*k^{2^2}=0$$
So long as
$$a+b+c+k^{2^2}=0$$
Other than this being proof, if there are no mistakes, then you can also prove that a vector has two directionalities, (one negative and one positive), any power of $$i$$ can be either $$\sqrt{1}$$ or $$\sqrt{-1}$$ so long as the exponent is only ever a multiple of $$2$$, so we can have what we call ‘’odd powers,’’ since they are strictly given only as $$\sqrt{1}$$ or $$\sqrt{-1}$$ since $$-i=i^3$$.
Out of further analysis, going back to $$a^2-a(i^{0}_{(j)i}) \pm b >< (-1)=-i^2=\sqrt{-1}$$, we can allow (by geometrical methods and mathematical disciplines), that $$\sqrt{-1}=i^{n}$$, so long as it has a remainder of $$2$$, so as far as I have deducted,
$$i^{n}=i^{18}=i^{2}=\sqrt{-1}$$
So the $$i^0$$ in $$a^2-a(i^{0}_{(j)i}) \pm b >< (-1)=-i^2=\sqrt{-1}$$ is $$i^{40}$$ if there is absolutely no remainder at all, so $$i^0=(a^{2}-a(i^{40}))$$.
Let is know what you think, because if there are any errors, I would like to see why…Ta Ta 4 the now!
Last edited: