The Tales of 1 (One) to 2 (Two)

[TriggerGrimm]

This is probably an already discovered mathematical canundrum since there are so many mathematical geniuses out there but I bring forwith to present to you none the less to show the true value of the first numbers in order of chronological sense. Bottom up logistical analysis.

First we start with 1 to show that 1 alone is not a real number. There must always be two or for that matter both sides of an equation.


[math]\sqrt{1} = 1[/math]

Error!

Now moving onto [math] \sqrt{2} = 1.414213562[/math]

Actual answerable result. This first value answered as a result above of 1.41423562 is the constant found in the square root of 2, an actual number so I defer to call it at this time, and is such the hypotenuse of [math]2^2{/math]

Now let's do [math]3^2[/math] and the resultant answer is 4.242706

Now let's do [math]4^2][/math] and the resultant answer is 5.65694248

However, let's look at the difference between the constant or hypotenuse between [math]2^2{/math] and [math]3^2[/math]

What we find is if we multiply the found constant of /sqrt of 2 by 1.5 hypotenuse value we get the same value as [math]3^2[/math].

This Proves as far as I can tell that 1.5 is acutually the True number 1 because it can not only extend infinitely into 1.99999999999999999...... onward but it is also the value of which is the half way point or the balance point at which we round up or down numbers being that it is 1.5 (up) or 1.5 (down)... and so with that saying every true number has a .5 attached to it to make it true to standard frameworks of mathematics.

Of course as you inscrease the values that are being square rooted the larger the hypotenuse value is calculated to be in perportion to the increasing numerical value being square rooted, but lest we not forget that the value of [math]/sqrt{1} = 1[/math] shows that because the answer is 1 it is an errored result because the closest you can get to the value of 1 is in-toe-factoe- 1.00000000001..... persay etc.

This Means 1 is infinite in nature both + and - of sides of zero which is only a divider and not a representation of a value because zero is also exactly the same as infinity or nothing either way.

I'm sure these mathematics are farewell understood as I mentioned before but I present them here nonetheless and look for well broughtup responses about what I am digging into here with these mathemetics.

In case this is something new then I will title these mathematics the same at which Einstein said about Quantum Physics, and that is "Spooky Mathematics at a distance" (sure noticably adding Mathematics in the paraphrased quote).

What say you?

 

Edited May 20, 2021 by TriggerGrimm

[TriggerGrimm]

I should add that every number should have a .5 attatched to it is because we learned from quantum physics that there is always an extension of room for error in a numerical value 1 -999,999 (for example). And being that .5 can be equally rounded up or down but we're taught initially to always round up a number 5 at least in Canada that is, it really isn't always this way is it? It's perfecly okay to round down from say 5.5 to 5.0 however it is more liklely than not that there will be some decimal value somewhere hidden in a lenghtly line of decimal places that teeters the 0.5 value upwards more towards 0.6 rather than 0.5 simply using that age old scientific method of making room for error in calculations.

I just realized I didn't quite write out the mathemtics quite right there so  I'll take a picture instead. Lock this mother trucker down.

 

[TriggerGrimm]

Wait it took me another try to reproduce the same results I had starting this post:

Where c= 1.414213562 the square root of 2 constant for the 2^2 square or triangle.

How this applies to physics? I have a number of idea is in mind.... yes...

 

Edited May 20, 2021 by TriggerGrimm

[TriggerGrimm]

Cx3 = C 2 X 1.5

C's can cancle out can't they? Oh I don't think you can cancle things out when its represented as a constant. They must stay and be on both sides of the equation I am assuming to say the least. If you did cancle out the C's youd be left with an equation that said 3 = 2 x 1.5  Which holds true. Maybe now I see what has happened here with these equations. I think I had a mix up but none the less the number theory still holds true. I discovered this constant a long time ago and figured it could be used to convert spherical or circle style physics (which uses infinite types of numbers due to the implication of pie) into that of square physics with a 45 Degree constant (force) for an atom model, which is paramountly interchangeable with a 90 degrees constant which we use in common graphing anyways. Am I just catching on to modern physics or onto something new and different. It's been years since I have posted here but there was always a lot of good help so respond in due time.

[HallsofIvy]

Sorry but I have no idea what you are trying to say.   You say "hypotenuse of 2^2"  but I feel  sure you mean the hyptenuse of a  2 by 2 right triangle" is \sqrt{2^2+ 2^2}= \sqrt{8}= 2\sqrt{2}.  That makes your next comments,

"Now let's do 3^2 and the resultant answer is 4.242706" and "Now let's do 4^2 and the resultant answer is 5.65694248" very difficult to understand!  3^2= 9 and 4^2= 16.  You mean that the hypotenuse of a 3 by 3 right triangle is \sqrt{3^2+ 3^2}= \sqrt{18}= 4.242641, approximately, and \sqrt{4^2+ 4^2}= \sqrt{32}= 5.656854, approximately.

[quote]However, let's look at the difference between the constant or hypotenuse between [math]2^2{/math] and [math]3^2[/math]

What we find is if we multiply the found constant of /sqrt of 2 by 1.5 hypotenuse value we get the same value as [math]3^2[/math].[/quote]

No, that is also wrong.  If you multiply your value for\sqrt{2}, 1.414213562, by 1.5 we get 2.121320343. not 4.242641.

What is true is that \sqrt{n^2+ n^2}= \sqrt{2n^2}= n\sqrt{2} and \sqrt{2}= 1.414213, approximately, NOT 1.5.

What you have is really just a lot of bad arithmetic.

Edited May 20, 2021 by HallsofIvy