Is this real maths or somebody winding me up?

You want a question ? I have asked lots of questions and the failure to answer is not on the person asking the question.

Quite clearly this thread that the moderator has moved to the correct forum section has a paradox.

The questions are :

1) Can two opposite polarities occupy the same point space?

2)Can an Electron exist without a Proton?

I have already give you the answers to my own question.


1) yes P=1

2) no P=0

 

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Like normal the scientists ignore the content of the posts.....

 

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Here's a question:

Have you stopped beating your wife yet? Acceptable answers are yes and no.

Failure to answer the question lies on you.

Not all questions are sensical.


Anyway, Tiassa is right. I am simply exacerbating this.

 

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I do not have a wife to beat, I am a live in carer. I look after a women who is disabled, for real.

That is why I can quick reply, the computer is always on in the living room.

p.s I would never hit a women or a man.

 

You have failed to answer the question.

Instead, you tried to tell me how my question was improperly constructed such that it couldn't be properly answered the way I wanted.

That is what has been happening to you for about nine pages. Your questions cannot be answered except to say that they are gibberish.


This is not a school. Pick up a text book on the foundations of the math that will give you a basis for eventually learning the advanced maths that you want. Set aside your vision of The New Gravity till you have the tools.

That's the only answer anyone can give you.

 

I do see your point honestly, I understand I need to learn the tools . But, there is nothing that says the existing tools would even explain The New Gravity. Considering that the gravity mechanism is not known at this time, why not make some new tools like I am trying to do , to explain it?

As long as I explain the process, explain what I am using the tools for, why is this not enough to understand it?

 

Ok, I have had some good advice and help in PM elsewhere, now I am going to change my approach . The new discussion will start with :


For the purposes of this discussion consider an array to be a rectangular arrangement of coordinate values, similar to, but not exactly like the classical "matrix". My first question is , can a mono-pole retain form in a single array?

Now quite clearly all the coordinate points of a mono-pole array, would be repulsive points to all other points of the same array.

By the laws of Physics and Coulomb's laws , the mono-pole array should always be in a state of expansion.

The array would have no strong nuclear force or gravity. All the force would be ''centrifugal'' (outwards from a central point) .

I then considered free electrons and free protons, meaning ones that are not paired.

How could an electron exist when considering the first array?

In a similar fashion to a Matrix, all electron points in the array of an electron would be repulsive to each other.

I then considered the proton , what seems true for the electron, must also be true for the proton.

So in my first array A, I want to describe this part first. I want to describe that array A cannot retain form and is always in a state of expansion.

 

We shall define array A's length to be 10, that we shall represent the indices with numbers and place two likewise polarity elements in position 5 and 6.

A={1,2,3,4,5,6,7,8,9,10}/t

A={1,2,3,4,5,6,7,8,9,10}/t

A={1,2,3,4,5,6,7,8,9,10}/t

A={1,2,3,4,5,6,7,8,9,10}/t

A={1,2,3,4,5,6,7,8,9,10}/t

A={1,2,3,4,5,6,7,8,9,10}/t


We can observe from the above consequence in order of displacement , both elements leave our array.

 

%B={1,2,3,4,5,6,7,8,9,10}/t1

%B={1,2,3,4,5,6,7,8,9,10}/t2

%B={1,2,3,4,5,6,7,8,9,10}/t3

%B={1,2,3,4,5,6,7,8,9,10}/t4

%B={1,2,3,4,5,6,7,8,9,10}/t5

%B={1,2,3,4,5,6,7,8,9,10}/t6


The improved %B array shows in a similar fashion to a Matrix, all proton points in the array of a proton would be repulsive to each other and also leave our array.

 

In the final array if we merge the two arrays , we can observe the stable state of the position 5,6 elements, while they simultaneously occupy the same array position

%A,B={1,2,3,4,{5,6,}7,8,9,10}/t1

%A,B={1,2,3,4,{5,6,}7,8,9,10}/t2

%A,B={1,2,3,4,{5,6,}7,8,9,10}/t3

%A,B={1,2,3,4,{5,6,}7,8,9,10}/t4

%A,B={1,2,3,4,{5,6,}7,8,9,10}/t5

%A,B={1,2,3,4,{5,6,}7,8,9,10}/t6

Any comments?

 

Hmmm, why are you saying my opinions and matrix ideas are nonsense when it show you two matrices merging here:

https://en.wikipedia.org/wiki/Matrix_(mathematics)#/media/File:Matrix_multiplication_diagram_2.svg

 

https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=72234.0;attach=24910;image

My model and equation.

 

Snap. Is this you at NakedScientist? Your ''Thebox'' there?

https://www.thenakedscientists.com/forum/index.php?topic=72234.msg532754#msg532754

 

I have been the Box on the naked scientist for several years , this is where I have learnt the most about science. Their patience is commendable.


In the thread over there , you can see diagrams ,

 

I am writing someone is pulling your leg. Example! Opener alone should read 3>u>-2

 

Could you please explain in words what 3>u>-2 means?

 

Three is greater (the vertical distance the three side is greater) than u, which is greater than minus two. ☺

 

You may even attempt every combination of mathematics.

1<A<10
1<B<10
1<C<10
1<D<10
1<E<10

((((A+B)×C)-D)÷E)
((((A+B)×C)÷D)-E)
...etcetera...

When running combinations, personally I switch the last two operators and then move the previous operator forwards one position...

1234
1243
1324
1342
1423
...etcetera...

The above method has the disadvantage of not providing an answer (it is without an equals operator.) ...you could write a program in an Excel macros (toolbar, name macros) to combat this problem...

 

...OR You could add an equals operator...

((((A+B)×C)-D)÷E)=I (1)
((((A+B)×C)÷D)-E)=I (2)
...
((((A÷B)-C)×D)+E)=I (24)

 

This is a beautiful thing. Amber and Counter.