E(c2)=M additional formula for hyperspeed

[Aethelwulf]

Hi, Laurie.

 

sorry, you have me lost. Why quote me?

[maddog]

Poor old Craig, I agree too. Why would anybody ever consider using a field equation (of one cycle) as a basis for a universal model?

Actually, it was poor Qfwfq. Though we add all of us, for having to read through this whole damn thread. :angry: For any meaning at least. :eek:

 

Hi Laurie... :D

 

maddog

[maddog]

I just want to know of what value is any kind of math where both of these equations apply.

 

(from ryan)::

 

2 x 3 = 6 & 2 = 3 x 6

 

what does this say about the operator 'x'.

 

I can also answer some of the above crap about the '=' as an operator.

 

In Abstract Algebra, it is shown how '=' is an "Equivalence Relation". Look that up on wiki. I am not prepared to show the "proof" at the moment (I will if I have to). Because of this property, it would prevent both of the above equations holding for the binary operator 'x' known as multiplication on the given set "the Integers" which form a group.

 

Ipso factor, "YOU CAN'T DO THAT!".

 

Enough said. :o

 

maddog

[ryan2006]

There were only 135 experts on special and general relativity. What makes you people experts and the whole idea of science is based on a hypothesis which thus becomes a synthesis. I took college statistics and the first principle of science is A) the hypothesis is true or B) the hypothesis is false so for those of you who profess E(C2)=M is false then let's see how exactly you arrived at your hypothesis so until you can do that complex task I regret that your hypothesis that the equation is False is wrong. And to add I don't care for you people to attack my character if you can't do the math don't post. Thank you.

[JMJones0424]

E(c²)=m is false because E=mc² is true. What is your source for there being only 135 experts on SR and GR? What is it that you don't understand about your equation being obviously false? As you've said, this is grade school math.

[maddog]

I have to take issue with this post.... PLEASE!!!

I will analyze it sentence by sentence...

 

There were only 135 experts on special and general relativity. What makes you people experts and the whole idea of science is based on a hypothesis which thus becomes a synthesis.

I would like to know how you arrived at "135" as the Exact number? You sure it wasn't 136 or 134. I know we can count Einstein (1), maybe Lorentz(2), & Minkowski (3). That leaves 133 more obscure ones. Maybe you could list them please???

 

I took college statistics and the first principle of science is A) the hypothesis is true or B) the hypothesis is false so for those of you who profess E(C2)=M is false then let's see how exactly you arrived at your hypothesis so until you can do that complex task I regret that your hypothesis that the equation is False is wrong.

Got it you took "college statistics". Did you take High School Algebra, as Statistics depends on it? I also doubt you took Logic in college either (we wouldn't be having this silly conversation if so.

 

So you say the first principle of science that

A) a hypothesis is True (T) or False (F)...

What I was looking for was the rest of the principle?!?!? You had an "A)" and yet no "B )" and no Conclusion either.

Maybe what you misunderstood was what I posted in #71 above. Laws and Theorems of Abstract Algebra gave us the perogitive to claim your formula False when Einstein's formula is True. As I quote from Thomas Hungerford, "Algebra", p6.

 

An Equivalence relation is "A subclass R of A x B is called a relation on A x B...."

 

[since we are only interested in the equivalence part '=', I skip a bit]

 

A relation R on A x A is an equivalence relation provided

 

[this relation is]

 

Reflexive: (a, a) element of R for all a element of A or simply "a = a"

Symmetrifc: (a, b ) element of R ==> "if a = b then b = a"

Transitive: (a, b ) element of R & (b, c) element of R ==> (a,c) element of R or simply "if a =b and b = c then a = c"

 

Then Relation R is an Equivalence relation.

 

Now to show that '=' is an Equivalence Relation under Multiplication '*' on the set of integers is as follows:

 

Let a, b be in the set of Integers.

 

We can see for any a in Z (Integers) that

 

Reflexive: a = a is True (T)

Symmetric: if a = b then b = a is True (T)

Transitive: if a = b and b = c then a = c is True (T)

 

Thus the operator Equal Sign "=" is an equivalence operation under multiplication on the integers.

This can also be show for the Reals which I will leave as an exercise for the Reader.

 

And to add I don't care for you people to attack my character if you can't do the math don't post. Thank you.

I don't see how anyone here has "attacked" your "character". What is issue here is apparently your ability to reason. Form logical statement one after another. You appear to be definitely "math challenged".

 

You throw the Gauntlet down, I will take up the challenge and meet it! This I have done. So please don't waste our time with this silly drivel.

 

You see, I told you I would prove it if I had to.

 

Now Ryan, I suspect you will retort back something "this doesn't prove my formula false!" I would expect nothing less.

 

By the way, this has already been done multiple times in this thread

 

Your formula is E(C2)=M right?

Einstein's formula is E = Mc^2.

 

According to the notion of an equivalence relation and the operations of multiplication you can do the following

 

[math]M/c^2 = Mc^2[/math]

 

Simplifying yields

 

[math]c^4 = 1[/math] or [math]c = +- \sqrt(i)[/math]

 

Thus [math]c = (i, -i)[/math] of possible values.

 

Bad enough that you are saying if both are true formulas that you are saying in effect that the speed of light ( c ) has two values (Nonsense). Even worse they are Imaginary!!!

 

Here [math]i = \sqrt(-1)[/math].

 

So either you formula or Einstein's formula is inaccurate representation of the real world. Since Eintsein's has been tested multiple times to high accuracy. Guess which one we here think is the Silly one...

 

To quote an Author who disagrees with Einstein, "this theory of yours is Not Even Wrong!!!"

 

maddog

Edited August 31, 2012 by maddog

[Aethelwulf]

I have to take issue with this post.... PLEASE!!!

I will analyze it sentence by sentence...

 

 

I would like to know how you arrived at "135" as the Exact number? You sure it wasn't 136 or 134. I know we can count Einstein (1), maybe Lorentz(2), & Minkowski (3). That leaves 133 more obscure ones. Maybe you could list them please???

 

 

Got it you took "college statistics". Did you take High School Algebra, as Statistics depends on it? I also doubt you took Logic in college either (we wouldn't be having this silly conversation if so.

 

So you say the first principle of science that

A) a hypothesis is True (T) or False (F)...

What I was looking for was the rest of the principle?!?!? You had an "A)" and yet no "B )" and no Conclusion either.

Maybe what you misunderstood was what I posted in #71 above. Laws and Theorems of Abstract Algebra gave us the perogitive to claim your formula False when Einstein's formula is True. As I quote from Thomas Hungerford, "Algebra", p6.

 

An Equivalence relation is "A subclass R of A x B is called a relation on A x B...."

 

[since we are only interested in the equivalence part '=', I skip a bit]

 

A relation R on A x A is an equivalence relation provided

 

[this relation is]

 

Reflexive: (a, a) element of R for all a element of A or simply "a = a"

Symmetrifc: (a, b ) element of R ==> "if a = b then b = a"

Transitive: (a, b ) element of R & (b, c) element of R ==> (a,c) element of R or simply "if a =b and b = c then a = c"

 

Then Relation R is an Equivalence relation.

 

Now to show that '=' is an Equivalence Relation under Multiplication '*' on the set of integers is as follows:

 

Let a, b be in the set of Integers.

 

We can see for any a in Z (Integers) that

 

Reflexive: a = a is True (T)

Symmetric: if a = b then b = a is True (T)

Transitive: if a = b and b = c then a = c is True (T)

 

Thus the operator Equal Sign "=" is an equivalence operation under multiplication on the integers.

This can also be show for the Reals which I will leave as an exercise for the Reader.

 

 

I don't see how anyone here has "attacked" your "character". What is issue here is apparently your ability to reason. Form logical statement one after another. You appear to be definitely "math challenged".

 

You throw the Gauntlet down, I will take up the challenge and meet it! This I have done. So please don't waste our time with this silly drivel.

 

You see, I told you I would prove it if I had to.

 

Now Ryan, I suspect you will retort back something "this doesn't prove my formula false!" I would expect nothing less.

 

By the way, this has already been done multiple times in this thread

 

Your formula is E(C2)=M right?

Einstein's formula is E = Mc^2.

 

According to the notion of an equivalence relation and the operations of multiplication you can do the following

 

[math]M/c^2 = Mc^2[/math]

 

Simplifying yields

 

[math]c^4 = 1[/math] or [math]c = +- \sqrt(i)[/math]

 

Thus [math]c = (i, -i)[/math] of possible values.

 

Bad enough that you are saying if both are true formulas that you are saying in effect that the speed of light ( c ) has two values (Nonsense). Even worse they are Imaginary!!!

 

Here [math]i = \sqrt(-1)[/math].

 

So either you formula or Einstein's formula is inaccurate representation of the real world. Since Eintsein's has been tested multiple times to high accuracy. Guess which one we here think is the Silly one...

 

To quote an Author who disagrees with Einstein, "this theory of yours is Not Even Wrong!!!"

 

maddog

 

This is getting worse. No you can't do this:

 

[math]\frac{M}{c^2} = Mc^2[/math]

 

The units are wrong, again.

Edited September 3, 2012 by Aethelwulf