Is 2+2=5 Possible in Physics?

[sancho2007]

some people says that for the large values of 2
2+2=5
is it really true. I mean what is the physics behind this equation.
yours sincerely,
sancho,

 

[cristo]

Large values of 2?? Huh?? 2+2\neq 5

 

[Xezlec]

some people says that for the large values of 2
2+2=5
is it really true. I mean what is the physics behind this equation.
yours sincerely,
sancho,

That is an old joke. It's not true. "Large values of 2" means nothing.

 

[sancho2007]

but 2 has large value sometimes. why not?
sometimes some experimental parameters also have large values?

 

[Office_Shredder]

sometimes some experimental parameters also have large values?

Yes, and if it has a value greater than two, it just might sum with itself to 5. The actual equation is meaningless though

 

[cristo]

But 2 is not a "parameter"-- it is a number, and thus takes only the value 2!

 

[Integral]

There is no physics nor is there any math behind that statement. It is simply an old joke.

 

[Cyrus]

Well, when you take the cross product of the transformation vector in R^n and assume a linear time invariant system then the approximation that 2+2=5 holds in the limit that alpha approaches infinity.

 

[radou]

but 2 has large value sometimes. why not?

Why not? Because 2 isn't a balloon in the shape of number 2 which gets large sometimes because we blow it up some more.

Well, when you take the cross product of the transformation vector in R^n and assume a linear time invariant system then the approximation that 2+2=5 holds in the limit that alpha approaches infinity.

You forgot about the key assumption about the invariant approximation tensor and about the uniform convergence of the gamma-series generated by non-uniform hybrid Laplace members.

 

[dontdisturbmycircles]

2.3+2.3=4.6 = 5 for 1 sig dig? lol is that maybe what he is getting at? In any case I don't think there is any physics behind this equation

(This is a desperate attempt to try to understand what he meant by "large values of 2" lol)

 

[Cyrus]

Why not? Because 2 isn't a balloon in the shape of number 2 which gets large sometimes because we blow it up some more.



You forgot about the key assumption about the invariant approximation tensor and about the uniform convergence of the gamma-series generated by non-uniform hybrid Laplace members.

Ah yes, of course. Only when the skew-symmetric, non-invertible mass-matrix is in place, or det(A)=cross(J,F).

 

[cristo]

Well, when you take the cross product of the transformation vector in R^n and assume a linear time invariant system then the approximation that 2+2=5 holds in the limit that alpha approaches infinity.

Damn.. forgot about that. Good spot, cyrus.

 

[dontdisturbmycircles]

lol

text

 

[radou]

Ah yes, of course. Only when the skew-symmetric, non-invertible mass-matrix is in place, or det(A)=cross(J,F).

Which implies an obvious isomorphism between Schmidt's dihedral group and the group od positively definite inertia matrices spanned by Van der Haagen's dual basis.

This would be the complete frame-set of the problem.

Now we're talking.

 

[Jimmy Snyder]

I think what they are getting at is that since 3 is a large value of 2 and 3 + 3 = 5 (for small values of 3), therefore 2 + 2 = 5.

 

[G01]

Well, when you take the cross product of the transformation vector in R^n and assume a linear time invariant system then the approximation that 2+2=5 holds in the limit that alpha approaches infinity.

 

[radou]

I think what they are getting at is that since 3 is a large value of 2 and 3 + 3 = 5 (for small values of 3), therefore 2 + 2 = 5.

Exactly. A typical example of diophantine isoparallelism induced by general invariancy.