WhyNot?

Do not agree with ∞× 0 = 0

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Responses: 3 (1 in support, 0 neutral, 2 in opposition)
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Description.

The axiom

∞× 0 = 0

is accepted in mathematics but also often used in physics.

However from the point of view of mathematics, this equation can not be proved. This is the reason why I named this generally accepted equation - axiom.

I would rather suggest the following:

∞ × 0 = 1

Proof.

2 × 1/2=13 × 1/3=1 …

…and so on. n ×1/n =1

∞ × 0 = 1 1/0= ∞, 1/∞=0 Example: (if 2× 1/2=1, 1:2= 1/2, 1:1/2=2)

The product of multiplication of two mutually inverse numbers will always equal 1.

Epilogue.

The majority of physics reference books and manuals state that the radius of the gravitational and electromagnetic interactions equals infinity.

It is also considered that the mass graviton and photon in rest (particles of corresponding interactions) equals zero.Therefore from the formula (1), which determines the radius of the interactions, we get the result that the radius equals infinity, as a result of division by zero. (1) WhereR - radius of the interaction m - mass of the fraudulent particle c - speed of the light in vacuum ħ=h/2 - where h- Plank’s constant

(2) R because ∞× 0 = 1

From the above discourses and proof follows the conclusion that there can not be fraudulent particles which mass equal zero since the product of the mass of the fraudulent particle multiplied by the radius of the interaction must equal 1.On the other hand however, presence of the mass in interacting particles (photon and graviton) contradicts with the statement of A. Einstein about the fact that when moving at the speed of light any particle with the mass other than zero, obtains the mass equal to infinity (see Laurence’s Transformations).

This is one of the variants of solving these equations, which were described by Einstein.It seems to me, there are two other ways to solve Laurence’s equations, with positive, and more importantly, with negative signs.

As a result, if we are to solve for mass of the particle, which moves at a speed of light, we get:

As a result one of the solutions will be negative, which may seem absurd for the notion of mass. However I would not rush to conclusions and disregard this result.

Possibly, this result will be taken into account for a different hypothesis.

I am thinking mainly about the hypothesis of Mach which speaks about the presence and reason of inertia, which A. Einstein took into consideration in his own discourses.

But this is a different topic. P.S.In one of the telecasts, the academician, laureate of Nobel Prize, V.L. Ginzburg, has said that in recent years he does not remember anybody from the field of non professional, self taught physics who would invent or at least suggest anything new and valuable in the field of physics. "Time of self-taught scholars and dilettantes have passed, and presently science can promote only upper classes of highly qualified, professional specialist".It is hard not to agree with this statement. Multiple academies, institutes and other scientific enterprises work on the development and advancement to scientific thought.The Theory and experiment in physics became so much more complicated that it seems, there can not be any interference from the outside in this complex entanglement of ideas, theories and experimental data. But perhaps there are still exceptions to the rules even in our time?

During my thirty-year -fascination with physics I have found several contradictions in its "graceful body". And thus I have offered the decisions of these contradictions and would love to share more of them if you find my ideas somewhat interesting.

Several words about myself. I am an engineer. Physics has always been my hobby ( for more 30 years already.) All findings were born during my readings and reflections. If they are of any interest to you, I will be honored to send the rest of the findings about alternative perception of some of the physical processes. Sincerely,Gennadiy Gibskiy,

Kharkov, Ukraine.

schiecsek, Jul 18 2008

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You have this incorrect. If you have anything zero times you have zero of it.

gattmott, Jul 19 2008

You're multiplying a real number (zero), by a "not-number" (infinity). It would probably be a simple matter to show that anything times zero is zero using calculus (infinity is an approachable limit).

The graph of f(x) = x × 0 is a flat line along zero, and remains zero as x approaches either extreme. If you were to do something similar with infinity (i.e. f(x) = ∞ × x), you'd have a line along positive infinity for x > 0, a line along negative infinity for x < 0, and a discontinuity at 0 (which should be a point at 0,0).

nayhem, Jul 22 2008

Also, dividing by zero (used in your proof), is either impossible, or something you absolutely do not want to do if you like the universe as it is, according to who you believe.

nayhem, Jul 22 2008

I'm not a mathematician but this has been bugging me so I have to weigh in, if only to have a mathematician explain why I'm wrong.

It occurs to me that since you cannot have half an infinity or twice as much infinity that you either have infinity or you don't. And if there is a measurable point of some kind there must be measurable points on either side of that point that would go on infinitely. This would mean there could be no zero infinity. The value of infinity times zero would have to be something since it can't be zero.

Hyenuf, Jul 23 2008

Sorry, but I don't how to type an infinity sign, so I'm going to use ~ as infinity.

There is a glaring error with your axiom:

2 x 1/2 = 1

thus:

~ x 0 = 1

1/2 is the inverse of 2. Your are implying that 0 is the inverse of ~. But this is wrong. The inverse of ~ is 1/~, and 1/~ does not equal 0.

Dwane Anderson, Jul 30 2008