What does dividing by zero really mean?
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WhiningKhan
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K.I.L.E.R said:
Here I come to wreck your day:
Physics is a mathematical model of laws of nature.
If it's meaningless in mathematics, it has to be meaningless in physics.
Division by zero is never valid. Divider can, of course, go arbitrarily close to zero, but not to zero.
Another way to look at it is to consider both zero and infinity as being limits defined in terms of each other where infinity is 1/0 and zero is one divided by infinity. It's interesting that people seem have more trouble with infinity than with zero, though.
I've never really studied general relativity, but I'd imagine there's some kind of singularity in the equations at the center of a black hole. They occur in much more mundane mathematical situations so that seems as good a place as any.
I've never really studied general relativity, but I'd imagine there's some kind of singularity in the equations at the center of a black hole. They occur in much more mundane mathematical situations so that seems as good a place as any.
There's a much used divide by zero in physics called a delta function. It can be defined(usually is) as a normal distribution where the standard deviation is 0. So at the center of the distribution you get infinity.
If you where to integrate f(x) * (delta function) over some interval the delta function will pick out the value of f where the delta function goes to infinity.
In other words the delta function tends to infinity in such a way that it is normalized so that the area under it is unity and it is non-zero only in a single point.
It's all just a trick of course. It allows you to use the same machinery for calculating things which behave like delta functions without having to resort to special cases. (Say, we can look at the wave function of a particle with an exactly known position. The momentum of said particle in quantum mechanics is related through a fourier transform, so that means we have NO idea what the momentum of the particle is. If you have any wave train shorter than infinitely long it does not have an exact frequency, this is what leads to heisenbergs uncertainty principle).
If you where to integrate f(x) * (delta function) over some interval the delta function will pick out the value of f where the delta function goes to infinity.
In other words the delta function tends to infinity in such a way that it is normalized so that the area under it is unity and it is non-zero only in a single point.
It's all just a trick of course. It allows you to use the same machinery for calculating things which behave like delta functions without having to resort to special cases. (Say, we can look at the wave function of a particle with an exactly known position. The momentum of said particle in quantum mechanics is related through a fourier transform, so that means we have NO idea what the momentum of the particle is. If you have any wave train shorter than infinitely long it does not have an exact frequency, this is what leads to heisenbergs uncertainty principle).
Lux_ said:Maybe it's the other way around? Sqrt(-1) is so meaningless in mathematics they had to came up with special symbol to deal with reality
I'm not going to claim to be an expert, but every time we used complex numbers in applied formulas, it was simply done "to make the math easier." All the calculations could be completed using strictly real numbers (and we were taught that method first), but it was a much longer process. So, in those situations at least, reality didn't require complex numbers, but complex numbers were used out of convenience.
YeuEmMaiMai
Regular
I dunno but it seems to me that if you have nothing in quantum mechanics and you divide it by something you still have nothing and if you have something and you devide it by nothing you still have something.
Bumpyride said:
Ironic, since you're wrong. 1/0 is not infinity, it's undefined.
You can prove it in so many ways. For example, consider f(x) = 1/(1/x) and g(x) = 1/(-1/x) As x-> inf, both f(x) and g(x) approach 1/0, only f(x) approaches to 0+ and g(x) approaches to 0-. Yet, if you simply the sequences, you get f(x) = 1,2,3,...inf and g(x) = -1,-2,-3,...-inf
Hence, the limit does not exist, and therefore 1/0 is undefined. If you assume otherwise, you are faced with a contradiction.
You can also try to use the relation you stated (1/0 = inf, or 0 * inf = 1, etc). If you use this relation, algebra breaks down.
For example, assume 1/0 exists and call it X. 1/0 = X. Or 1 = 0 * X. But we know that zero times anything is zero.
If you made a special exception for 0 times infinity, you are left with nonsense. Let 0 * X = 1 be written as (X-X)*X = 1. Now, we have X^2 - X^2 = 1. So infinity squared minus infinity squared = 1? Umm, ok.
But what about 2/0? Clearly we have X^2 - X^2 = 2, so infinity squared minus infinity squared = 2!?! What you end up with is complete nonsense. No sensible definition of 1/0 exists for algebra.
Now, in analysis there are all sorts of ways to deal with infinity and different kinds of infinity that can be consistent, but none of them assert that 1/0 = infinity.
soylent said:
Yep, but amazingly you get the same result if you use wavelet analysis, which allows you to analysis the wavetrain in chunks smaller than infinity.
As for sqrt(-1), there are processes in physics which rely on various properties of field extensions. Even in classic physics, wave mechanics are heavily dependent on trig functions, and trig functions are elegantly related together via exponential complex numbers.
But sqrt(-1) isn't really special. For any polynomial equation for example, that does not have a desired zero at point X, we can extend the algebraic field with a new number, that takes the value zero at that point.
Thus, if you have x^2 + 1, we know that there is no solution for reals. However, we can extend the number system with a new number i such that i^2 = -1, and therefore -i and +i are zeros of the equation.
Likewise, if we consider polynormials over Q, the rational numbers, we can create lots of polynomials that have no zero, such as x^2 - 2. There is no rational number for X in Q such that x^2 - 2 = 0. One solution of course is to just use the reals. However, in abstract algebra we can extend the field of Q to contain a solution for the above polynomial (and others), but still preserving all of the properties of the rationals.
One such field is Q(SQRT(2)).
evil said:Wow, all this math just to say you can't divide something into 0 pieces
I was thinking the same, i mean how many words and f( ) and bloody roots do we need to know that we just CAN'T do something? I thought the point of science was efficiency. The simplest theory is the best theory. "You can't do that" seems to be the best theory in this case
Division by zero is undefined in maths, because you cant define it without breaking some very basic laws/axioms. In special cases you can find useful definitions, but you must restrict the context of its useage ( ex. you know 0 is going to be iterated from positive numbers xn, so 1/0 := lim 1/xn = infinity. if iterated from negative numbers 1/0 := -infinity ). That doesnt mean though division through zero is possible, because it still isnt, you just wrote 1/0 for lim 1/xn.K.I.L.E.R said:
I dont know what string-theory would help, physics is nothing else than finding mathematical formulas (and their context) which represent or approximate reality.
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How do you know you can't do it? Mathematics is based on proof. People's intuition is often wrong. The reason why you need all the symbol manipulation is to show exactly why it can't be done.
For centuries people searched for a way to square the circle, or trisect an angle with compass and ruler, or to find a formula for solving quintic polynormials like the quadratic formula. Many people had the intuition that it could be done, but the reality is, it's impossible, but it take some heavy Galois Theory math to show it.
Is fermat's last theorem true? We thought so, but only recently proved it. However, there are an infinity of other diophanine equations like Fermat's in which we don't know the answer.
How about twin primes? Are there an infinite number of primes separated by one number? (such as 3,5) ? Intuitively it would seem so, but we have been surprised before.
One just can't assert that something is impossible, or claim something is impossible via flowery language. Math has axioms and defined rules, and the challenge is to show how a given conjecture cannot be reached with those axioms and rules (or can be) via a chain of logical statements.
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