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Regarding the expression $a/0$, according to Wikipedia:

In ordinary arithmetic, the expression has no meaning, as there is no number which, multiplied by $0$, gives $a$ (assuming $a\not= 0$), and so division by zero is undefined.

Is there some other kind of mathematics that is not "ordinary", where the expression $a/0$ has meaning? Or is the word "ordinary" being used superfluously in the quoted statement?

Is there any abstract application of $a/0$?

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    $\begingroup$ Sure. Look up projective geometry. $\frac{1}{0}$, or just $\infty$, is a convenient name for the "point at infinity" on the projective line. $\endgroup$ – Qiaochu Yuan May 3 '16 at 2:33
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    $\begingroup$ Something that may be of interest: $$\lim_{x,y \to 0} x^y$$ $\endgroup$ – zerosofthezeta May 3 '16 at 2:36
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    $\begingroup$ I don't know if the first comment covers this, since I'm not good at projective geometry, but if we compactify the real line using one point (i.e. "turn it into an infinite circle" - see Alexandroff compactification), then we can make the function $1\over x^p$ continuous on the ENTIRE real line, not just at $x \not= 0$ by defining $1/x=\infty$ in accordance with "intuition". $\endgroup$ – Chill2Macht May 3 '16 at 2:38
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    $\begingroup$ @LorryLaurencemcLarry If you asked Euler, he would have said that $0/0$ can be anything; see Chapter 3 paragraphs 84 and 85 of imcs.dvfu.ru/lib.int/NEW/Math/MC_Calculus/… $\endgroup$ – lastresort May 3 '16 at 7:20
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    $\begingroup$ There is no problem in defining the result of $a/0$ to be $\infty$. The problem is that $\infty$ won't have properties we expect from a number. $\endgroup$ – Curd May 6 '16 at 9:45
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You said you wanted an application. Inspired by the example from Exceptional Floating Point, consider the parallel resistance formula:$$ R_{total} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}} $$This formula tells you the effective electrical resistance of a path when the current can choose two routes to take.

Let's pretend that $R_1=0$. Then we have:$$ R_{total}=\frac{1}{\frac{1}{0}+\frac{1}{R_2}}=\frac{1}{\infty+\frac{1}{R_2}}=\frac{1}{\infty}=0 $$The resistance being zero is indeed the correct answer; all current flows along the single wire that has no resistance.

Naturally, you need to make appropriate definitions for arithmetic on $\infty$ (i.e., use the projective reals). For well-behaved applications like this, that's fairly straightforward.

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    $\begingroup$ I don't think $0$ resistance is technically possible---should the wire itself not have some, albeit negligible, resistance? This is a somewhat idealized situation. $\endgroup$ – MathematicsStudent1122 May 3 '16 at 21:16
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    $\begingroup$ @MathematicsStudent1122 Superconductors have exactly zero electrical resistance and have many practical applications. $\endgroup$ – In silico May 3 '16 at 21:28
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    $\begingroup$ @MathematicsStudent1122 Wire has a resistance of 0 in the same way that a spherical cow falls in a vacuum. It's an abstraction used by practitioners in the field because it's Close Enough. $\endgroup$ – fluffy May 4 '16 at 4:20
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    $\begingroup$ @Insilico, I find it hard to believe that anything has exactly zero resistance. Presumably, what's really going on is that our understanding of superconductivity is based on a model, an idealization, and in that model, the resistance is exactly $0$. But I bet you one day we'll have a better model, and the predicted resistance will be small-but-nonzero in that better model. $\endgroup$ – goblin May 4 '16 at 5:26
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    $\begingroup$ @goblin Our understanding of superconductivity is based on measurement, not models; when it was discovered, we had no clue how to explain it. Zero resistance was based on multiyear experiments. Maybe you're right, and it's not zero, but "I find it hard to believe" is not an appropriate justification for ignoring a large body of experiments on the subject. $\endgroup$ – prosfilaes May 4 '16 at 19:52
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In complex analysis, we talk about the value of a function at infinity. To evaluate $f(z)$ at infinity, compute $f(1/z)$ then plug in $0$. This allows us to talk about things like the order of zeros and poles at infinity.

Example: $$f(z) = \frac{az+b}{cz+d}$$ with $ad-bc \neq 0$ and say $a,c \neq 0$. $$f(1/z) = \frac{\frac{a}{z}+b}{\frac{c}{z}+d} = \frac{a+bz}{c+dz}$$ So $f(\infty) = \frac{a}{c}$.

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    $\begingroup$ Couldn't you do the same thing with limits? Why do you need to define division by zero here? $\endgroup$ – jpmc26 May 3 '16 at 11:11
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    $\begingroup$ @jpmc26: Why would you go through all the trouble with limits when you can just do arithmetic? $\endgroup$ – Hurkyl May 3 '16 at 17:07
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    $\begingroup$ @jpmc26: It's difficult to use limits here since we are in the complex plane, and thus you would have to show that the limit of $f$ is independant of the path used to go to infinity... $\endgroup$ – Alexandre C. May 3 '16 at 18:36
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    $\begingroup$ @Hurkyl Doesn't all the math here still hold if you use $\lim_{z\to0} f(\frac{1}{z})$? Alex: Isn't that important even if you try to define $\frac{1}{0}$? If the function has multiple values as you approach infinity, I'm not sure how you can handwave that away by declaring $\frac{1}{0} = \infty$. $\endgroup$ – jpmc26 May 3 '16 at 20:35
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    $\begingroup$ @Hurkyl I'm asking the opposite question, though. Why redefine arithmetic when you can use limits? I assume that doing so leads to other complications. Also, how is $\frac{1}{x}$ continuous? It wouldn't be even if you do define $\frac{1}{0}$ since it hops between $-\infty$ and $\infty$ when you cross from $x < 0$ to $x > 0$. It doesn't seem like defining $\frac{1}{0} = \infty$ simplifies anything here. I feel like there's something major I'm missing, but I really don't know what. But that's my point: this answer doesn't point out where the knowledge gap is. $\endgroup$ – jpmc26 May 5 '16 at 7:21
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The algebraic structure "wheel" is an algebra with division by zero. The one point compactification of the complex plane into the Riemann sphere almost produces a wheel (one still needs to adjoin the element $0/0$).

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This may not be what you would consider an "application', but for computer floating point arithmetic, division by zero is useful for setting up three special values: 1.0/0.0 gives +inf, which is a valid floating point value satisfying the usual extended number line properties. -1.0/0.0 gives -inf, and my all-time favorite is 0.0/0.0, which gives NaN (not-a-number). In spite of the name, NaNs are also valid floating point numbers in the IEEE 754 standard, which almost all modern computers implement. A few languages like MATLAB let you directly specify those (e.g., a statement like x = inf is allowed) but others do not allow this - requiring the divide by zero syntax.

Possibly irrelevant to you but NaNs are useful in some numerical computing applications and highly useful in debugging some codes. The plural "NaNs" is valid because there are many of them: at least $2^{51}$ of them in 64-bit arithmetic. If you are interested, look for papers written by W. Kahan (father and grand inquisitor for the IEEE 754 standard).

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    $\begingroup$ I would suggest that floating-point math doesn't really have positive and negative zeroes so much as it has positive and negative infinitesimals, but most implementations lack a true zero or unsigned infinitesimals and substitute a positive infinitesimal instead. If x is a positive number which is small enough that xx would underflow, then I would suggest that 1/(xx) should yield "+INF" [since x*x is clearly positive] but 1/(x-x) should yield "NaN" [because x-x should be neither positive nor negative]. $\endgroup$ – supercat May 4 '16 at 15:37
  • $\begingroup$ supercat, Floating point implementation in computers has representation for zero. " if every bit is zero (the sign bit being irrelevant), then the number is considered zero." From here -> cprogramming.com/tutorial/floating_point/… $\endgroup$ – r12 May 5 '16 at 5:14
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    $\begingroup$ @supercat This answer is basically saying that it's useful to define division by zero in discrete mathematical systems that attempt to approximate the real numbers. In such systems, zero can be used as an approximation for values too small for the system to represent, and so it's useful to have a definition for what happens when you divide by it. Whether that's a good idea or there are better ideas is, of course, up for debate, but it is a real world system that's widely used. So +1. $\endgroup$ – jpmc26 May 6 '16 at 0:21
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    $\begingroup$ @r12: In trying to have the sign of 1/(x*y) remain constant as the product becomes too small to represent, IEEE-754 severely damaged other aspects of the structure of floating-point numbers. For example, x==y should imply 1/x==1/y, but in IEEE-754 it doesn't. x-y should be equivalent to -1*(y-x), but in IEEE-754, it doesn't, etc. $\endgroup$ – supercat May 6 '16 at 4:14
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Yes, in projective geometry or hyperbolic geometry for example, you can see applications or geometric entities that are $\frac{a}{0}$ or just $ \infty$ . Generally, non-euclidean spaces have such type of entities or applications.

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Theory of relativity. $$m = \frac {m_0}{\sqrt{1 – (\frac vc)^2}}$$ This means that as your velocity (speed) increases, and gets closer and closer to the speed of light, your mass increases (therefore, mass is related to velocity). It also proves, that it is impossible to travel faster than the speed of light. If an object were to do that, its mass would reach infinity, and that is impossible, so travelling faster than the speed of light is impossible.

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    $\begingroup$ no, actually you reach infinity when $v = c$, and there are particles that do what (they have $m_0 = 0$). Going faster then $c$ would mean that the square root becomes complex. I don't know about physics but my guess is that in that case this equation can no longer be used. (and the existence of this equation per se does not imply that $v > c$ is impossible, only that this equation makes sense only when $v < c$) $\endgroup$ – Ant May 4 '16 at 20:49
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    $\begingroup$ Traveling faster than light is perfectly well-defined (if complex) in that formula. All it really says is that your velocity can't pass through the speed of light. $\endgroup$ – Mark May 4 '16 at 21:27
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    $\begingroup$ This doesn't answer the question. The question is asking about forms of math that do allow for division by zero. The conclusions you mention require that division by zero be undefined. You reach these conclusions by taking a limit and showing that it approaches infinity (which is just a shorthand for "is unbounded" when discussing limits). If you define division by zero, then the mass becomes whatever value you defined $\frac{m_0}{0}$ to be (which may or may not be some kind of "legitimate" infinity). Bottom line: defining $\frac{1}{0}$ doesn't help here; it makes things harder. $\endgroup$ – jpmc26 May 6 '16 at 0:03
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    $\begingroup$ @Mark If you allow $m$ to have complex values, then yes, it is well defined. This equation is only useful for particles with real positive mass, since other values don't let us draw any physical conclusions (as far as I know, anyway). So you can rule out $v > c$ on that basis. $\endgroup$ – jpmc26 May 6 '16 at 0:14
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Though not actually correct, I've seen it used to "trick" people with a proof that 1=0

Consider two non-zero numbers x and y such that

$x = y$

Then $x^2 = xy$

Subtract the same thing from both sides:

$x^2 - y^2 = xy - y^2$

Dividing by $(x-y)$, obtain

$x + y = y$

Since $x = y$, we see that

$2 y = y$

Thus $2 = 1$, since we started with y nonzero.

Subtracting 1 from both sides,

$1 = 0$

Su, Francis E., et al. "One Equals Zero!." Math Fun Facts. http://www.math.hmc.edu/funfacts

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    $\begingroup$ You just divided numbers by (x - y) which is zero because x and y are different from zero and x = y, you can't divide a number by zero, it's against the laws of mathematics. $\endgroup$ – GingerHead May 4 '16 at 9:16
  • $\begingroup$ @GingerHead That's the point, and why it's a "trick" (that's explained in the link). Still, it is an application of dividing by zero $\endgroup$ – Xen2050 May 4 '16 at 9:23
  • $\begingroup$ OK, I get your point, we used to do this at highschool $\endgroup$ – GingerHead May 4 '16 at 9:26
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In two dimensional space, if we want a vector corresponding to a given angle $t$, we can do:

$(x, y) = ( \cos(t), \sin(t) )$

But what about the reverse? We have a vector, and want to know the angle. A common solution is to use arctan:

$t = \arctan( \frac{y}{x} )$

Here $\frac{y}{x}$ represents the gradient of the vector. So it has a meaningful value even when $x = 0$.


Trivia: Many program languages will fail to perform that calculation, because division by zero will produce an error. They usual address this by offering an alternative function which takes two arguments: atan2(y, x).

But a notable exception is Javascript, which has the concept of +Infinity and -Infinity as numbers, and can perform this calculation even when x = 0.

$ node
> Math.atan( 1 / 0 )
1.5707963267948966           (Pi/2)
> Math.atan( -1 / 0 )
-1.5707963267948966          (-Pi/2)

Another language which can perform this calculation is Haskell.


Inspired by Qiaochu Yuan's comment.

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I gave an answer to a similar question some time ago here.

To put it briefly, if you are ever solving an equation in one variable by just addition, multiplication, subtraction, and division, and you end up with a contradiction, then you can conclude with certainty that the one time you divided both sides by an expression involving a variable, the expression you divided by was actually $0$ (because having multiple answers can only happen if you had a $\frac 00$).

Also, when computing limits, if you get a $\frac a0$ for $a\neq0$, you can directly conclude the limit does not exist (as a finite number).

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