Why is something divided through zero not treated similar like an imaginary number?

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So in grade 7 math we learned that you can’t take the square root of negative numbers because any number squared is always going to be positive. A few years later we learn that you can actually calculate with the square root of negative numbers. You replace the square root from -1 with i. So why aren’t we replacing something divided by zero also with a letter?

In: Mathematics

8 Answers

Anonymous 0 Comments

Can this be answered in ELI5 terms? I don’t know enough math to be able to address this, but did stumble over this link to a wikipedia article on wheel theory ( https://en.wikipedia.org/wiki/Wheel_theory ) which says
> Wheels are a type of algebra, in the sense of universal algebra, where division is always defined. In particular, division by zero is meaningful.

Anonymous 0 Comments

You can do that, but the unfortunate consequence is that whatever that letter is now starts making your equations break really bad. Imaginary numbers don’t have that problem. The math with them still has a consistent set of rules you can follow to get sensible answers, but this “divide by zero number” doesn’t. Thus, it’s generally better to just leave it as undefined.

Anonymous 0 Comments

> You replace the square root from -1 with i.

There is quite a bit more to it than just making up a new symbol for √-1 and calling it i.

Imaginary (and complex) numbers are a two-dimensional extension of the one-dimensional number line, and i = √-1 is a natural consequence of that. In two dimensions, you can think of multiplying by -1 as a 180^(o) rotation. Multiplying by i is a 90^(o) rotation, so doing it twice is the same as 180^(o). That means i x i = -1, and √-1 = i.

There is no similar extension, at least none that I am aware of, that cleanly replaces 1 / 0 with some discrete value.

Anonymous 0 Comments

Because imaginary numbers actually work a certain way and have certain properties this doesn’t allow.

To better think you can show it with algebra. A÷0=C, then implies by multiplication that A=C×0. But if A doesn’t equal 0 then the universe explodes

Unfortunately dividing by 0 doesn’t give us imaginary numbers because imaginary was basically saying we know what the answer should be but we can’t get the right sign ( positive or negative). Dividing by 0 causes two numbers that aren’t equal to be declared equal. It’s also the reason we have calculus

Anonymous 0 Comments

If you want to divide by zero, you may be interested in [Riemann Spheres](https://en.wikipedia.org/wiki/Riemann_sphere), where division by zero is defined.

Anonymous 0 Comments

“Limits” is close to what you’re talking about.

The actual “divide by zero” is undefined but you can examine what happens to the equation as it approaches zero. This is generally useful when the thing you’re dividing also looks like it’s equal to zero at the same time.

Eg: (x^2 + x ) / x

As x approaches zero the result approaches one, as the x^2 approaches zero much faster than the x values, so in practice it approaches x/x, which cancels to 1 as x tends to zero.

Anonymous 0 Comments

There are certain rules that we want certain kinds of number systems (what a mathematician would call “fields”) to satisfy. Namely, on a set *F* we want two operations + and * such that

* Addition and multiplication are commutative and associative.
* There is an additive and a multiplicative identity, usually called *0* and *1* respectively. We require that *0* and *1* not be the same.
* Every element *a* has an additive inverse, usually called *-a.*
* Every element *a* except *0* has a multiplicative inverse, usually called *a****^(-1)****.*
* Multiplication distributes over addition.

We want these properties because we consider them “essential” to our usual field of rational numbers or real numbers. In particular, we want *0* and *1* to be distinct because then every number in our field would be equal to *0* (a fact which I prove at the end). There are lots of different fields, some of which are very weird. The complex numbers, in which negative numbers have square roots, and finite fields, in which I can add *1* to itself enough times and eventually get *0*, are some examples.

Now, it follows from the bullets above that anything times *0* is *0*. Indeed, if I have any *x*, then *x*0=x*(0+0)=x*0+x*0*, and subtracting *x*0* from both sides gives me *0=x*0.*

Now let’s see what happens if we allow *0* to have a multiplicative inverse in one of these fields. Say *x* is the inverse of *0*, so that *x*0=0*x=1*. Well, then I could multiply this equation by zero and get *0=1*0=(x*0)*0=x*(0*0)=x*0=1*. So in this case, *0=1*! Since we explicitly required that *0* not equal *1*, this shows that the inverse for *0* which we called *x* cannot exist if we want all of the bulleted properties.

Well now the question is, what if we don’t want all of the above properties? Well, mathematicians study a bunch of other objects which satisfy only some of the above properties (try googling “rings”, “groups”, “semigroups”, etc.). It turns out that none of these can have inverses for *0* either (assuming they even have a *0*). I have heard of one object, called a “wheel,” which tries to formalize divison by zero. I played around with the concept in undergrad once but I quickly became uninterested. I’m not sure how useful this concept is. Google “wheel theory” to learn more (although it’s not very easy to understand).

The point here is that if you want to invent a new number system where you allow an operation that is usually not allowed, then you need to prove (in the precise mathematical sense) that your number system doesn’t have any contradictions like the one we showed above. Introducing a number *i* such that *i**^(2)**=-1* indeed introduces no such contradictions, so mathematicians consider it to be perfectly valid, hence the field of complex numbers.

EDIT: For those curious, if *0=1*, then if we have a number *x*, then *x=x*1=x*0=0*. Hence every number is equal to zero.

Anonymous 0 Comments

I thought any number divided by zero equaled infinity? Wouldn’t it just be an extra infinity in the formula?