# Why can one multiply by zero

Dividing by zero isn't allowed because it results in the same answer (infinity) for every input and therefore is considered "undefined." Multiplying by zero is allowed, even though it results in the same answer for every input (zero). It also allows me to start with the assumption (1 != 2), multiply both sides by 0, and prove that 0 != 0. Why is zero not treated like a limit, so that multiplying by zero would not become a black hole for information. Please forgive the fact that I didn't reach calculus, and haven't touched math in years, and explain to me where understanding is wrong or my reasoning went astray.

Edit/Update:

Good answers, thanks. Can anyone include an example of something that can't be solved easily without multiplying by zero, to show me that it's necessary to allow the operation, rather than just not contradictory? I know of proofs in formal logic that require assuming "x = X", so I imagine there's an obvious example of the need to multiply by zero?

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–  NasuSama Feb 22 at 19:40
@NasuSama - Although answers at those linked questions are relevant, I strongly disagree that this is a duplicate. Both of those questions were essentially asking "what's the problem with dividing by zero?" The title question here is not why can't one divide but why can one multiply. –  Ben Blum-Smith Feb 22 at 20:21
Also, +1 I love this question. –  Ben Blum-Smith Feb 22 at 20:24
If you define $a/0$ as infinity then it isn't undefined, it's defined as infinity. Now you have to give a meaningful definition of infinity that is consistent in the framework of arithmetic, and this is where the trouble lies. The reason that $a/0$ is called undefined is precisely because no definition is given for it. –  R R Feb 22 at 23:16
I can also ask what it means to divide or multiply by cat, but that begs the question of what the heck I mean by cat and whether operations with cat are consistent with operations on numbers. –  R R Feb 22 at 23:18

## 5 Answers

If you have operations of addition and multiplication which both have a group structure and interact via the distributive law and where $-ab=(-a)b$ you have $$0=ab-ab=ab+(-a)b=(a-a)b=0b$$

These are very natural conditions in a wide variety of circumstances, and lead to fruitful and rich mathematical structures, so we take their effects on the chin.

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This is due to the fact that the mapping $\mathbb{R}\ni x\mapsto 0\cdot x\in\mathbb{R}$ is not injective and thus there is no inverse, that means $0^{-1}$ doesn't exist.

EDIT: (sorry for the many edits but i think this is an important point)

this misunderstanding comes from the following calculation:

$$0=\lim_{n\in\mathbb{N}}\frac{1}{n}=\frac{1}{\lim_{n\in\mathbb{N}}n}=\frac{1}{\infty}$$ so one could get the idea that $$0^{-1}=\frac{1}{0}=\frac{1}{\frac{1}{\infty}}=\infty$$

the point is: the limit theorems do not apply!

when we look closely at $\lim_{n\in\mathbb{N}} n=\infty$ we must note that this is actually a misuse of notation. it does not mean, that the series tends to infinity. tending to infinity would mean $\forall \epsilon > 0 \exists n_{0}\in\mathbb{N}:\forall n\geq n_{0}: \left|n-\infty\right|\leq\epsilon$ which is complete nonsens because $\left|n-\infty\right|$ always equals $\infty$ and $\infty$ is never smaller than $\epsilon\in\mathbb{R}$.

What people actually mean by $\lim_{n\in\mathbb{N}} n=\infty$ is, that the series does not converge at all, but it gets arbitrary big. But the limit theorems do not always apply to this misused notation. (not always, sometimes it works)

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I came a little closer to understanding it with your edit (again, you might have to dumb yourself down a bit here). I'd rephrase my question like this: Why do we treat 500*0 as though it actually reached 0, and destroyed the information, when it seems like the question really being asked is "what happens as the function gets arbitrarily small"...sorry for all the incorrectly used terms I'm sure I've dropped –  TheIsolatedNetwork Feb 22 at 20:11
yes, sorry... i edited again now because i've seen what i've done after my first edit and noticed there is a whole stream of thoughts missing... now it should be understandable –  Max Feb 22 at 20:23
Again, never studied limits, so forgive the nonsense, but am I closer to understanding by saying this?: limits apply to zero because zero is a discrete point upon which functions can coverge; limits do not apply to infinity because infinity is a diverging horizon from which you cannot take relative measurements, and therefore functions cannot "grow closer" to infinity by definition? –  TheIsolatedNetwork Feb 22 at 20:29
$500\cdot0$ is not treated to reach zero it is zero. maybe your question is also a question about exchanging the order of limits. –  Max Feb 22 at 20:30
yes, what you stated above is indeed correct. zero is a real number and you can treat it as one, so you can multiply by it. division of a number by infinity is always a misuse of notation and means: limit of the sequence "your number devided by a sequence growing arbitrary big". –  Max Feb 22 at 20:36

Mathematicians' policy toward definitions and rules of deduction tends toward "if it's not logically contradictory, it's allowed". Particularly, "allowable" operations do not result in false statements being deduced from true statements. In this sense, multiplication by zero is perfectly acceptable. Even if you multiply a falsehood (such as $1 = 2$) by zero (obtaining the true statement $0= 0$), you haven't made a logical error; you've merely made a vacuous deduction.

The excellent accepted answer to this question explains why, by contrast, division by zero is undefined. It's not because the result is "infinity". :)

As for the benefits of "allowing" multiplication by zero, Mark Bennet's answer gives the concise algebraic rationale. However, I'd go farther: An equation of the form $U = 0$ is almost always easier to work with (both theoretically and in practice) than an equation of the form $X = Y$.

As a simple example, consider solving $x^{3} - 2x^{2} - 5x = -6$ in the real numbers. Factoring the left-hand, obtaining $x(x^{2} - 2x - 5) = -6$, is not much help: Knowing that a product of numbers equals $-6$ says almost nothing about the factors. By contrast, moving everything to one side and then factoring brings to bear the powerful fact that a product of real numbers is zero if and only if some factor is zero. Here, $$0 = x^{3} - 2x^{2} - 5x + 6 = (x - 1)(x + 2)(x - 3),$$ which immediately gives all solutions.

More substantive examples abound. To give just one, let $A$ be a real square matrix, and consider the eigenvector problem, finding a non-zero vector $x$ and a real number $\lambda$ satisfying $A\mathbf{x} = \lambda \mathbf{x}$. This is all but hopeless to solve without using the fact that $0 \cdot x = 0$ for all real $x$.

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### Summary

Allowing multiplication by $0$ requires that $0=-0$. That is essentially never a problem. In contrast, even if we permit arithmetic operations with $\infty$, allowing division by $0$ would require that $\infty=-\infty$. This is usually considered unacceptable because it is nonintuitive and conflicts with the way we use the $\infty$ symbol in calculus (and for other reasons). In situations where we accept $\infty=-\infty$, division by $0$ may be allowed.

### Full Explanation

Dividing by zero isn't prohibited because it always gives the same answer. There is nothing wrong with an operation that always gives the same answer. It's prohibited because there's always more than one answer or zero answers, and for division to be useful it is typically expected to produce exactly one answer.

Understanding why dividing by zero is really not allowed makes it possible to quickly see why multiplying by zero is allowed, since the problematic situations that arise from trying to divide by zero do not apply to trying to multiply by zero.

• When $x\neq0$, "$x/0$" cannot evaluate to anything that is a number in the ordinary arithmetic sense of "number." The product of any number and $0$ is $0$, after all, and thus will fail to equal $x$ when $x\neq0$.

• When $x=0$, "$x/0$" can be anything that is a number in the ordinary arithmetic sense of "number." The product of any number and $0$ is $0$, after all.

Allowing infinities as well as finite numbers doesn't actually improve the situation much; by itself, using infinities as though they were numbers doesn't fix the problems that prevent division by zero. Suppose $x/0=\infty$. Since $0=-0$,

$$\frac{x}{0} = \frac{x}{-0} = -\frac{x}{0} = -\infty.$$

Then we have $\infty=-\infty$.

To state the matter intuitively, no matter how many times you add $0$ to $0$, even infinitely many, there's no reason to think the propensity for reaching a larger value is greater than the propensity for reaching a smaller value. $\infty$ sometimes means "a very big positive number" (with $-\infty$ taking on the meaning "a very big negative number"). And $0$ sometimes means "a very small number." But $0$ doesn't specifically mean "a very small positive number."

There are three ways of dealing with the problem that naive division by $0$ to yield $\infty$ makes $\infty=-\infty$ a theorem.

1. Don't allow dividing by $0$, even when thinking of $\infty$ and $-\infty$ as numbers--for example, even when using the (affinely) extended real line. This is the typical approach. This doesn't require that we stop multiplying by $0$, since while it's often problematic to say $\infty=-\infty$, it's essentially always fine that $0=-0$.

2. Accept that $\infty=-\infty$. This is the approach of the projective real line and the Riemann sphere.

3. Reject negative numbers. If there are no negative numbers, then it's meaningful and natural to allow arithmetic operations with $\infty$ while considering $-\infty$ to be meaningless or undefined. Then $0$ can only be approached from the right (i.e., the positive side of the number line) and quotients like $1/0$, like all other quotients in such a system, are guaranteed to be positive if they are defined.

We rarely actually want to divide by $0$. After all, $0/0$ will still be undefined (or will be permitted to be any number), so the goal of being able to divide anything by anything is still not achieved. But we very often want numbers to be able to be negative. And we usually like to distinguish between $\infty$ and $-\infty$ (so that, for example, we can think of a limit that "diverges to infinity" and one that "diverges to negative infinity" as "equaling" distinct infinite values).

A final note: To see the symmetry between division by $0$ requiring $\infty=-\infty$ and multiplication by $0$ requiring $0=-0$, imagine you didn't know $0=-0$ but did know that anything multiplied by $0$ is $0$ (and some other rules of arithmetic). $$0(1) = 0 = 0(-1) = -0.$$

### When would one need to multiply by $0$?

Often. Consider some word problems:

• $375$ means $5 \times 10^0 + 7 \times 10^1 + 3 \times 10^2$. What does $1001$ mean?
(This should illustrate the ubiquity of multiplying by zero.)

• The only things in Sam's closet are 5 bags and the things in the bags. There are no apples in any of the bags. (Or: each bag contains no apples.) How many apples are in Sam's closet?

• Mary and Billy are driving forward on a straight road, starting at the "10 miles" highway marker. They arrive side-by-side at the "70 miles" highway marker. In front of them are mile markers with higher values. Mary continues driving at 60 mi/hr, while Billy continues at a speed so slow you may assume he is not moving at all. How far from where they started are Mary and Billy, three hours later? (You may assume all mile markers are accurate.)

• Frannie has found that no matter how many times she adds $0$ to itself, she gets $0$ as the sum. In particular, for all the positive integers $n$ she has tried, $0 + 0 + ... + 0$ with $n$ terms has summed to $0$. Is there a general statement about multiplying with $0$ that you could give Frannie, to explain (or characterize) her results?

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Here are some general rules

1. Multiplying both sides of an equality by any number is permitted, though there is no guarantee that this would be helpful or even meaningful. Multiplying by zero is okay but it totally destroys any information that was in the original equality. If this is what you want, then so be it!
2. Multiplying both sides of an inequality by an number is generally not okay. If you multiply by a positive quantity, the inequality remains the same. If you multiply by a negatve quantity, the inequality reverses ($<$ bcomes $>$ and $>$ becomes $<$). If you multiply by zero, then the inequality becomes an equality.
3. Division by zero is never okay. By the way, most mathematicians do not consider $\infty$ to be a number but more a concept. That is another story altogether!

Good luck. Hope this helps.

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I suppose my real question is "why can I multiply by zero" even though causes many silly problems, like arithmetic "destroying" information. Perhaps a better way to phrase it would be "Why is zero considered a real number?" Why isn't it treated like the inverse of infinity? –  TheIsolatedNetwork Feb 22 at 19:55
As I mentioned in my answer, infinity is not considered a number. Zero is that special number that does not have an inverse. But for this, zero is just like any other number. –  user44197 Feb 22 at 20:01
I understand infinity is not a number. I'm asking why zero is, when treating it as such results in contradictions and tons of "special cases." Why not treat zero like we treat infinity - with limits? –  TheIsolatedNetwork Feb 22 at 20:04
This is a philosophical question! However, only two exceptions that arise often is $0^{-1}$ and $0^0$. The first is the characteristic feature of zero. Second is more out of trying to define what $x^0$ is. There is no consistently meaningful way to define it. –  user44197 Feb 22 at 20:09
Squaring also "destroys information", namely whether the input was positive or negative; arithmetic doesn't generally set out only to study invertible functions. $x/0$ isn't undefined because it collapses too much information, it's undefined because there's no such thing. –  Malice Vidrine Feb 22 at 20:09