Why can one multiply by zero

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.


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:

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

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)


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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Limits