Why doesn't $1/x=0$ have any solution?

Question

Just out for curiosity ! Why $1/x=0$ doesn't have any solution?

Or is it that the solution takes you to $1=0$ situation which would nullify mathematical principle that we stood for years

Educate me...And thanks

Answer 1

Asking why $\frac{1}{x} = 0$ has no solution in the real numbers is the same as asking if there is a real number $x$ that solves $1 = 0\cdot x$. Clearly if we multiply any real number by $0$, we will get $0$ back and so there cannot possibly be a solution. We are not limited to considering real numbers, but the argument doesn't change much when you consider other structures known as fields, of which the real numbers are a great example.

If you were to leave the realm of a field (e.g. stop restricting yourself to only the real numbers), then things get more interesting and more complicated (algebraically). If you were to include some symbols $\pm\infty$ (plus and minus infinity) with the real numbers and define them by $\frac{1}{\pm\infty} = 0$, things get a little weird. Notice that even though we say $\frac{1}{\pm\infty} = 0$, we do not want to say that $1 = 0\cdot\pm\infty$. The reason we don't is that multiplication would no longer be associative.

When multiplying real numbers, it doesn't matter how we multiply them: $a(bc) = (ab)c\,$. This is called the associativity of multiplication. If we introduce plus and minus infinity like we just did, this breaks down if we try to multiply by infinity haphazardly. Let's take our equation $\frac{1}{\pm\infty} = 0$ and multiply both sides by $\pm\infty$ and pretend things work out like they used to with strictly real numbers (meaning that we assume our multiplication is associative), we would have:

$$ 1 = 0\cdot\pm\infty.$$

Now, let's just see what happens when we multiply both sides by $2$. We would get

$$ 2 = 2(0\cdot\pm\infty).$$

If we try to rearrange our multiplications, we get

$$ 2 = (2\cdot0)\cdot\pm\infty = 0\cdot\pm\infty.$$

But this tells us that $1=2$ and this is very bad. Even though we said that $\frac{1}{\pm\infty} = 0$, we do not want to say that $1 = 0\cdot\pm\infty$. For not too dissimilar reasons, we also do not want to attempt to attribute any meaning to $\infty + (-\infty)$ or $-\infty + \infty$. Returning to the issue of multiplying infinity by $0$, we could just define it to be given by $0 = 0\cdot\pm\infty$ (and one does this in some contexts; see: measure theory). This definition is perfectly fine and does nice things for you, but note that we no longer obey the relationship $\frac{1}{x} = y \implies xy = 1$ (since $\frac{1}{\pm\infty} = 0$ but $0\cdot\pm\infty = 0$).

Even though we can kind of successfully introduce infinity as a "solution" to $\frac{1}{x} = 0$, it does give rise to some very quirky things. For these reasons we try to avoid it as much as possible, or when we can't, we try to limit ourselves to situations where the algebra we do with infinity is entirely consistent and gives meaningful results.

Answer 2

In the setting of a field it is because $\frac{1}{x}=x^{-1}$ has an inverse (namely $x$ itself) - while $0$ has none.

Answer 3

Simply because if you multiply left and right member for $x$ you get: $x\times0=1$. Now, every number multiplied by $0$ is $0$.

Answer 4

It all depends on the domain of $\frac{1}{x}$; that is, where $x$ can come from. The solution to $\frac{1}{x}=0$ is usually left undefined, but this is not always the case. It has a solution in the extended complex plane $\bar{\mathbb{C}}=\mathbb{C}\cup\{\infty \}$: $\frac{1}{\infty }=0$.

The reason why this makes sense in $\bar{\mathbb{C}}$ is that the rules for using $\infty$ are restricted in such a way that contradictions are avoided. Things like $\infty -\infty$ are omitted.

This fact is important for things called Möbius transformations, which underpin many mathematical concepts.

Answer 5

It does, it the zero ring $\{0\}$. In fact, this is the only ring for which $1/x$ has a solution (multiply both sides by $x$). It also has a solution in various other systems like the extended real line as mentioned somewhere here, but those are not rings, of course.