As other users have noted, it doesn't make sense to have infinitely many nines to the left of the decimal point. This is because the sequence $9, 99, 999, \ldots$ doesn't converge to anything, unlike the sequence $0.9, 0.99, 0.999, \ldots$, which converges to $1$.

However, you have touched on something interesting. Are there number systems where this does converge? And if so, does it converge to $-1$? The answer to both is yes.

For an integer $k$, there's a ring called the $k$-adic numbers*. Let's use $k = 10$. It's similar to the regular real numbers in base $k$, but a few things are backwards.

$$ \small{\begin{array}{c|c|c} & \textrm{Reals (base }k\textrm{)} & k\textrm{-adic numbers} \\ \hline \textrm{number of digits after the point} & \textrm{finitely many} & \textrm{infinitely many} \\ \textrm{number of digits before the point} & \textrm{infinitely many} & \textrm{finitely many} \\ \textrm{two numbers are close if} & \textrm{they match in the leftmost digits} & \textrm{they match in the rightmost digits} \end{array}} $$

That's not really the way that people like to construct the $k$-adics, but it's easier than all that "completion wrt a metric" or "inverse limit" stuff.

Anyways, in this system, $9, 99, 99, \ldots$ does have a limit, and it's $\ldots 999.0$. And if you add $1$ to that, you have the sequence $10, 100, 1000, \ldots$, which converges to $0$. So in that system, $\ldots 999 = -1$.

*Usually people like $k$ to be prime, so they call it $p$. So if you wanna find out more, look up "$p$-adic numbers", not "$k$-adic numbers".

As other users have noted, it doesn't make sense to have infinitely many nines to the left of the decimal point. This is because the sequence $9, 99, 999, \ldots$ doesn't converge to anything, unlike the sequence $0.9, 0.99, 0.999, \ldots$, which converges to $1$.

However, you have touched on something interesting. Are there number systems where this does converge? And if so, does it converge to $-1$? The answer to both is yes.

For an integer $k$, there's a ring called the $k$-adic numbers*. Let's use $k = 10$. It's similar to the regular real numbers in base $k$, but a few things are backwards.

$$ \small{\begin{array}{c|c|c} & \textrm{Reals (base }k\textrm{)} & k\textrm{-adic numbers} \\ \hline \textrm{number of digits left of the point} & \textrm{finitely many} & \textrm{infinitely many} \\ \textrm{number of digits right of the point} & \textrm{infinitely many} & \textrm{finitely many} \\ \textrm{two numbers are close if} & \textrm{they match in the leftmost digits} & \textrm{they match in the rightmost digits} \end{array}} $$

That's not really the way that people like to construct the $k$-adics, but it's easier than all that "completion wrt a metric" or "inverse limit" stuff.

Anyways, in this system, $9, 99, 99, \ldots$ does have a limit, and it's $\ldots 999.0$. And if you add $1$ to that, you have the sequence $10, 100, 1000, \ldots$, which converges to $0$. So in that system, $\ldots 999 = -1$.

*Usually people like $k$ to be prime, so they call it $p$. So if you wanna find out more, look up "$p$-adic numbers", not "$k$-adic numbers".

As other users have noted, it doesn't make sense to have infinitely many nines to the left of the decimal point. This is because the sequence $9, 99, 999, \ldots$ doesn't converge to anything, unlike the sequence $0.9, 0.99, 0.999, \ldots$, which converges to $1$.

However, you have touched on something interesting. Are there number systems where this does converge? And if so, does it converge to $-1$? The answer to both is yes.

For an integer $k$, there's a ring called the $k$-adic numbers*. Let's use $k = 10$. It's similar to the regular real numbers in base $k$, but a few things are backwards.

$$ \begin{array}{c|c|c} & \textrm{Reals (base }k\textrm{)} & k\textrm{-adic numbers} \\ \hline \textrm{number of digits after the point} & \textrm{finitely many} & \textrm{infinitely many} \\ \textrm{number of digits before the point} & \textrm{infinitely many} & \textrm{finitely many} \\ \textrm{two numbers are close if} & \textrm{they match in the leftmost digits} & \textrm{they match in the rightmost digits} \end{array} $$

That's not really the way that people like to construct the $k$-adics, but it's easier than all that "completion wrt a metric" or "inverse limit" stuff.

Anyways, in this system, $9, 99, 99, \ldots$ does have a limit, and it's $\ldots 999.0$. And if you add $1$ to that, you have the sequence $10, 100, 1000, \ldots$, which converges to $0$. So in that system, $\ldots 999 = -1$.

*Usually people like $k$ to be prime, so they call it $p$. So if you wanna find out more, look up "$p$-adic numbers", not "$k$-adic numbers".

As other users have noted, it doesn't make sense to have infinitely many nines to the left of the decimal point. This is because the sequence $9, 99, 999, \ldots$ doesn't converge to anything, unlike the sequence $0.9, 0.99, 0.999, \ldots$, which converges to $1$.

However, you have touched on something interesting. Are there number systems where this does converge? And if so, does it converge to $-1$? The answer to both is yes.

For an integer $k$, there's a ring called the $k$-adic numbers*. Let's use $k = 10$. It's similar to the regular real numbers in base $k$, but a few things are backwards.

$$ \small{\begin{array}{c|c|c} & \textrm{Reals (base }k\textrm{)} & k\textrm{-adic numbers} \\ \hline \textrm{number of digits after the point} & \textrm{finitely many} & \textrm{infinitely many} \\ \textrm{number of digits before the point} & \textrm{infinitely many} & \textrm{finitely many} \\ \textrm{two numbers are close if} & \textrm{they match in the leftmost digits} & \textrm{they match in the rightmost digits} \end{array}} $$

That's not really the way that people like to construct the $k$-adics, but it's easier than all that "completion wrt a metric" or "inverse limit" stuff.

Anyways, in this system, $9, 99, 99, \ldots$ does have a limit, and it's $\ldots 999.0$. And if you add $1$ to that, you have the sequence $10, 100, 1000, \ldots$, which converges to $0$. So in that system, $\ldots 999 = -1$.

*Usually people like $k$ to be prime, so they call it $p$. So if you wanna find out more, look up "$p$-adic numbers", not "$k$-adic numbers".