You're noticing a pattern that holds when you have an even number of items that are arranged in a mirror pattern:
$$A B C D | D C B A$$
Here, you can put a dividing line like a mirror, separating the figure into a left half and a right half. Also, every item appears twice: once on the left side, and once on the right.
But notice what happens when you have an odd number of items arranged in a mirror pattern :
$$A B C D \hat{E} D C B A$$
In the odd case, there is a "middle" item which falls between the left and right sides. As a result, you can't place a dividing line neatly between the left and right sides. Instead, the "middle" item marks the mirror point.
I think you see decimal numbers as an even-like case, with the decimal point $\LARGE .$ dividing the left side from the right side. As such, you expect each item to appear twice, once on the left and once on the right.
But in fact, decimal numbers are an odd-like case (!). The units (or "singles") position marks the midway point in the way that $\hat{E}$ does in the diagram above —the decimal $\Large .$ is not part of the picture and should be ignored for now.
Look at it this way, when you ignore the decimal point:
$$\ldots\text{thousands}\quad \text{hundreds}\quad \text{tens}\quad \hat{\text{units}}\quad\text{tenths}\quad\text{hundredths}\quad\text{thousandths}\ldots$$
Do you see the mirror symmetry now? Because it is odd-like symmetry, there is no place to put a dividing mirror at the halfway point.
So where can we put the decimal point? We choose to put the decimal point on the righthand side of the units position. This has the feature of putting all of the whole number positions to the left of the decimal point, and all the fractional number positions to the right of the decimal point. The decimal point therefore breaks the symmetry — but the notational advantages are worth it.
But why does the decimal have odd-type symmetry in the first place, instead of even-type symmetry? Why couldn't it be something like:
$$\ldots\text{hundreds}\quad \text{tens}\quad \text{units}\quad| \quad{\color{blue}{\text{uniths}}}\quad\text{tenths}\quad\text{hundredths}\ldots$$
One answer is that we are listing all the integer powers of 10 in order, and so the odd-type symmetry of the decimal system comes from the odd-type symmetry of the integers:
$$\ldots 10^{3},10^2, 10^1, 10^0, 10^{-1}, 10^{-2}, 10^{-3},\ldots$$
where here $10^0$ is the midpoint of the odd symmetry. So now the question becomes: why do the integers have odd-type symmetry like this:
$$\ldots +3 \quad +2 \quad +1 \quad 0 \quad -1 \quad -2 \quad -3 \ldots$$
One answer is that our number system does not have two different numbers called, say, $0^+$ and $0^-$. If we did, we would have even-type symmetry exactly the way we'd like:
$$\ldots 10^{+2}\quad 10^{+1}\quad 10^{0^+} \quad {\color{blue}{10^{0^-}}} \quad 10^{-1}\quad 10^{-2}\ldots$$
And the reason why our number system does not have two different numbers $0^+$ and $0^-$ is because then you'd have to tell me how to do arithmetic such as $0^+ + 0^-$—and it's hard to give a satisfying definition. For example,
- If $0^+ + 0^-$ is just $0^+$, then $0^+$ isn't "really" like zero. We know that zero should obey a law like $0 + a = a$ for any number $a$.
- $0^+ + 0^-$ is just $0^-$, then $0^-$ isn't "really" like zero. We know that $0^-$ should obey a law like $a + 0 = a$ for any number $a$.
- If $0^+ + 0^-$ is some new number $0^\star$, then neither $0^+$ nor $0^-$ is is like 0.
And now it seems we've run out of options, but it's good to keep thinking.
