In base 10, if we add 1 and 1, we get the next digit, 2.

In base 2, if we add 1 and 1 there is no 2, thus we increment the next place by 1 getting 10.

We can expand this to numbers with more digits: 111(7) + 1 = 112 = 120 = 200 = 1000

In base 10, with A representing 10 in a single digit: 199 + 1 = 19A = 1A0 = 200

We could do this with larger carryover too: 999 + 111 = AAA = AB0 = B10 = 1110 Different orders are possible here: AAA = 10AA = 10B0 = 1110

The "carry the 1" process only starts when a digit exceeds the existing digits. Thus 192 is not 2Z2, nor is 100 = A0. The whole point of carryover is to keep each digit within the 0-9 range. Furthermore, by only processing individual digits, we can't start carryover in the middle of a chain. 999 doesn't carry over to 100-1, and while 0.999 does equal 1 - 0.001, (1-0.001) isn't a decimal digit. Thus we can't know if any string of 9s will carry over until we find a digit that is already trying to be greater than 9.

This logic is how basic binary adders work, and some variation of this bitwise logic runs in evey mechanical computer ever made. It works great with integers. It's when we try to have infinite digits that this method falls apart, and then only in the case of infinite 9s. This is because a carry must start at the smallest digit, and a number with infinite decimals has no smallest digit.

Without changing this logic radically, you can't fix this flaw. Computers use workarounds to speed up arithmetic functions, like carry-lookahead and carry-save, but they still require the smallest digit to be computed before the result of the operation can be known.