>>11482630I think it's worth mentioning that there are 2 ways to look at the natural math here. (A)What intuitively "seems right" and (B) that which is mechanically computable.
Both approaches make no attempt to validate infinity and simply recognize decimal repetition as a matter of error in translating fractional to decimal, and both treat decimal repetition as something that doesn't "literally" continue on forever and ever, but rather only continues for as many significant digits in the decimal which are needed for a given application.
In (A), we have 1÷3 = 0.333...
If we need 5 significant digits, we would answer as
>1÷3 = 0.33333Thats the intuitive way to look at it as you solve the long division, write out five 3's, and move on. Whatever happened to remainder math, anyway? The other easily comprehended way to write it is 1÷3=0.3r1, but I suppose remainders don't account for much without more information regarding the problem when looking to change the order of the equation.
0.3r1×3 = 0.9r3 doesn't make much sense by itself.
The only issue arises when we do
>1÷3 = 0.33333>3 × 0.33333 = 0.99999For we already know (1÷3)×3 = 1, but this too is merely an error explained by stating that a repeating decimal result can not finalize(end), so is LESS than the fraction division. "1÷3 > 0.33333" rather than "1÷3 = 0.33333"
Then in (B), we have an extra "hidden" significant digit and assume rounding. If we need 5 significant digits, we get 6 and use the sixth to round as
>1÷3 = 0.33333(3 >1÷3 = 0.33333which also allows
>1÷3 = 0.33333(3 × 3 = 0.99999(9>9 is greater than 4, round up and carry>0.99999(9 = 1thereby accounting for the predetermined knowledge that (1/3)*3 = 3/3 = 1
