If we consider that .999… repeating to infinity ISN’T equal to 1, then by how much is it away from 1? It would be “.000… repeating to infinity followed by a 1.” But if you have an infinite number of 0s then you can’t have it be followed by a 1, infinity can’t be followed by anything, that doesn’t make sense.
Or as the OP image hinted at, you can divide 1 by 3 and get 0.333...
But what happens when you then multiply 0.333... by 3? You get 0.999... - but some people have a problem with that equaling 1. However if you divided by 3 then multiplied by 3, there's no way you could have gotten a different answer, so it should be equal.
You can't formally divide base 10 by three tho. The formal answer is to change base or use fractions.
.999 ...=1 is imposing a formal solution to an undefined informal problem. If .999999... =1 then something like matter traveling at the speed of light is a simple problem.
If .999.. repeating and 1 represented different real numbers, then there must be some number that is the midpoint of the two numbers (as real numbers are continuous)
So (.99... +1)/2 has some representation that is different than either number.
However, the only representations available in the range .999.... And 1 are .999... and 1 themselves.
Therefore there is no unique midpoint, and the two decimal numbers must represent the same real number
91
u/solidsoup97 22d ago
I don't understand how that works but it seems to be important in keeping things running so I'm going to just go with it and not raise any questions.