But you are assuming 0.3... is the representation of 1/3. We don't have to make this assumption, it's just the one we are usually taught. Math doesn't really break from making different assumptions, quite the opposite.
Let's make some different assumptions, not following high school math: When I divide 1 by 3, I always get a remainder. So it would just be as equally valid to introduce a mathematical object representing this remainder after I performed the infinite number of divisions. Then
1/3 = 0.3... + eps / 3
2/3 = 0.6... + 2eps / 3
3/3 = 0.9... + 3eps / 3
and since 0.9... = 1 - eps, we get 3/3 = 0.9... + eps = 1
It's all still sound (I haven't proven this, but so far I don't see any contradiction in my assumptions). And it comes out where 0.9... is not equal to 1. Just because I added a mathematical object that forces this to come out.
Edit: Yes, I am breaking a lot of other stuff (e.g. standard calculus) by introducing this new eps object. But that is not an indicator that this is "wrong", just different from high school math.
LiKao|9 months ago
Let's make some different assumptions, not following high school math: When I divide 1 by 3, I always get a remainder. So it would just be as equally valid to introduce a mathematical object representing this remainder after I performed the infinite number of divisions. Then
1/3 = 0.3... + eps / 3
2/3 = 0.6... + 2eps / 3
3/3 = 0.9... + 3eps / 3
and since 0.9... = 1 - eps, we get 3/3 = 0.9... + eps = 1
It's all still sound (I haven't proven this, but so far I don't see any contradiction in my assumptions). And it comes out where 0.9... is not equal to 1. Just because I added a mathematical object that forces this to come out.
Edit: Yes, I am breaking a lot of other stuff (e.g. standard calculus) by introducing this new eps object. But that is not an indicator that this is "wrong", just different from high school math.