3 ms·

Thanks for clarifying the strict superset part, I'm not actually a mathematician. To clarify my convention comment; it is only convention that mathematicians h

by esrauch 14y ago

Thanks for clarifying the strict superset part, I'm not actually a mathematician. To clarify my convention comment; it is only convention that mathematicians have decided that any given set of properties are useful or interesting to be used pervasively and alternate number systems are not. To any person who is asking why .9 repeating is 1 those reasons are entirely outside the scope of their knowledge and so entirely unrelated to the question; is it not true that the answer is "based on the axioms of which this number line is created due to complex reasons that you can't possibly know at this stage, this is effectively decided to be true as an axiom". If one of these alternate number lines was in common use and reals were only uncommonly used then I'm pretty sure there would be tons of people saying "why is .99.. NOT equal to 1 in ($reals-replacement)" and all the people who have been taught it would sigh condescendingly because they believe it is just an intuitively obvious feature of numbers and not simply an axiom of the number line they are choosing to use. I'm interested in whether you disagree with that sentiment. Re: your proof, trivially could be refused based on the other number line not being closed over division, then there would simply be no such number Z=(X+Y)/2 for X=.999.. and and Y=1. Then the proof where be like those silly ones where someone uses a /0 and arrives at a contradiction. Actually it seems that reals themselves already are not closed over division since x/0 doesn't result in a real.

gizmo686 14y ago
My point was that .999=1 is a direct result from the fact that the number line is continuas. Imagine that you had a number system which does have holes in it (for example, the difference between .999... and 1). Now, take 2 object, one at mass .999..., and the other one of mass 1. In this hypothetical system, these objects would have a different mass. Now, take a third object, whose mass is between the 2 of them. I consider it a critical property of our number system that we can be guaranteed to have a number representing the third mass. As a historic note, the Greeks believed that any 2 numbers could be expressed as a multiple of some unit. For example 2 and .333... could both be expressed as multiples of 1/3. Eventually, they proved that this system could not accuratly moddel their world (the diagonal of a square for example). I suspect that if we did have a number system such as you described, we would eventually discover that it did not work well enough for us, and switch to a continuas one.

by esrauch 14y ago

gizmo686 14y ago

My point was that .999=1 is a direct result from the fact that the number line is continuas. Imagine that you had a number system which does have holes in it (for example, the difference between .999... and 1). Now, take 2 object, one at mass .999..., and the other one of mass 1. In this hypothetical system, these objects would have a different mass. Now, take a third object, whose mass is between the 2 of them. I consider it a critical property of our number system that we can be guaranteed to have a number representing the third mass. As a historic note, the Greeks believed that any 2 numbers could be expressed as a multiple of some unit. For example 2 and .333... could both be expressed as multiples of 1/3. Eventually, they proved that this system could not accuratly moddel their world (the diagonal of a square for example). I suspect that if we did have a number system such as you described, we would eventually discover that it did not work well enough for us, and switch to a continuas one.