Greta wrote:With help earlier in the thread I see the situation as this:

0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999

is definitely less than 1.

However 0.999... must be equal to 1 because the 9s are infinite. Infinity is not only probably not present in the physical world (unless nothingness is infinite) but infinity is also beyond mathematics, so the standard mathematical logic of "it must be just a little less than one" cannot work. Any such number that is less than one cannot be 0.999.... It instead would be a finite decimal, as above, with a finite number of repeated 9s.

I can prove they're not equal with bases:

Assume we're working in base 16

0.99999... in base 16 = 9/16 + 9/256 + 9/4096 + .... in base 10 = (9/16) / ( 1 - 1/16 ) in base 10 = (9/16) / ( 15/ 16) in base 10 = 144 / 240 = 0.6 in base 10 which does not equal 1 in base 10

There's some interpretation involved when answering the question particularly pinning down what equals and real numbers are.

I'm still trying to figure out whether 0.9999... is a real number or not. I know the order properties of real numbers states adding two real numbers or multiplying two real numbers produces a real number. In that case:

9/10 is real, 9/10 + 9/100 + 9/1000 is real, but I don't know about an infinite series like 9/10 + 9/100 + 9/1000 + 9/1000 + .... this doesn't seem to help since we know an infinite series like 1+2+3+4+... does not produce a real number either.

Any theorems of axioms that I've looked at to determine whether 0.9999... = 1 or whether 0.9999... is a real number I run into circularity issues like here

https://en.wikipedia.org/wiki/Construct ... _sequences
Synthetic approach

The synthetic approach axiomatically defines the real number system as a complete ordered field. Precisely, this means the following. A model for the real number system consists of a set R, two distinct elements 0 and 1 of R, two binary operations + and × on R (called addition and multiplication, respectively), and a binary relation ≤ on R, satisfying the following properties.

(R, +, ×) forms a field. In other words,

For all x, y, and z in R, x + (y + z) = (x + y) + z and x(yz) = (xy)z. (associativity of addition and multiplication)

For all x and y in R, x + y = y + x and xy = yx. (commutativity of addition and multiplication)

For all x, y, and z in R, x(y + z) = (xy) + (xz). (distributivity of multiplication over addition)

For all x in R, x + 0 = x. (existence of additive identity)

0 is not equal to 1, and for all x in R, x*1 = x. (existence of multiplicative identity)

For every x in R, there exists an element −x in R, such that x + (−x) = 0. (existence of additive inverses)

For every x ≠ 0 in R, there exists an element x^−1 in R, such that x*x^−1 = 1. (existence of multiplicative inverses)

(R, ≤) forms a totally ordered set. In other words,

For all x in R, x ≤ x. (reflexivity)

For all x and y in R, if x ≤ y and y ≤ x, then x = y. (antisymmetry)

For all x, y, and z in R, if x ≤ y and y ≤ z, then x ≤ z. (transitivity)

For all x and y in R, x ≤ y or y ≤ x. (totalness)

The field operations + and × on R are compatible with the order ≤. In other words,

For all x, y and z in R, if x ≤ y, then x + z ≤ y + z. (preservation of order under addition)

For all x and y in R, if 0 ≤ x and 0 ≤ y, then 0 ≤ xy (preservation of order under multiplication)

The order ≤ is complete in the following sense: every non-empty subset of R bounded above has a least upper bound. In other words,

If A is a non-empty subset of R, and if A has an upper bound, then A has a least upper bound u, such that for every upper bound v of A, u ≤ v.

The rational numbers Q satisfy the first three axioms (i.e. Q is totally ordered field) but Q does not satisfy axiom 4. So axiom 4, which requires the order to be Dedekind-complete, is crucial. Axiom 4 implies the Archimedean property. Several models for axioms 1-4 are given below. Any two models for axioms 1-4 are isomorphic, and so up to isomorphism, there is only one complete ordered Archimedean field.

Using "For all x and y in R, if x ≤ y and y ≤ x, then x = y. (antisymmetry)" as an example

if 0.9999... <= 1 and 1 <= 0.9999... then 0.9999... = 1. 0.9999... <= 1 is true but proving 1<=0.9999... would involve proving 0.9999... = 1 first, it's circular

My real analysis book tries to incorporate the concept of limits with the equal sign called infimum or greatest lower bound without saying whether they are the same thing as far as I can see.

it says:

If S is bounded below, then w is said to be an infimum (or greatest lower bound) of S if it satisfies the conditions:

1) w is a lower bound of S, and

2) if t is any lower bound S, then t<=w

For example if I picked the subset S of all reals greater than 0, every negative would qualify as a lower bound but 0 would be the greatest lower bound even though that bound is not part of the subset. The same would be true if subset S of all reals greater than or equal to 0.

I found this theorem a Corollary of the Archimedean Property

If S := {1/n : n∈N}, then inf S = 0

Which at first I thought could have been helpful to prove whether an 0.00000...1 = 0 or 1 - 9/10 - 9/100 - 9/1000 + ... = 0, a similar problem to the 0.99999... problem. But because "2) if t is any lower bound S, then t

<=w" both 0.00000...1 < 0 or 0.00000...1 = 0 could be true and still not contradicts this Corollary

Another question that I thought could distinguish 0.9999... and 1 is the rounding down function

Does floor(0.9999...) = 0 or floor(0.9999...) = 1? It's pretty much the same dilemma.

Here's the definition of the floor function:

floor(x) = max{m∈Z | m<=x}

but floor(0.99999...) cannot be evaluated without first knowing whether 0.99999...>=1 aka whether 0.99999...=1 beforehand. I suspect I'll run into this same problem of not being able to pin down "=" in real numbers and both interpretations may work within the reals unless someone can point out something that I haven't looked at?