Zero point nine repeating is equal to one

 A note on the real number system The real numbers are an ordered field which satisfies the completeness axiom (the least upper bound property). This means that the real numbers are complete, algebraic operations like addition and multiplication are defined and behave normally, and for any real numbers X and Y exactly one of the following holds: X > Y , X = Y , or X < Y. And if X > Y, then X + Z > Y + Z. [1] Hence, if 0.9r ≠ 1, then |1 - 0.9r| > 0. I will argue that there are no non-zero infinitesimals in the real number system and therefor 1 - 0.9r must equal 0 which implies that 0.9r = 1.  The difference between 1 and 0.9r is 1/∞ Consider the following summation:  For n < ∞, one minus S has real, non-zero solutions and as n tends towards infinity, S tends towards 0.9r. At infinity S = 0.9r and the difference one minus S is 1/∞. Either |1/∞| > 1 or 1/∞ = 0.  The real numbers do not contain non-zero infinitesimals; therefor 1/∞ must be 0 Recall that the real numbers must satisfy the least upper bound property. This means that for any set of the real numbers [X, Y] there is a smallest number which is greater than or equal to Y called the supremum. For example, consider the set of real numbers between and including 0 and 3: [0, 3]. There are many numbers which are upper bounds to this set (which are greater than or equal to the largest number in the set); 4, 5, and 300 are all upper bounds. The supremum to this set (the number which is greater than or equal to all numbers in the set and less than or equal to all upper bounds to this set) is 3. [1] Let us assert that the real numbers contain the set of completed rationals and the set of infinitesimals. An infinitesimal is defined as a number δ such that |δ| < 1/m for all natural numbers m = 1, 2, 3, ... , n. The set of completed rationals are the set rational numbers combined with the set of irrational numbers. [2] If the real numbers contain non-zero infinitesimals, then there must be a supremum for the set of infinitesimals [-δ, δ], and since some infinitesimals are positive the least upper bound to the infinitesimals must be positive. Either the supremum of the infinitesimals is a member of the positive infinitesimals, or it is a member of the completed rationals. r is a positive member of the completed rationals, hence r is an upper bound to the set of infinitesimals. If r is the supremum to the set of infinitesimals, then it must be less than or equal to all other upper bounds. However, r/2 is also a positive member of the completed rationals so r/2 is also an upper bound to the infinitesimals. r is not greater or equal to than r/2, therefor there is no member of the completed rationals which is the least upper bound to the set of infinitesimals. δ is a member of the positive infinitesimals. If δ is an upper bound to the set of infinitesimals, then all other infinitesimals must be less than or equal to δ. Since δ is less than all numbers 1, 1/2, 1/3, …, 2δ is also a positive infinitesimal. 2δ is not less than or equal to δ, thus there is no member of the infinitesimals which is an upper bound to the set of infinitesimals. Consequently, the inclusion of non-zero infinitesimals in the real numbers violates the completeness axiom. The completeness of the real numbers is crucial to mathematical analysis, enabling us to take limits and do calculus. The real numbers cannot contain non-zero infinitesimals; therefor 1/∞ must be 0.  Hence, the difference between 1 and 0.9r is 0; 0.9r is equal to 1 If 1 > 0.9r, then 1 + Z > 0.9r + Z. If Z = -0.9r, then 1 - 0.9r must be greater than 0. As established above, this is false; 1 - 0.9r = 0, therefor 1 = 0.9r. The resolution is affirmed. My opponent must prove either that the value of the difference 1 - 0.9r is a real number greater than 0, or that the real numbers contain non-zero infinitesimals.  References: [1] http://www.math.umaine.edu...[2] http://www.britannica.com...;