RickG
I have a counter-proof (inductive) that (in base 10) .99999… is not equal to 1. I would appreciate your bringing your opinion to bear on it.
Let me present first your (i.e., the classic)deductive proof just a bit more formally, plus one classic inductive proof, and then my counter-proof. Naturally, if my proof is sound, and it contradicts the classic proofs, then I will have to demonstrate something specific that is wrong with both of them.
Classic Deductive Proof
Axiom 1: For every real number, A, A = A.
Axiom 2: If A = B, then AN = BN.
Axiom 3: If A = B, then A-N = B-N.
Axiom 4: A = .99999…
Premise 1: 10A = 9.99999…
Premise 2: 10A-A = 9.99999…-A
Premise 3: 9A = 9.99999… - .99999…
Premise 4: 9A = 9
Premise 5: 9A/9 = 9/9
Premise 6: A = 1
Conclusion: Since A = A, by Axiom 1, and A = 1, by Premise 6, and A = .99999…, by Axiom 4, then 1 = .99999…
Classic Inductive Proof
Axiom 1: For every real number A, A = A.
Axiom 2: If A > B and B > C, then A > C.
Observation 1: If .99999… is not equal to 1, then there must be some number, A, between .99999… and 1, such that .99999… < A < 1.
Conclusion: Since no such number exists, 1 must be equal to .99999…
Inductive Counter-proof
Axiom 1: For every real number, A, A = A.
Axiom 2: If A = B, then A-B = B-A = 0.
Axiom 3: A = 1
Axiom 4: B = .99999…
Observation 1: 1 - .9 = .1
Observation 2: 1 - .99 = .01
Observation 3: 1 - .999 = .001
Observation 4: 1 - .9999 = .0001
Observation 5: 1 - .99999 = .00001
Observation 6: In general, it appears that 1-.9(N digits) = .0(N-1 digits)…1
Observation 7: Though the trailing 1 of .0(N-1 digits)…1 moves further and further to the right with each additional digit in .9(N digits), and is therefore of less and less magnitude, it does not ever disappear, and therefore never attains the value of 0.
Conclusion: Since 1-.99999… > 0, then by Axiom 2, .99999… does not equal 1.
Note that if Observation 6 is proved, the entire proof becomes deductive.
Summary
I have named the number representing 1-.99999… “eta”. But before I present its properties, I am beholden to show flaws in the classic proofs that 1 does equal .99999…
Flaw 1
The flaw in the classic deductive proof occurs early, in Premise 1. The step ignores the fact that, for N 9s to the right of the decimal, there will be N-1 9s after multiplying by 10. For example, .99999*10 = 9.9999. It does not equal 9.99999.
Flaw 2
The conclusion assumes only one possible interpretation for there being no number between .99999… and 1, because the assumption is made that the set of real numbers is infinitely large, an assumption not presented as an axiom and not proved.
(Yes, I am aware of Cantor’s work, but save that for another post.)
The existence of eta implies the existence of a smallest non-zero real number. In other words, eta is exactly the distance between .99999… and 1 on the real number line, or, put another way, eta is beside zero on the real number line.
We can use the same extrapolation process to discover another number, the multiplicative inverse of eta, and therefore the largest number in magnitude, thus: 1/.9 = 10; 1/.99 = 100; 1/.999 = 1000; etc; therefore, 1/.99999(N digits) = 10(N digits). That number, I named “omega”.
There are also some practical arithmetic applications for the same extrapolation process, as solutions to heretofore perplexing problems. For example, what is 4*.77777…?
Well, 4*.7 = 2.8; 4*.77 = 3.08; 4*.777 = 3.108; 4*.7777 = 3.1108; 4 * .77777 = 3.11108; therefore, 4*.7 = 3.1(N digits)07.
What do you think?
“It is lucky for rulers that men do not think.” — Adolf Hitler