I know it is generally accepted and proven that 0.9 repeated is equal to 1. However I got an idea for a proof against it, and decided to write the proof to see where I run into a flaw or contradiction. As of now I am unable to see the flaw in my logic. Can somebody else run through it and tell me where it is? I apologize for the complex notations I added.
Hypothesis: 0.9 repeated does not equal 1
Proof:by mathematical induction
Let the notation 0.(x) mean zero with x amount of 9's after the decimal point and the notation 0.(#) also mean zero with some amount of 9's after the decimal place dictated by the number in place of #
Let the notation 0.#[x1] #[x2] #[x3] mean that the digit in # is in the decimal place as defined by x1, x2, or x3. For example, 0.5 61 is 0.005061
Let the zero'th decimal place be the ones decimal place.
Let P(x) be the predicate “0.(x) is less than 1 if x >= 0”
Step1: Prove P(0)
P(0)= “0.(0) is less than 1.”
As defined above, 0.(0) = 0
0 is less than 1.
Thus P(0) is true.
Step2: Prove that if P(n) is true, then P(n+1) is true.
By P(n) we know that 0.(n) is less than 1.
We must show that 0.(n+1) is less than 1.
0.(n+1) is equal to 0.(n) + 9 X 10 ^ -(n+1) by separating out the last digit added.
We must prove the inequality 0.(n) + 9 X 10 ^ -(n+1) < 1
=9 X 10 ^ -(n+1) < 1 – 0.(n) by subtracting 0.(n) from each side
=0 < 1 – 0.(n) - 9 X 10 ^ -(n+1) by adding 9 X 10 ^ -(n+1) to each side.
But 1– 0.(n) - 9 X 10 ^ -(n+1)
=0.1[n] - 9 X 10 ^ -(n+1)
So we have 0 < 0.1[n+1] which is true because 0.1[n+1] is a positive number and is greater than 0.
Therefore we have proven that if P(n) is true, then P(n+1) is true.
By mathematical induction we have shown that P(n) is true for all integers n >= 0. There fore P(positive infinity) is true or in other words, a 0 with an infinite number of 9's after it is less than 1.