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

by mathematical inductionProof:

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[3] 6[5]1[6] 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)

=0.1[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.