# Thread: .9 repeated does not equal 1

1. ## .9 repeated does not equal 1

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[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.

2. ## Re: .9 repeated does not equal 1

I think I see one potential flaw. By mathematical induction i have proved this over the interval x = [0, infinity) and this interval does not include infinity (which of course is a concept, not a number.)
Therefore my predicate cannot be evaluated at infinity, only at numbers with arbitrarily large magnitudes. What do you guys think?

3. ## Re: .9 repeated does not equal 1

You are confusing the statement "0.(n) < 1 for infinitely many n" with the statement "0.(∞) < 1". 0.(n) is a terminating decimal whereas 0.(∞) isn't.

4. ## Re: .9 repeated does not equal 1

I think what you are saying was similar to what I replied just above. Thank you for the clarification.

The basis of the proof is that at no point does adding another 9 on make it one, but we can never reach 0.9 repeated by this method, whether by induction or not.

5. ## Re: .9 repeated does not equal 1

Yes, we can. $\displaystyle 0.(n)$ is $\displaystyle \sum_{r=1}^n\frac9{10^r}$ whereas $\displaystyle 0.(\infty)$ is $\displaystyle \sum_{r=1}^\infty\frac9{10^r}$ – the latter is, by definition, the limit of the sequence of partial sums $\displaystyle \sum_{r=1}^n\frac9{10^r}.$ A sequence of real numbers less than $\displaystyle 1$ can indeed converge to $\displaystyle 1.$

6. ## Re: .9 repeated does not equal 1

Originally Posted by Sylvia104
You are confusing the statement "0.(n) < 1 for infinitely many n" with the statement "0.(∞) < 1". 0.(n) is a terminating decimal whereas 0.(∞) isn't.
I will add to the above. You proved the for every n $\displaystyle \sum\limits_{k = 1}^n {\frac{9}{{{{10}^k}}}} < 1$.

BUT $\displaystyle \sum\limits_{k = 1}^\infty {\frac{9}{{{{10}^k}}}} = {\lim _{n \to \infty }}\sum\limits_{k = 1}^n {\frac{9}{{{{10}^k}}}} = 1$

7. ## Re: .9 repeated does not equal 1

very cool. I see now how that works. I thought the point of induction is that it proved the predicate all the way up to infinity, but I can see now that this is not necessarily the case.

8. ## Re: .9 repeated does not equal 1

This problem reminds me of the paradoxes of motion posed by the ancient Greek philosopher Zeno of Elea twenty-five centuries ago.