0.99999 recurring does not equal 1

All solutions that I have seen only demonstrate a 'limit towards 1 answer'. They do not demonstrate that 0.9999 recurring is exactly 1.

How can they be the same if they're different?

Only by 'definition'.

0. \dot{9} = \displaystyle\sum_{r=1}^{r= \infty} 0.9 \cdot 0.1^{r-1}

Actually...I'll concede that I'm not qualified to comment on this topic because my knowledge of maths doesn't go beyond A2 Level, and it's not something I've ever encountered.

But definition is all that matters, if nothing is defined then you can't categorically say that anything is true

As I have mentioned earlier. The method relies on 'limits' and not equivalence.

0. \dot{9} = \displaystyle\sum_{r=1}^{r= \infty} 0.9 \cdot 0.1^{r-1}

Ah-ha so if you have studied A Level maths you will know how to find the sum to infinity of a geometric series.

If we wanted to sum $\frac{9}{10} + \frac{9}{100} + \frac{9}{1000} ...$ we would note that this is a geometric series with first term $a=\frac{9}{10}$ and common ratio $r=\frac{1}{10}$.

Now substitute these numbers into the formula $S_{\infty} = \frac{a}{1-r}$.

What's wrong with taking limits? We can define 0.999999 is by taking limits, so it makes sense that limits should come up in your proof

We can also prove it by using the construction of the real numbers (slightly beyond my level so I probably won't be able to explain, but IIRC it involves taking Dedekind cuts)

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As I have mentioned earlier. The method relies on 'limits' and not equivalence.

Had you asked me this question in 2008/2009, I would have answered it without hesitation...unfortunately, I've completely forgotten that part of the curriculum, I have no idea how to solve that! I would be interested in the answer.

I had a brief look at the Wikipedia page for this topic and there are some very basic algebraic and arithmetic summaries which clearly show that the two numbers are equal, so I'm convinced that that's the case (it's impossible to dispute, therefore it must be true). It's so strange though...my immediate instinct was to reject that, but as you said, there is no number between 0.9 recurring and 1...it's all quite odd.

Actually...I'll concede that I'm not qualified to comment on this topic because my knowledge of maths doesn't go beyond A2 Level, and it's not something I've ever encountered.

Actually...I'll concede that I'm not qualified to comment on this topic because my knowledge of maths doesn't go beyond A2 Level, and it's not something I've ever encountered.

It's not hard to understand. If you want some intuition think about what value you would give to $\dfrac {3}{3}$ ?

Considering that $\dfrac {1}{3}=0.333....$

I am nearing the end of my physics degree at a top university and I have not seen an error free proof that that 0.9999 recurring and 1.0 are one and the same.

Given that the way any real number (Such as your 0.9 recurring) is defined is as the limit of a sequence of rationals, your objection doesn't really make sense. As has been mentioned, your assumption that real numbers exist relies on limits.

However, I have a method which (whilst implicitly using limits) does not require taking one. Let's turn the question on its head: What is the absolute difference between the two? More specifically, what is:

$|1-0.999 \cdots|$?

As this is an absolute difference, we clearly have $|1-0.999 \cdots| \geq 0$. It cannot be greater than 0; suppose the difference is some $\epsilon > 0$, then truncating the recurring decimal at the nth decimal digit leaves a difference of $10^{-n}$ which for sufficiently large n we can take smaller than $\epsilon$. Thus the only possibility is the absolute difference is 0, thus the two are the same.

Couple of addendums:

1) In particular, take $n>\frac{-log\epsilon}{log10}$

2) The "implicitly uses limits" part is that we can take $10^{-n}$ as close to 0 as we like, giving it limit 0. You never actually need to take a limit though; you just see the difference between the two is $\geq 0$ and any non-zero difference fails, leaving 0 as the only option.

﴾͡๏̯͡๏﴿ O'RLY?

I suspect the OP will argue that three thirds of a cake isn't the same as a whole one because some of the filling always gets stuck to the blade of the knife when you cut it.