Why does the same equation have different results?

Question

I bring the equation in (1) in order to ilustrate what I mean.

Since

(1.1) $12 - 6$

(1.2) $(4*3) - (2*3)$

(1.3) $(4-2) * (3)$

(1.4) $(2) * (3)$

(1.5) $6$

So...

(2.1) $9.999... - 0.999...$

(2.2) $9.000...$

Or

(2.1)' $10*(0.999...) - 1*(0.999...)$

(2.2)' $(10-1) * (0.999...)$

(2.3)' $(9) * (0.999...)$

(2.4)' $8.999...$???

How is that possible???

Answer 1

What you need to know is that, in the real numbers, $$0.999\ldots = 1$$

How can this be? Well, when you work with the reals, you must assume that you are working with a kind of infinity (like having infinite decimal digits), which forces you to introduce limits (or some other kind of limiting device, e.g. least upper bounds) if you want your reasonings to be rigorous.

So, what we really mean by $0.999\ldots$ is $$\displaystyle \lim_{n\rightarrow\infty} 1-10^{-n},$$ because $0.9=1-10^{-1}$, $0.99=1-10^{-2}$, and so on. But as you probably know, $$\displaystyle \lim_{n\rightarrow\infty} 1-10^{-n} = \lim_{n\rightarrow\infty} 1 - \lim_{n\rightarrow\infty} 10^{-n} = 1 - 0 = 1$$ (because both limits exist).

Therefore $$9.99\ldots - 0.99\ldots = 10-1=9=8.99\ldots.$$

Answer 2

Long story short 0.999...=1

Any positive integer $n$ has two equally valid representations: $n$ and $(n-1).9999\ldots$: $$1=0.999\ldots\\ 2=1.999\ldots\\ 35894=35893.999\ldots$$

This is not an approximation, these are exact equalities.

Answer 3

The answer is interesting: they aren't different numbers!

I should be careful how I word this here - they aren't different real numbers, despite being different decimal expansions. If what I'm saying is to make any sense, then I must mean something different by 'real number' than what I mean by 'decimal expansion'.

By "real numbers", I essentially mean an ordered field (there's another property called completeness that separated the real numbers from, say, the rational numbers, but I won't dwell on that). What do I mean by ordered field? Well, I basically mean a set of numbers with operations $+$ and $\times$ which satisfy all the happy normal rules of algebra (numbers have additive inverses, non-zero numbers have multiplicative inverses, etc) that you use in your proof.

By "decimal expansion", I mean a collection of digits $a_{1}a_{2}....$ (integers between $0$ and $9$) that represent a real number when we write them like this: $$0.a_{1}a_{2}a_{3}\ldots$$ (where I could have chosen some finite number of the digits to come before the decimal point, but that would look messy). More precisely, we have written them as an infinite sum $$\sum_{n=1}^{\infty}10^{-n}a_{n}$$ The fact that every decimal expansion represents a real number has an easy proof (once you know the definitions!) that you'll see in any introductory analysis course.

Now for the key fact: we can have two decimal expansions that are different (that is, two sequences $a_{1},a_{2},\ldots$ and $b_{1},b_{2},\ldots$ that differ in at least one position), such that the real number they represent (that infinite sum) is the same. And, in fact, this is essentially the case you point out. Consider the two representations:

• $a_{0}=1$, $a_{n}=0$ for all $n>0$.
• $b_{0}=0$, $b_{n}=9$ for all $n>0$.

That is, the two numbers $0.999\ldots$ and $1.000 \ldots$. The first sum is a geometric series with common ratio $0.1$ and first term $0.9$, so its sum is $$\frac{0.9}{1-0.1}=1$$ and the other representation is trivially $1$.