# Why does the same equation have different results?

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???

• $10\cdot0,999\neq9,999$. And BTW, $9\cdot0,999\neq8,999$ either!!! Apr 11, 2016 at 20:15
• $10\cdot0,999=9,990$. And BTW, $9\cdot0,999=8,991$. No trick, just a couple of (extremely) bad computations. Apr 11, 2016 at 20:17
• The calculation is correct assuming you are using a comma instead of a decimal point and you intend that all decimals are repeating . there is no problem because 8.9999... = 9
– WW1
Apr 11, 2016 at 20:23
• @WW1: I've assumed the comma as a "thousand" separator. If that's not the case, then my comments above are obviously wrong (I was not aware of the usage of comma as a decimal point). Apr 11, 2016 at 20:28
• comma for decimal point is common in the French and Spanish speaking world
– WW1
Apr 11, 2016 at 20:30

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

• Beautifulll!!! xxx Jose Brox! Apr 12, 2016 at 14:51
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• I'm sorry, but I did click on the upward yet I recieved a message saying "you can not do that because you don't have 15 reputation score yet" I appreciate your concern in talking about the score, but do not be too indelicate at belittle my "xxx" :) It means "I read it and I could understand it because you were awesome, however I couldn't give you up vote yet" Apr 12, 2016 at 20:14

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.

• I ve liked that short representation n(n-1).999... xxx Wouter! Apr 12, 2016 at 14:55
• Is every integer actually a short form in the Limit Theorem? Apr 12, 2016 at 15:03

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 $1$.

• i.e. Should I say the equation (2)' is another way to show how limits works? xxx Daniel! Apr 12, 2016 at 14:58
• I would word it the following way: Let $x$ be the decimal expansion $0.999...$ Then $10x-x=9x$ as real numbers. But the decimal expansion represented by $10x-x$ is $9.000...$, while the decimal expansion represented by $9x$ is $8.999...$. While these decimal expansions are different, they must represent the same real number. Apr 12, 2016 at 17:39