# Upper and lower bounds in regards to 0.(9) [duplicate]

Possible Duplicate:
Does .99999… = 1?

I'm only doing this at GSCE and I'm really only asking here because of an interesting email conversation between my Grandfather and I regarding the fact that 0.(9) equals 1, so I'd appreciate it if you could make any explanation as simple as possible.

Basically, I have proven to my Grandfather that 0.(9) must equal 1, using the following method:

Let x = 0.(9)

So, 10x will equal 9.(9); 10x - x is 9x which is the same as 9.(9) - 0.(9) = 9, and therefore 9 / 9 is 1!

However, he has questioned the fact that 0.(9) * 9 equals 9, as he rightly stated that it equals 8.(9). I do remember learning in my maths lesson a rule regards to upper and lower bounds that meant that 8.(9) was actually the same as 9, or something along those lines, but I can not remember the correct statement to inform my Grandfather - so any suggestions would be appreciated.

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## marked as duplicate by Nate Eldredge, Ｊ. Ｍ., Srivatsan, Henning Makholm, Asaf KaragilaNov 30 '11 at 17:39

the equation 10x = 9.(9) - x is false, instead, write 10x=9.(9)=9 + 0.(9) = 9+x therefore 10x=9+x –  sdcvvc Nov 30 '11 at 12:22
@sdcvvc I don't see why the equation is false; this is what I have been taught at school, and surely 9 + 0.(9) is the same as 9.(9)? –  Andy Nov 30 '11 at 12:25
Andy: Sorry, my mistake, the equation is correct. –  sdcvvc Nov 30 '11 at 13:14
@sdcvvc: Thanks, I'm glad it is correct. –  Andy Nov 30 '11 at 13:22
As was pointed out by @sdcvvc, the equation $10x=9.(9)-x$ is false, it leads to $11x=10$. Alternately (and you explicitly say this yourself) it gives $10x=9x$, which, if $x\ne 0$, implies that $10=9$. Of course, that's not what you mean. But that is precisely what is written. –  André Nicolas Nov 30 '11 at 15:06
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Your "10x = 9.(9) - x = 9x = 9" is a confusing shorthand so let's try:

$x = 0.(9) \implies 10x=9.(9) \implies 10x-x = 9.(9) - 0.(9) \implies 9x=9 \implies x=1.$

Nowhere do we try to multiply out $9 \times 0.(9)$. Instead we calculate $9.(9) - 0.(9)$, and I would be surprised if your grandfather thought that this was not $9$.

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I did think that there was no multiplication involved. However, my subtracting we end up 9x = 9, which is surely the same as saying 9 lots of 0.(9) is equal to nine - which it is not!? –  Andy Nov 30 '11 at 12:46
@Andy: In fact $9 \times 0.(9)$ is both $9$ and $8.(9)$, as they are the same thing, but the point here is that you do not have to attempt that multiplication as part of this proof. –  Henry Nov 30 '11 at 19:53
Yes, I have recently discovered the fact that 8.(9) and 9 are equal because of 8.(9) < x < 9! I know that you don't have to attempt multiplication here, but my Grandfather being my Grandfather was eager to try and 'unprove' this "theory", but at last it appears that mathematics has the upper hand here! –  Andy Nov 30 '11 at 20:30
The definite response to the 0.(9) issue is this: By definition, $0.(9)=\sum_{n=1}^\infty\frac{9}{10^n}$. Using the formula for infinite geometric sum, you get $0.(9)=1$. Every other "proof" is simply an appeal to some basic intuition that allows one to avoid going through the formal definition, but when that intuition hits the wall it's time to use the definition.
Now, in general if a real numbers ends in a sequence of digits the form $a999\dots$ where $a\ne 9$ you can replace it by $b000\dots$ where $b=a+1$. This leads to $8.999\dots$ being equal to $9.000\dots$. This is also true in other bases: in base $d$, every number whose representation ends in a sequence of the digit $d-1$ can be changed similarly (adding 1 to the last digit different from $d-1$ and changing the rest of the digits afterwards to 0).
0.(9) equals the geometric sum whether or not the sum is equal to 1. The derivation of the formula for the sum gives $10x = 9 + x$. The infinite decimal solutions of this equation are $x=0.(9)$ and $x=1.(0)$. This shows only that the numbers could be equal. The assertion that they are equal involves other concepts, such as imposing algebraic axioms or defining equality to mean distance less than $1/n$ for all integer $n > 0$. –  zyx Nov 30 '11 at 13:40