1=0.999 again

1. Jun 9, 2005

LittleWolf

I was thinking about x=0.999.. so 10x=9.999... therefore 10x-x=9 when I asked myself does the distributive property apply to an infinite number of terms since 0.9999....= sum(9/10^(-k), k=1,2...)? Maybe I'm confused about what "=" means.

2. Jun 9, 2005

mathwonk

you are right. but it needs proof of course.

3. Jun 10, 2005

Zurtex

It is of course wrong to assume the distributive property applies to an infinite number of terms. Once you however learn the maths to prove it, you'll find in fact it does (under more strict cirumstances).

Last edited: Jun 10, 2005
4. Jun 10, 2005

NateTG

Actually, the distribution is only necessary accross two terms:
$$x(9.\bar{9})=x(9+0.\bar{9})=x9+x0.\bar{9}$$

The original question, which I assume is really, "Is it correct distribute across $\sum_{i=1}{\infty}\frac{9}{10^i}$ as if it were a real number." Is still hard to answer well without resorting to higher math.

5. Jun 11, 2005

vsage

I'll admit I never was good at proofs but I think the fact that 9/10^k for each integer k is a real number in that sequence mentioned above because it is rational and that the reals are closed under addition would be enough to prove the distributive property can be applied.

6. Jun 11, 2005

mathwonk

this way of evaluating an infinite decimal was shown to us in 8th grade. i never forgot it. so the moral is: this is the reason it is true, and is the easiest way to see it, regardless of how rigorous it is.

7. Jun 14, 2005

Night Owl

I thought this topic was beaten to death.

Twice.

Right?

8. Jun 14, 2005

Zurtex

This topic has been beaten to death at least a few dozen times on this forum, I remember one thread that went on for over 400 posts.

9. Jun 14, 2005

NateTG

Horses are directional. You point the legs in the direction you want them to go, and line up twenty pounds of explosives along the spine.

10. Jun 14, 2005

Curious3141

We need a "beating a dead horse icon" in our smilie list.

11. Jun 14, 2005

cronxeh

oH you wont believe just where I've found it :rofl:

On catholics forums! :rofl:

Here you can either use their link or mine:

Speaking of distributive properties:
1 = 0.999.. can be represented as:

$$\begin{document} \displaystyle\sum_{i=1}^\infty \frac{9}{10^i}. \end{document}$$

So when you multiply by 10 it makes perfect sense since you can divide or multiply the Sum by any number, therefore both multiplication and division will hold for distributive property

Last edited by a moderator: Apr 21, 2017
12. Jun 14, 2005

pallidin

Hey, I have a new number system. Ok. probably not new.

Anyway, here it goes:

Instead of 1,2,3,4, etc..., it should be .999~, 1.999~, 2.999~, etc..
And .5 should be .4999~

What my theory posits is that any number can not be an absolute quantity, rather that is must be held in an infinite form.

13. Jun 14, 2005

jcsd

Well no, though it's often called a 'sum to infinity' and shares some properties of 'finite sums' it clearly doesn't share all properties, for example an infinite sum whose terms are all real numbers does not necessarily have a real number even though the reals form a field. If the sum to infinfity doesn't presevre such fundmanetal properties as closure why should we expect it to preserve distribuitivity? To check whether or not distrbuivty applies we have to examine the defintion of a sum to infinity which is the limit of the sequence of partial sums).

Last edited by a moderator: Apr 21, 2017
14. Jun 14, 2005

Zurtex

All real numbers are represented in infinite form, e.g:

$$1 = \ldots 0000000000001.000000000000 \ldots$$

0.5 and 0.49999... are the same number in real numbers so it does't make any difference.

15. Jun 14, 2005

Hurkyl

Staff Emeritus
There are times when one prefers to define the decimals in such a way that a decimal cannot end in an infinite string of zeroes. Or, to define it so that a decimal cannot end in an infinite string of nines.

16. Jun 14, 2005

Zurtex

Would that be when one wants to fit it on the page? Or perhaps when one is a silly physicist and can't get their head round the real number line?

I think I've had too much caffeine tonight :!!)

17. Jun 18, 2005

Bob3141592

Could you elaborate on this please? Under what circumstances would one want to do that either way? And most importantly, what are the consequences of doing it? What's to be gained by prohibiting a number ending in an infinite string of nines, when they have to be allowed to end in an infinite string of threes?

This might relate to a different question I'm struggling with, so additional information might provide the wedge I'm searching for. Thanks.

18. Jun 18, 2005

Hurkyl

Staff Emeritus
The fact that some numbers are represented by two different decimal strings can make the minor details of some proofs more difficult. I think one example where it would be an issue is if you were trying to do Cantor's diagonal argument in binary.

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