Does .9-repeating equal 1?

[Lul Thyme]

Originally posted by Will9


As someone already said .1 repeating.

In the last part of my post I meant to say that 1 is an acceptable, but inaccurate substitute for .9 repeating (instead of the way around). You can use 1 instead of .9 repeating, but is slightly inaccurate. It is the same replacing 999,999,999,999,999,999,999 with 1,000,000,000,000,000,000,000. You will replace the first one with the second because it is easier to say and use, but is slightly inaccurate.

Ok let's talk about the "inacurracy".
Let call 1-0.99999... =e
You agree that e is the error or inacurracy.
Yes this e has the property that it is smaller then any strictly positive number.
(Think about it).
(Please, actually do convince yourself that this is true, or explain why you think it's not, don't just post the same thing yet again)

Ok now the real number system is such that the only non-negative number that is smaller then any strictly positive number is 0.
so e is 0.


By the way your example shows that you don't understand the issue at all.
Nobody is saying that
.9=1 or
.99=1 or
.99...9=1!!!
BUT
.99...=1 notice the difference in the last two.
All your examples have been with a finite sequence of 9, thus non-zero error.
So basically you are not even talking about the right thing.

[KrazyHorse]

Originally posted by Will9


As someone already said .1 repeating.

Wait. 1 - 0.99999... = 0.1111....

?

[loinburger]

Well duh. 1 - 0.999 = 0.111, so by induction we can prove that 1 - 0.999... = 0.111...

[KrazyHorse]

[KrazyHorse]

Originally posted by Ben Kenobi
They aren't the same.

n(n)^x where n is 1 is always one.

lim x -> inf will be 1.

What is the value for

n(n)^x where n is 0.9 repeater?

what about the lim x -> inf? that should be 0.

Makes absolutely no sense

a) What the **** is n(n)? Is that a function or a product? Is n(n)^x = n^(x+1)? What are you trying to say?

b) Why are you assuming that (0.9999...)^x is less than one? Prove it.

This is pretty sad coming from somebody who is supposed to have at least some math background.

[aneeshm]

And by the way - this thread is comedy gold . I think I'll save it once it's done .

[KrazyHorse]

Is this what you were trying to say?

Let a(n) = 0.999...9 (where there are n 9s)

0.9999... = lim(n->inf) a(n)
=> lim(x->inf) (0.9999...)^x =
lim(x->inf) (lim(n->inf) a(n))^x =
MISTAKE OCCURS NOW
lim(n->inf)(lim(x->inf)(a(n)^x)) =
lim(n->inf)(0) = 0
=> 0.999... < 1
QED



If so, then who the **** taught you limits?

YOU CAN'T PASS LIMITS OVER EACH OTHER WITH NO FEAR OF THE CONSEQUENCES

Example: lim(m->inf)(lim(n->inf) m/n) = lim(m->inf)(0) = 0
but lim(n->inf)(lim(m->inf) m/n) = lim(n->inf)(inf) = inf

[Pekka]

KH, dude I already did that very clearly earlier Except it was shorter and more easy to understand!"!!

[KrazyHorse]

This should be the question they ask you before you get to vote. If you answer "no", then they brand you with a giant red "I" on your forehead and you are forever barred from taking part in the electoral process again.

[Donegeal]

I'm so very glad KH is Canadian.

[Tim_Augustus]

Originally posted by Impaler[WrG]
A some what simpler explination I heard someware, you dont need to have even the fogyest idea what calculus is to get it.

.999... * 10 = 9.999... (just moved the decimal point up)

9.999... - .999... = 9.0 (infinite 9's all canceld out)

Thus 9 x .999... = 9 and 1 x .999... = 1

.999... = 1

All you really need is an understanding of the concept of "infinity" and a bit of rational and deductive thinking ability. Everything else is just fluff.

[Jon Miller]

Originally posted by aneeshm
And by the way - this thread is comedy gold . I think I'll save it once it's done .

It's gotten a lot better.

I don't know if it topped the last one yet though.

JM

[rah]

Yes, they're mathmatically the same, but to a programmer, you have to be careful because it may cause you problems since some programming languages deal with some of these issues differently. So I can say yes and sometimes NO.
example.
In SAS, (a statistical language) SAS WILL see .9999...(to as many as the max precision in the language) as equal to one and if you base triggers off the comparison, they will work just fine. But if you do .33333.... and compare it to 1/3, it will not consider them equal. (which was a common proof used in many examples here.) If I then multiplied it by 3, it did recognize it as 1. Hmmmm
Now granted the programs is not set to handle infinite precision.

So the final point is, yes, they're the same, but in THE REAL WORLD, you have to verify that the tools that you're using also believe this truth.

Computers can't be wrong

[SlowwHand]

Originally posted by Donegeal
I'm so very glad KH is Canadian.

If he wasn't, you'd have another crime scene investigation.

[Pekka]

rah, yes but that doesnt' really apply, we could always argue that Java uses round ups totally against the 'reality', so we can't really use that as an argument to what is mathematically correct.