Does .9-repeating equal 1?

[Pekka]

Ohh, Aeson is trying to be cute.

Listen, if you don't like it, you'll be sent to the hole. That's isolation to you, prisoner 93F83UK33K3E2D, and maybe after that you will show some respect to the Gods of math.

[Aeson]

Me? Naw... I would never try to be cute in prison!

[Impaler[WrG]]

It's a stupid question, since numbers have no real existance, they are concepts that only exist in our heads used to help understand the world

NO, numbers in a platonic sense dont have any physical existence but they and mathamatical principles definatly are real parts of our world and our literaly in the fabric of reality.

If people feel like debating this we should probably start a new thread for it.

[Nubclear]

Pekka enjoys prison porn

[Ben Kenobi]

So what have I done wrong here?

what is the lim x-> inf ( n(n)^x )where n = 0.9 repeater

If the equality is true, this should equal one, but I can't see how that is. I keep thinking the limit ought to be 0.

[Deity Dude]

We were taught this in 5th grade or something.

Express a fraction as a decimal:

1/3 = .333 inf

Fractional approach:

1/3*3 = 3/3 = 1

Decimal approach:

.333 inf * 3 = .999 inf = 1

[aneeshm]

Originally posted by KrazyHorse

Anyhow, your answer is obviously correct (by the definition and existence of transcendentals)

Could you explain the solution to me ?

[loinburger]

Originally posted by Ben Kenobi
So what have I done wrong here?

what is the lim x-> inf ( n(n)^x )where n = 0.9 repeater

If the equality is true, this should equal one, but I can't see how that is. I keep thinking the limit ought to be 0.

That's because you keep thinking that .9-repeating is less than 1.

[Spec]

Originally posted by Deity Dude
We were taught this in 5th grade or something.

Express a fraction as a decimal:

1/3 = .333 inf

Fractional approach:

1/3*3 = 3/3 = 1

Decimal approach:

.333 inf * 3 = .999 inf = 1

Best and clearest explanation. There is no way to not understand this and still say .999..... is not 1.

.9999....is one. I am convinced.

Spec.

[Will9]

Originally posted by KrazyHorse


If 0.999... does not equal 1 then |1 - 0.999...| = x > 0

What is x?

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.

[civman2000]

Originally posted by Ben Kenobi
So what have I done wrong here?

what is the lim x-> inf ( n(n)^x )where n = 0.9 repeater

If the equality is true, this should equal one, but I can't see how that is. I keep thinking the limit ought to be 0.

The flaw in your logic is that the function f(a)=lim a^x is not continuous. It is 0 for |a|<1, 1 for a=1, and infty for a>1. Thus even as you approximate a=1 from below, you will not approximate f(a): lim_{a-->1} f(b) \not= f(1). That is, the numbers f(.9), f(.99), f(.999), ... do not approach f(1) even though .9, .99, .999 ... do approach 1.

[loinburger]

Originally posted by Will9
As someone already said .1 repeating.

.1-repeating is equal to 1/9, so you're saying that .9-repeating equals 8/9. Plug 8/9 into a calculator and let us know what you get. (Plug 1/9 into a calculator if you don't believe that it's equal to .1-repeating.)

Originally posted by Will9
You can use 1 instead of .9 repeating, but is slightly inaccurate.

Precisely how inaccurate is it?

[Lul Thyme]

Originally posted by Drogue

The dots go above the numbers that are recurring, not afterwards

Oh really?
never seen that actually.
That's LulThyme approved then.

[Lul Thyme]

Originally posted by civman2000

What, Dedekind doesn't get a mention? He had more to do with the definition of real numbers than any of the three you named!

Cauchy constructed the real numbers the same year as Dedekind so he should also get a nod.
I say kick out Russell.

[Lul Thyme]

Originally posted by Ben Kenobi
So what have I done wrong here?

what is the lim x-> inf ( n(n)^x )where n = 0.9 repeater

If the equality is true, this should equal one, but I can't see how that is. I keep thinking the limit ought to be 0.

You keep thinking that because you assume that n=0.9... < 1
but in fact n=1 and the limit is 1.

This whole line of reasoning is not going to prove it one way or another because the value of the limit depends wheter n is 1 or not in the first place.