.9 repeated = 1?

Does .9 repeated = 1? Here's the logic:

1/9 = .1 repeated
2/9 = .2 repeated
3/9 = .3 repeated

...

8/9 = .8 repeated
9/9 = ?

everything from 1-8 over nine = that number repeated. So when you get to 9/9 (1), it should be .9 repeated, assuming the pattern continues.

So what's up with this? .9 repeated CANNOT equal 1, but does the pattern just magicaly dissapear?

do a search....this debates already gone on

Quote:

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Originally posted by Flip
.9 repeated CANNOT equal 1

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Here we go again...
:p

YES! My legacy lives on :) :D :cool:


http://maya.effectonline.com/~vbforu...ghlight=.99999

What flip said earlier was pretty cool!

Quote:

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Originally posted by SilverSprite
What flip said earlier was pretty cool!

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Thats extremely interesting, but I tend to disagree since it seems to amount to a limited viewpoint upon the reality of the situation.
But, Mr. Sprite, I can see how you would think that such statements are pretty cool, with your track record.

:D
just taking the piss outa you

Ouch NotLKH! I'm sorry if i'm not a 4 time university graduate like you guys. I'm only 15. I just happen to think it interesting.

And you still havent told me what you meant about the similar triangles in the winding function thread.

Hey NotLKH! How old are you? Do you go to school? And if you do where?

lol, what winding function thread? :D

9/9 = 1

Moron.

From that page:

Quote:

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Prove that 0.9999(recuring) = 1

Here goes:
let x = 0.9999(recuring)
then 10x = 9.9999(recuring)
then 10x-x = 9.9999(recuring) - x
then 9x = 9
then x = 1

Well....?

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Infinity is an idea, repeated 9's would be an infinty. Infinity cannot be expressed as a number.

x=.9
10x=9
10x-x=9-.9
9x=8.1
x=.9

Your flaw is this there is no such thing as infinity, when multiplied by 10 the length of the numbers would remain constant, like .9999 becomes 9.999. Where as you'd have .9999 = 9.9999. You are inventing numbers. This is unaccurate(ever hear of significant digits?). As such your result is not accurate.

It's as simple as that. You're wrong. Precise, sure! Accurate? Nope!

x=.9
10x=9.9
10x-x=9.9-.9
9x=9
x=1

When I invent numbers I can make it = 1 too.

x=.9
10x=9.99
10x-x=9.99-.9
9x=9.09
x=1.01

I can also make it greater than 1. I'm not accurate, but I'm precise.

Yeah, except .(9) * 10 = 9.(9) whereas .9 * 10 != 9.9 or 9.99

Infinity is not a number, therefore when you have an infinite number of digits you can't use the standard rules, which are a special case of mathematical laws for when there are a finite number of digits. You can't have "1 less than infinity" number of digits precisely because infinity is not a number.

And "there is no such thing as infinity". Wow. The ignorance there is unbelievable.

By that I meant infinity is simply an idea, it is not applicable to reality.

There is nothing in this universe which is infinite except space. Space being the vacum of our universe.

Everything in life is finite. In no case would you have .9999... recuring for ever, that is an accuracy which is impossible.

So, while the solution is technically correct, the applicability of it is irrelevent.

Unless you can have absolute infinite accuracy. Which is an oxymoron.

An arguement for the equivalence of 1 and 0.9 recurring is as someone said:

Here goes:
let x = 0.9999(recuring)
then 10x = 9.9999(recuring)
then 10x-x = 9.9999(recuring) - x
then 9x = 9
then x = 1

A more shifty little argument though:

Given a positive real number x, find the positive square root of it (calling the positive square root Root(x)). Examining the behavior of Root(x) with different groups of values for x, it seems that:
If x > 1, Root(x) > x > 1
If x = 1, Root(x) = x = 1
If x < 1, Root(x) < x < 1

Now let's say that Root(x) = 0.9 recurring. It's pretty safe to say that x < 0.9 recurring, but we just defined x as being between Root(x) and 1, therefore:
0.9 recurring < x < 1

This pretty much proves that 0.9 recurring = 1.

Quote:

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Originally posted by DiGiTaIErRoR
9/9 = 1

Moron.

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Moron? you got some guts coming here and calling other people morons! apparently, after reading your post, i think the moron here is you. I guess pi and e will no longer have infinite amount of decimals, its gotta be terminating right?

1/3 no longer becomes 0.33333..., because according to you there is no infinity, so 1/3 actually = 0.333...3333

LOOK AT WHAT YOU ARE SAYING!!!!!!!

pfft and he calls himself a mathematician!lol(joking around)

Infinity is an idea, which only means it is NOT a number.
However, it CAN be used when working with numbers.
0.9 repeating=1
It's a known fact.
Live with it.

YA! BELY DAT!lol

.9 repating may be similar to 1 but it is not one.

There is no real-world application where .9 repeating can exist. It would be infinite accuracy, there's no such thing. All real-world measures will terminate, if you go far enough. ;)

So in the real-world .9 repeating != 1. Maybe in Happy Land on Lolli-pop Lane .9 repeating will = 1.

Consider this, what would you add to .9 repeating to get 1? By your definition it would be 0. It would really be .0 repeating then a 1, but can you define a 1 after infinite 0's? Oh yeah! That infinity plus one stuff you used to use when arguing with your sibblings.

x=.0...1
y=.999...
x+y=1
10x=.0...1
10x-x= .0...1-.0...1
9x=0
x=0
x+y=.999...

so mr digitalerror here claims there there is an end to inifinity as he added a 1 in the end of the infinite 0s to get 0.000...1

so apparently he recognizes inifinity as a finite number. does anyone want to tell him whats wrong with that?

and yet strangely, he himself proved that x+y=1=0.999...

...

I do! Mr. DigitalError, your wrong with that!

I'm just pointing out the error in .999=1.

Your right about one thing. .999 doesnt = 1. but .9(recurring)does!

In fantasy land.

:rolleyes:

Maybe you've got things in reverse. This is the real life, not 'fantasy land'. And that place in your head, thats fantasy land.

.999... recuring is not a 'real' number. It can only be imagined.

ok and.... therefore it equals 1

it does not equal 1, it is similar to 1

you cannot have a real number = infinity

.999... repeating is like saying .9*infinity

1*infinity=27*infinity

By your logic I can eliminate infinity.

so 1=27

as 9.999... - .999... you're basicly subtracting infinity from infinity. Since repeating 9's cannot be expressed, as infinity. Repeating 9's is an idea, like infinity.

So .999 <> 1. If you believe so you're a moron.

what in the name of hell is 27*infinity? That has nothing to do with .9(recurring). Your saying that 27*infinity is the same as .9*infinity? The only moron here is you. Who comes into a forum calling people morons? Since youve started it i'm gonna call you one too. YA MORON

Express .9 repeating as a real number.

I have nothing more to say to you. I dont talk to moronic people like you! MORON!(9/9)

You're the moron. You can't prove me wrong.

Shut up Moron!

Have you ever heard of checking your solution?

let x = 0.9999(recuring)
then 10x = 9.9999(recuring)
then 10x-x = 9.9999(recuring) - x
then 9x = 9
then x = 1

put your x back in the original equation

1=.9999(recuring)

The solution doesn't check therefore 1 is not in the solution set of x.

Moron.

I dont talk to morons. I will have nothing more to do with you!!!MORON MORON MORON thats infinite

Just like I an have:

let x=2
then x^2=4
then -2^2=4
so x=-2

therefore 2=-2

I can be a moron like you too!

right "0.999... doesn't exist you can't prove me wrong!" whats your reason? infinite accuracy doesn't exist?

well newsflash, it does. if you are in a room with 5 other people, you'd say that there are EXACTLY 6 people in the room, and the number 6 is INFINITELY ACCURATE. --- we can't have 5 and 3/4 people, we can't have 6 and 1/7 people! and apprently you think infinity doesn't exist! let me ask you something, what is infinity?

do you even know what you talk about before you say it?? I've had it with you and your attitude. I see that the biggest moron here is you!

You're confusing infinite with absolute.

an absolute value has infinite accuracy.

DigitalError previously stated the following:

Have you ever heard of checking your solution?

let x = 0.9999(recuring)
then 10x = 9.9999(recuring)
then 10x-x = 9.9999(recuring) - x
then 9x = 9
then x = 1

put your x back in the original equation

1=.9999(recuring)

The solution doesn't check therefore 1 is not in the solution set of x.

However, the final conclusion that 1 is not in the solution set of x is false. There is no proof that 1 cannot=0.9 recurring, which means that you cannot put in the last statement and claim it valid. It is simply your word against ours, except for that fact that we have proof and you don't.

And that your a moron!

It's not my fault your brain cannot fanthom the infinitely small difference between .999 repeating and 1.

Anyone know of a fraction representing .999 repeating?

Infinity doesn't measure accuracy.

Absolute can.

Illiteracy is a big problem is this society.

Come on guys, can you see that Digitai is just jerking you? =).

The trick is to ignore until he posts something useful =).

Z.

Quote:

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Originally posted by DiGiTaIErRoR
Express .9 repeating as a real number.

---

1
:D

marnitzg knows what hes talking about!

Alright then, if 0.9(recurring) does NOT equal 1, explain this:

Given a positive real number x, find the positive square root of it (calling the positive square root Root(x)). Examining the behavior of Root(x) with different groups of values for x, it seems that:
If x > 1, x > Root(x) > 1
If x = 1, x = Root(x) = 1
If x < 1, x < Root(x) < 1
Let x = 0.9(recurring), the inequality must be satisfied:
0.9(recurring) < Root(x) < 1, so please do tell - exactly which real number (n) is between 0.9(recurring) and 1 such that n² = 0.9(recurring)???

Quite a nice example of proof by contradiction.

Quote:

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Originally posted by marnitzg
1
:D

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Close enough.

:rolleyes:

I will admit for any practical purpose .999... = 1.

But the truth is, it doesn't.

The difference is infininty small. But there is still a difference!

Express .999... as a fraction with result that will actually give you .999... no 1/1 stuff!

Quote:

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Originally posted by A$$Bandit
Alright then, if 0.9(recurring) does NOT equal 1, explain this:

Given a positive real number x, find the positive square root of it (calling the positive square root Root(x)). Examining the behavior of Root(x) with different groups of values for x, it seems that:
If x > 1, x > Root(x) > 1
If x = 1, x = Root(x) = 1
If x < 1, x < Root(x) < 1
Let x = 0.9(recurring), the inequality must be satisfied:
0.9(recurring) < Root(x) < 1, so please do tell - exactly which real number (n) is between 0.9(recurring) and 1 such that n² = 0.9(recurring)???

Quite a nice example of proof by contradiction.

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There would be nothing between .999... and 1.

Well, 1/9 = .111... and 8/9 = .888... then 1/9+8/9=9/9=1 but .111 and .888... = .999...

Inaccuracy of infinite.

I agree that our decimal system is incomplete in it's current form.

Hey Digital
you say
"The difference is infininty small."
and...
"Infinity doesn't measure accuracy"
I have the feeling those two contradict each other but oh well...

another thing you say is
0.8(recurring) + 0.1(recurring) = 0.9(recurring)
alright given this as a fact what can you say against this?

Code:

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    1/3 = 0.333(recurring)
    3/3 = 1
    3*1/3= 3/3 = 0.333(recurring) + 0.333(recurring) + 0.333(recurring)
    =0.990(recurring)=1

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How can you argument against this?

Well you can say that 1/3 is not 0.3(recurring), but you're gonna have to come up with some mathematical proof of it (i.e. not just saying "oh by the way infinite accuracy doesn't exist").

well I really can't discuss that with you to good here, since I took math in another language so my english math vocabulary is kind of limited... well for the 0.333...

isn't that proven?
I mean
1/3=0 remainder 1
so 10/3 = 3 remainder 1

shouldnt that be enough of a proof... I mean it shows that it't "perdiodic" right from the first step on... there can't be any change to that no matter how far you go....

Quote:

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=0.990(recurring)=1

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it's not .990...

/\/\isanThr0p, You're otherwise incoherent.

You're dealing with infinity. Infinty - infinity is undefined. So you're trying to define a 'close' solution for an otherwise undefinable one. Since you cannot define infinity into the real number system. It's an idea, that's all, a concept, if you will.

9.999 infinite...
- .999 infinite
= 9.? undefined

haha
well following my argumentation that's really easy

10-1=9

well I get what you are saying... you are probably a really smart person I really don't want to offend you, but your argumentation is what you hear from highschool kids a lot...

I am really no big mathwiz so I wont go on about this... I stated what I think, I get what you are saying even though I think you are wrong...

math is a system of axioms and I believe what I stated is accepted as true, I can't say any more about it...

I am a highschool student. What does that prove? You can discriminate against me using my age?

My questions still remains. What do you add to .999... recuring to get 1?

.999... + 0 = 1

You could say .999... + .000... = 1, but an infinite number of recurring zero's is zero.

So, 1 - 0 = .999...

-Lou

Quote:

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Originally posted by NotLKH
.999... + 0 = 1

You could say .999... + .000... = 1, but an infinite number of recurring zero's is zero.

So, 1 - 0 = .999...

-Lou

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The argument only gets more moronic.