Does 0.9999 repeating 9 equal 1?

Does 0.9999 repeating 9 equal 1?


as i see it 0.3 repeating 3 is one-third, right?
and 0.6 repeating 6 is two-thirds, right?

what is 0.3 repeating 3 plus 0.6 repeating 6? 0.9 repeating 9.
what is one third plus two thirds? 1.

no rounding. no calculus limits involved. 0.9 repeating 9 equals 1?

0.3333 is not exactly 1/3, but it's as close as we can get in decimal

It's like how if you looked at .8 repeating + .1 repeating most would assume that equaled .9 repeating...

But that is actually 8/9 plus 1/9 = 1

So no, .9 repeating != 1

Mrkai wrote:

no rounding. no calculus limits involved. 0.9 repeating 9 equals 1?

Sorry, but you are rounding and that's where you introduce the error.

Is a stop sign a circle?

(The answer is yes with the correct definition of a circle.)


For almost all applications the percent error by rounding .9repeating to 1 is negligible relative to the error caused by other factors.

If you want to get technical, your statements

Mrkai wrote:

as i see it 0.3 repeating 3 is one-third, right? and 0.6 repeating 6 is two-thirds, right?

are inaccurate. Thus making your question inaccurate.

corvettecrazy wrote:

Is a stop sign a circle? (The answer is yes with the correct definition of a circle.)

Not saying you're wrong but what definition would this be? It certainly doesn't fit the equation for a circle.

The plane will take off.

clutch1 wrote:

The plane will take off.

oh god not this debate again

I say they are equal. There's a funner geometric series proof, but I don't have time to post it now, so I'll update later. Here's one though I really like.

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

so 1 = .9 repeating


so try to find my fault in that^^^ theres isnt one.

.3 repeating is not 1/3. We just say it is one third as it is extremely close and easier to say.

.9 repeating is extremely close to 1. I don't understand how you can not understand something so simple.

Mrkai wrote:

I say they are equal. There's a funner geometric series proof, but I don't have time to post it now, so I'll update later. Here's one though I really like. x = .9 repeating 10x = 9.9 repeating 10x -x = 9.9 repeating - .9 repeating 9x = 9 x = 1 so 1 = .9 repeating so try to find my fault in that^^^ theres isnt one.

ummmm,
1 doesn't equal anything but 1
there is a fault.
you place the given in for x and it already solved you can't go on from there.

x = .9 repeating
10x = 9.9 repeating
10*.9 repeating=9.9 repeating
9.9 repeating = 9.9 repeating
MATH OWNED

if two numbers are not equal, there is a third number that is also unequal and that can fit in between them on the number line. Regardless of the type of real number or the difficulty in computing its value or representing its value ,there does exist a number that is larger than one but smaller than the other.
x<y<z

0.9999<y<1

If 0.9999... does not equal 1, then what number, y, could exist in between them such that math?

Since there is no conceivable number that can exist in between the two, they must be equal according to the definition of the real numbers as a continuum.

http://math.wikia.com/wiki/Proof:The_Decimal_0.999..._is_Equivalent_to_1

too bad 9.9 repeating doesn't equal 1 so its wrong.

Mrkai wrote:

funner

funner!? really? i don think im gonna take any advice that comes from the person that says funner.. lol and not to mention you could plug any number into that variable and it is true! 10 times any x then subtracting x will always give you 1.

Mrkai wrote:

if two numbers are not equal, there is a third number that is also unequal and that can fit in between them on the number line. Regardless of the type of real number or the difficulty in computing its value or representing its value ,there does exist a number that is larger than one but smaller than the other. x<y<z 0.9999<y<1 If 0.9999... does not equal 1, then what number, y, could exist in between them such that math? Since there is no conceivable number that can exist in between the two, they must be equal according to the definition of the real numbers as a continuum.

this is true but there is a number between .9 repeating and 1. it just is not a real number but it does exist its just a figure of rounding.

harris_23 wrote:

Mrkai wrote: funner funner!? really? i don think im gonna take any advice that comes from the person that says funner.. lol and not to mention you could plug any number into that variable and it is true! 10 times any x then subtracting x will always give you 1.

but he did the problem wrong, he didn't use order of operations.

^^^ obviously lol but i was just goin off of what he had posted.

Mrkai wrote:

I say they are equal. There's a funner geometric series proof, but I don't have time to post it now, so I'll update later. Here's one though I really like. x = .9 repeating 10x = 9.9 repeating 10x -x = 9.9 repeating - .9 repeating 9x = 9 x = 1 so 1 = .9 repeating so try to find my fault in that^^^ theres isnt one.

You're wrong in the second step. Your error comes in the basic understanding of rational and irrational numbers and how you can use arithmetic operations.

Without and approximation or rounding ever so slightly, you can NOT use an arithmetic operation (+, -, /, x, ^, etc.) on an irrational number. When you do use an arithmetic operation on an irrational number, you approximate the answer. Just as you can not multiply Pi by 10 w/out a little error, you can not do:
10*0.9 repeating

You are taking an irrational number, temporarily making it rational so you can multiply it by ten, and then rounding and making it irrational again.

The same holds true w/ the original post. You can not perform the operation 0.3 repeating plus 0.6 repeating w/out an approximation. 1/3 is a RATIONAL number and you can perform arithmetic operations w/ it. 0.3 repeating is an IRRATIONAL number and you can NOT perform arithmetic operations w/out approximations.

*EDIT* The "number" that would fall between 0.9 repeating and 1 would be:

0.9repeating + [0.1*10^(-infinity)]/2

But you can not compute that number because you can not use arithmetic operations on irrational numbers (ie, 0.9 repeating and -infinity). You can calculate the limit of the equation but that's just an approximation and not an exact number.

harris_23 wrote:

lol and not to mention you could plug any number into that variable and it is true! 10 times any x then subtracting x will always give you 1.

Let x = 3
10x = 10*3
10x-x = 10*3-3
9x = 27
x = 3

I fail to see how that give me 1.

Mrkai, if you have sat through any math or science type of class you would know that no matter what 0.9999999999999999999999999999999999 repeating is not equal to 1....If you were to take the limit as the variable or whatever tends towards infinity then it would come very close and most people might just say that it is 1 but they all are wrong....there is no way possible for 0.9999 repeating to be equal to 1.....At some point you have to round your number to get it equal to 1.....if you set up an equation like the ones on wiki about does 0.999 equal 1 then they say yes and give proofs, but i still say that it does not equal 1.....

jdusty914 wrote:

Mrkai, if you have sat through any math or science type of class you would know that no matter what 0.9999999999999999999999999999999999 repeating is not equal to 1....If you were to take the limit as the variable or whatever tends towards infinity then it would come very close and most people might just say that it is 1 but they all are wrong....there is no way possible for 0.9999 repeating to be equal to 1.....At some point you have to round your number to get it equal to 1.....if you set up an equation like the ones on wiki about does 0.999 equal 1 then they say yes and give proofs, but i still say that it does not equal 1.....

Limit of the infinite series DOES equal 1. The limit is not the answer, the limit is the number that the infinite series approaches and will come infinitely close to (but never technically reach).

ie. the lim(1/n) as n -> infinity = 0
In reality, no matter how large "s" becomes 1/n will never equal zero. It will approach zero, but it can never become zero w/out an approximation. But, the lim(1/n) as n -> infinity does equal zero.

The same thing is true if you take the sum/limit of the infinite series for 0.9 repeating. The sum of the infinite series will come to 0.9 repeating but the limit of the series as "n" approaches infinity does equal 1.

Skimming through this thread made me feel like this,

that picture fits perfect.

but 0.9999 repeating IS a limit.

0.9999 ... = %u03A3 (n=1 to k) 9/10^n (where k is the number of 9's)

0.9999 repeating is the limit of that series as k approaches infinity.

lim (k->%u221E) %u03A3 (n=1 to k) 9/10^n = 1

Therefore, 0.9999 repeating equals 1.

Mrkai wrote:

but 0.9999 repeating IS a limit. 0.9999 ... = %u03A3 (n=1 to k) 9/10^n (where k is the number of 9's) 0.9999 repeating is the limit of that series as k approaches infinity. lim (k->%u221E) %u03A3 (n=1 to k) 9/10^n = 1 Therefore, 0.9999 repeating equals 1.

A limit is NEVER going to be an irrational number. A limit will always be a rational number (hence the purpose of a limit).

I do know the equation you posted. It's the infinite series for 0.9 repeating. The ANSWER to that equation is 0.9 repeating. The LIMIT of that equation is 1. They are two seperate things. The answer is the exact number of the sumation when it's carried out to infinity. The limit of the equation is the number that the sumation approaches. If you continue to add them together as n approaches infinity, it will get ever closer to 1 but will never actually reach the number 1.

You can see my earlier post where I showed you an equation for the number that would fall between 0.9 repeating and 1. But you can not calculate it as 0.9 repeating is an irrational number.

Obviously you don't know what you're talking about. You seem to be Google searching and posting what other people say (or else your "equations" would be more than a series of symbols created from copying and pasting). I'm not going to argue w/ you about it b/c one can not prove a point to someone who doesn't have the basic subject knowledge to fully understand the subject at hand. You MUST understand that you can NOT perform arithmetic operations to an irrational number w/out approximating.

Mrkai is right.

This has long been accepted by the mathematical community.

.9 repeating denotes the same real number as 1

This reminds me of the brain teaser where your hand is a foot away from the door knob and so you reach for the knob by halving the distance between your hand and the knob, then halving it again, then again, and so on ad infinitum, never actually touching the knob .

But then someone suggests this is ample reason to assume door knobs can never be touched .

The halving of the distances between the hand and the knob is like the continuation of the .9 number, its purely theoretical . Its doesn't practically manifest itself in the 'real' world . In the real world we just open the darn door . Just as in the real world .9999.. is 1 . But since your asking the question as a theoretical exercise, I side with PwrRngr in that the real-world implications don't matter . Its never strictly identical to 1 in terms of pure mathematical theory . As he said, at some point during each of the proofs offered the number leaves the realm of pure mathematics and becomes 'real', which is what (falsely) gives the result of 1 .

MrKai, and Golfer are idiot.
nothing equals one except 1
1=1 thats it
324589i05243 doesn't equal 1 because i say so.

1/3 = 0.33333
0.33333 * 3 = 0.99999
1/3 * 3 = 1

Therefore 0.99999 = 1

Problem solved.

Brandon wrote:

1/3 = 0.33333 0.33333 * 3 = 0.99999 1/3 * 3 = 1 Therefore 0.99999 = 1 Problem solved.

Again, applying arithmetic operations to an irrational number results in an approximation. You can not multiply 0.3 repeating w/out introducing error.

1/3 is a RATIONAL number. 0.3 repeating is an IRRATIONAL number that's an approximation of the rational number, 1/3. They are not exactly equal in a theoretical sense.

I'm just using logic. If we assert that A = B and B = C, then A = C. If you say that 1/3 does not equal 0.3333 because its irrational, then this thread is worthless because 0.99999 is irrelevant.

Brandon wrote:

I'm just using logic. If we assert that A = B and B = C, then A = C. If you say that 1/3 does not equal 0.3333 because its irrational, then this thread is worthless because 0.99999 is irrelevant.

you dont use logic in math, you use math in math.
[/b]

Brandon wrote:

I'm just using logic. If we assert that A = B and B = C, then A = C. If you say that 1/3 does not equal 0.3333 because its irrational, then this thread is worthless because 0.99999 is irrelevant.

In the theoretical sense of the "proof" at hand, 1/3 does not equal 0.3 repeating. This is where the error comes into the problem and why the proof doesn't work.

Apparently I was wrong, we DO have smart ppl on Oznium..

Me not included.

justinwebb wrote:

clutch1 wrote: The plane will take off. oh god not this debate again

Hahaha

clutch1 wrote:

justinwebb wrote: clutch1 wrote: The plane will take off. oh god not this debate again Hahaha

vagina.

Suppose .9999999999 (repeating) is not equal to 1.

Then these are two distinct numbers, call them x and z. Then there exists a y such that x + y = z. i.e. (by the additive inverse property). So since x = .99999999 repeating, y = .00000000000 repeating and finally a 1. How many 0's are there?

If there are finitely many 0's (say n) then y = 10^-n. This contradicts that x has infinitely many nonzero digits since x + 10^-n = z

So y must have an infinite number of 0's before the 1. But if y has an infinite number of 0's then y = 0. Then it follows that x = y, i.e. .999999999 (repeating) equals 1.

the is no such thing as repeating zero. Idiot. Do you seriously beleive this ****?

Mrkai wrote:

Suppose .9999999999 (repeating) is not equal to 1. Then these are two distinct numbers, call them x and z. Then there exists a y such that x + y = z. i.e. (by the additive inverse property). So since x = .99999999 repeating, y = .00000000000 repeating and finally a 1. How many 0's are there? If there are finitely many 0's (say n) then y = 10^-n. This contradicts that x has infinitely many nonzero digits since x + 10^-n = z So y must have an infinite number of 0's before the 1. But if y has an infinite number of 0's then y = 0. Then it follows that x = y, i.e. .999999999 (repeating) equals 1.

You're still thinking finite. You simply don't have a good enough understanding to comprehend the problem. I gave you the equation for the number in between .9 repeating and 1. Ignore it or try to figure it out, it's up to you. This should throw your brain in a loop.

Suppose I add up all the prime numbers from 1 to infinity. The total of all these prime numbers will be infinity. Now, suppose I add up ALL the numbers from 1 to infinity. The total of all these numbers is also infinity. But, the infinity from adding up all the numbers is greater than the infinity from adding up all the prime numbers. How can this be? Does infinity not equal infinity?

ViciousBread wrote:

the is no such thing as repeating zero. Idiot. Do you seriously beleive this ****?

thats is my point

PwrRngr wrote:

Suppose I add up all the prime numbers from 1 to infinity. The total of all these prime numbers will be infinity. Now, suppose I add up ALL the numbers from 1 to infinity. The total of all these numbers is also infinity. But, the infinity from adding up all the numbers is greater than the infinity from adding up all the prime numbers. How can this be? Does infinity not equal infinity?

true.