# If 1/9=0.1111111..., 2/9=0.2222222..., 3/9=0.3333333..., and so on, then how come 9/9≠0.9999999...?

Asked by dxs (15019) April 14th, 2013

Why does this pattern not go on? How come 9/9=1 and not 0.9999999…. Is there an error in my information?

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Because the remainder is a infinitely long repeating decimal. So when you get to 8/9 = 0.8888888 you either take it to infinity or round up the last place to 0.8888889

Now add 0.8888889 + 0.1111111 and voila, you get 1. 9/9 = 1.

By the way, if you did take each repeating decimal to infinity, then 0.8888888… → ∞ + 0.1111111… → ∞ the answer is an endless strong of 9s approaching infinity, or infinitely close to 1

ETpro (34543)

It actually does! .999999… equals 1. If you don’t believe me, I can show a proof!

Mariah (25853)

Because 1/9 = .11111119
9/9 is essentially 9 x .11111119
Which equals 1.

Another way to think of it:

Say you have a basket that holds 9 apples. If you had 8 apples in that basket, you’d have .88888889 (88.89%) of a full basket. When you have 9 apples in it, you have 1 full basket (100%)

Silence04 (4708)

@Mariah That blows my mind. Please do show a proof.

@ETpro How is that accurate if you are estimating the last digit? There never truly is a “9” anywhere in 0.8888…. What you said at the end makes more sense, where there is an endless number of 9s, which is the closest number below 1. But not 1.

dxs (15019)

x = .999999….
10x = 9.99999999…..
10x – x = 9.99999…. – .99999
9x = 9
x = 1

There are more rigorous proofs but they aren’t as intuitive.

dxs (15019)

Math is cool!

Mariah (25853)

Another way of looking at it is to consider the difference between .9999… and 1.

1 – .9 = .1, so the difference is less than .1
We can keep extending the expansion of .999… and subtracting from 1, concluding that the difference is less than .01, less than .001 and is in fact less than any number you think of, no matter how small. The only way that can be true is if 1 and .9999…. are equal.

There are math theories that treat .9999… and 1 as different. They work with infinitesimals and say that 1 and .9999… differ by an infinitesimal. When Newton developed calculus, he also thought in terms of infinitesimals. In the 19th century, math was made more rigorous and infinitesimals were banished. Then in the 20th century, mathematicians found a way of reintroducing them.

I was pretty sure that when the numerator and the denominator are non-zero integers and they are equal then the quotient is exactly 1 (Integer).

Am I incorrect?

Dr_Lawrence (19711)

You are correct, but that does not help when dealing with fractions where the numerator and denominator are infinitely large. .999999… = 9999999…./1000000….

@Mariah Call me stupid, but why would you subtract x from one side of the equation and .99999 from the other side? Where did that come from? I always learned that what you do to one side must be done to the other – so if you wanted to subtract an x, it would be:

10x – x = 9.9999… – x

And the same for subtracting .9999.

To me, the simplest explanation is that there’s virtually no difference between 0.99999999… and 1, for the reason @ETpro explained.

Because x = .99999…., so by substitution,
9.99999… – x = 9.9999…. – .999999….

@LostInParadise That’s even more confusing, because if that’s the case it should be…

10x – .99999 = 9.99999…. – .99999

Or

10x – x = 9.99999…. – x

…which is clearly not the same thing mathematically. If we know x, why would we use x as opposed to the known value? That entire thing is basically:

.99999… = 1

I get that this is the point, but it makes the rest of the equation irrelevant if it doesn’t follow the order of operations.

I was taught that when the numerator and the denominator are the same value that they’re equal to 1.

Same reason that the windshield wipers occasionally cycle in harmony with the turn signal, then go out of phase, and then after a while go back in phase.

Nature’s eternal mystery

josie (30916)

@livelaughlove21 , Here is the proof in excruciating detail.
x = .99999….
10x = 9.99999…..
10x – x = 9.9999….. – x
9x = 9.9999….. – x (equation 1)

Setting aside equation 1 for a moment,
x = .99999….
-x = -.999999…..
9.9999…. – x = 9.9999….. – .99999….
9.999…. – x = 9 (equation 2)

Now we can combine equations 1 and 2 to get
9x = 9
x = 1

@livelaughlove21 “why would you subtract x from one side of the equation and .99999 from the other side?”
@LostInParadise‘s explanation covered this question, x and .999… are defined to be equivalent, so you can substitute one in for the other.

“If we know x, why would we use x as opposed to the known value?”
Because on the lefthand side we have 10x, so to maintain the necessary form (keeping it in terms of x) we subtract in terms of x.

You’ll note that 10x – x is exactly equal to 9.999… – .999… because 10x and x are defined to be 9.999… and .999… respectively. I was just attempting to show that in a sort of condensed form up in the proof.

Mariah (25853)

@LostInParadise Now that I understand. Thanks for clearing it up. Although I should probably feel pretty dumb for needed the “excruciating detail” version, I believe any math professor I’ve ever had wouldn’t think twice before counting off for leaving the second half out of the proof. In my experience, proofs are meant to be excrutiating.

Thanks again.

Glad that worked out. There is nothing to feel dumb about. Give yourself credit for being able to pinpoint where you felt more explanation was needed. The most frustrating thing is to give an explanation and the only thing the person can say is, “I don’t get it.”

@livelaughlove21 It’s good to question “obvious” steps in a proof – great thinkers do that. To @Mariah‘s GAs I would add that so long as we agree that .999… , whatever it is, has some unique real number value, then we can name it x. Once named, we use the variable as terse shorthand for the unwieldy infinite string of digits, so x may be manipulated like any other real number – an elegant mathematical trick that works in other contexts too.

Also, there’s no ambiguity in adding:
9 + 0.999… = 9.999…
It rigidly follows the rules of the decimal system.
Likewise
10 * (.999…) = 9.999…
because multiplying by 10 is accomplished by simply shifting the decimal point, no matter what the digits.

Btw there’s a wikipedia article… on this very topic, affirming that (.999…) is another name for the number 1.

@josie …windshield wipers occasionally cycle in harmony with the turn signal, then go out of phase, and then after a while go back in phase…Nature’s eternal mystery
No mystery really. They go in & out of synch periodically, at a frequency equal to the difference between individual frequencies of wiper & turn signal, akin to beat frequency heard with two slightly different tones. Nature’s eternal harmony – at least while waiting to make a turn in the rain.

gasman (11308)

Here is an article that talks about .9999… in terms of infinitesimals. The article is much more accessible than my original reference.

If you know calculus and have ever split apart the dy and dx terms from a dy/dx expression then you are implicitly using infinitesimals. Engineers do this all the time. It was part of the motivation for creating a mathematical framework for working with them.

It seems to me that if x=0.999999 then 9x=8.9999999 and not 9.999999

Showme (7)

^ @Showme, not sure where you see a claim that 9x = 9.9999. The claim is that 10x = 9.9999, which is true (simply move the decimal point).

Mariah (25853)

Try old fashioned multiplication and see the difference. Maybe there is an inconsistency in our math system.

Showme (7)

New to your system but here is where “9x=9.99999 is shown

@livelaughlove21 , Here is the proof in excruciating detail.
x = .99999….
10x = 9.99999…..
10x – x = 9.9999….. – x
9x = 9.9999….. – x (equation 1)

Setting aside equation 1 for a moment,
x = .99999….
-x = -.999999…..
9.9999…. – x = 9.9999….. – .99999….
9.999…. – x = 9 (equation 2)

Now we can combine equations 1 and 2 to get
9x = 9
x = 1

Showme (7)

@LostInParadise is saying that 9x=9.9999… minus x.
9x = 9.9999….. – x

dxs (15019)

So 9x=9.99999…-0.99999… or 9 if you multiply 9 by x (or 0,9999….) you get 8.9999…. now take x (or 0.9999….) away it comes out 8.0

Showme (7)

@Showme Yes… and taking x away from 9x is 8x, and dividing by 8 you get x=1 again. Which is what we were trying to prove.

Mariah (25853)

My question is still why is 9x-x (by moving the decimal point) apparently is not equal to 9x-x (being multiplied out)? {assuming x=0.9999….)

Showme (7)

Well the proof shows that .999 = 1. So 8.999 = 9.

Mariah (25853)

Still from Lost in paradise equation 1:
x = .99999….
10x = 9.99999…..
10x – x = 9.9999….. – x
9x = 9.9999….. – x (equation 1)
(9x = 9)
BUT multiply x (0.9999…) by 9, you get 8.9999.., then take away x (0.9999….) & you get 8
WHY the difference?

Showme (7)

Nobody is saying that you should take away x from 9x at any step. In LostInParadise’s explanation he takes x from 10x to get 9x, but we never subtract x from 9x. I’m not sure where you think you’re seeing this.

It makes sense that you would get 8 from 9x – x because that reduces to 8x = 8, or x = 1.

Mariah (25853)

I bow to the expert, but am uncomfortable with the idea that 0.99999….. is equal to one, should be a way to express it as rounded. Thank you

Showme (7)

Lots of concepts in math are uncomfortable and unintuitive. That’s why we need proofs to show us that our instincts are wrong! :)

Mariah (25853)

@Showme I was uncomfortable with it, too, and that’s why I made @Mariah prove it. And now I’m in awe of the amazing fact that 0.999…=1 .

dxs (15019)

A lot of people have trouble accepting that .999… is the same as 1. As I mentioned earlier, there is a branch of mathematics, called non-standard analysis, that treats the numbers as different. Non-standard analysis defines quantities called infintesimals that are infinitely small. In non-standard analysis, .999… and 1 differ by an infintesimal. I have not studied non-standard analysis and, as far as I know, it has not caught on very much.

How far out must 0,999 go for it to be equal to one , is 0.99 equal to one??

Showme (7)

No, only .999…. repeating infinitely.

Mariah (25853)

Thanks again for the info. How is 0.9999…... used in math, does it serve a useful purpose in solving equations? Thanks again, been a long ti.me since having fun with math

Showme (7)

It depends on what you mean by useful. For theoretical math, it is convenient to be able to replace an infinitely long sequence of digits by 1. For practical purposes, there is a limited number of significant digits that can be measured, certainly fewer than 20, so a number with infinitely many digits would never arise in the real world. If you had .999999 and only needed 4 places, you would just round up to 1.

Don’t like to assume, but can I assume that there is no fraction that can show 0.9999…. as l/9,2/9….... 8/9 , then in ordinary division it does not exist.

Showme (7)

@Showme That fraction is 9/9. 9/9=1/1=0.9999….=1

dxs (15019)

@Showme A repeating decimal is a rational number, which can, by definition, be shown as a fraction. See @dxs‘s answer.

Mariah (25853)

Somehow you will make me a believer, but am not quite there yet. Still not too old to learn and will sleep on this, again thank you.

Showme (7)

Love your curiosity! If there’s anything in particular that I can explain another way, let me know.

Mariah (25853)

Mariah, Pi is an irrational number (considered by some), but does it become rational if you divide it by 10. It fits the definition is a repeating rational number (I think)

Showme (7)

Found one answer—- Pi is a non-repeating decimal number , dividing it by 10 would not change its class. Learned a little

Showme (7)

Your self-correction is right. Simply moving the decimal point (which is what dividing by 10 does) will not cause repetition to appear where there is none. Pi is an irrational number.

Mariah (25853)

My head is going to esplode!

Dutchess_III (41921)

Math is fun!

dxs (15019)

It’s wonderful! It’s almost like magic.

Dutchess_III (41921)

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