# proof that 0.999... =1 how can this be true??i dont understand how this can be true

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There are a few different proofs for this. I think the easiest to understand is a simple algebraic proof:

Let x = 0.99...

10x = 9.99...

10x - x = 9.99... - 0.99... = 9

9x = 9

x = 1

For more in-depth info and other proofs, check out http://en.wikipedia.org/wiki/0.999...

Try adding 1/3+1/3+1/3. In fraction form this is obviously 1. In decimal form you have .33333...+.33333...+.33333...=.99999...=1

This is the first time I have been exposed to the concept that 0.99999... = 1.

The proofs given are hard to argue with.

It makes some sense to me if I consider the chain of 0.999999999....etc. going on to infinity. As one approaches infinity, the difference between the .999999...number and 1 approaches zero, and they become essentially identical. This is all hard to swallow, and I am wary that there is some sort of mathematical trickery going on in the proofs given that I have failed to recognize. Very interesting, though!

I just figured out the fallacy of the proof given by number 2. It is incorrect as follows:

if you let x = 0.999

then 10x = 9.99 (not 9.999)

10x - x = 9.99-0.999 = 8.991

9x = 8.991

x = 8.991 divided by 9 = 0.999

so the fallacy is created by using 0.999... as if the number of decimal places is irrelevant, which it is not. In other words, 0.999 is not the same number as 0.9999, nor is it the same as 0.999999999999....

If you obey standard rules of mathematics in the calculations, and don't change the number of decimal places, then the "proof" that 0.99999...= 1 won't work.

It does make a good party trick, though, and would fool most people most of the time.

I just figured out the fallacy of the proof given by number 2. It is incorrect as follows:

if you let x = 0.999

then 10x = 9.99 (not 9.999)

10x - x = 9.99-0.999 = 8.991

9x = 8.991

x = 8.991 divided by 9 = 0.999

so the fallacy is created by using 0.999... as if the number of decimal places is irrelevant, which it is not. In other words, 0.999 is not the same number as 0.9999, nor is it the same as 0.999999999999....

If you obey standard rules of mathematics in the calculations, and don't change the number of decimal places, then the "proof" that 0.99999...= 1 won't work.

It does make a good party trick, though, and would fool most people most of the time.

The proof I gave in #2 is correct.

When multiplying a number by 10, the digits do not change, but move one place to the left. Your analysis is correct for 0.999, but we are talking about an infinite number of decimal places, not three decimal places. Shifting an infinite number of digits one place to the left still leaves an infinite number of digits after the decimal. This is explained also at http://en.wikipedia.org/wiki/0.999...#Algebraic_proofs.

I disgree. Where is it shown that one can do standard calculations making use of an infinite number of decimal places? Moreover, where is it proven that one can use such calculations on a number with an infinite number of decimal places, and subsequently on a number with a number of decimal places that is infinity plus one? This is the reason the impossible "proof" seems to work.

Your proof should be written as follows:

Let x = 0.99

10 x = 9.9

10x - x = 9.9 - 0.99 = 8.91

9x = 8.91

x = 0.99

I don't think it is correct to use a number like "0.99..." in calculations.

I disgree. Where is it shown that one can do standard calculations making use of an infinite number of decimal places? Moreover, where is it proven that one can use such calculations on a number with an infinite number of decimal places, and subsequently on a number with a number of decimal places that is infinity plus one? This is the reason the impossible "proof" seems to work.

Your proof should be written as follows:

Let x = 0.99

10 x = 9.9

10x - x = 9.9 - 0.99 = 8.91

9x = 8.91

x = 0.99

I don't think it is correct to use a number like "0.99..." in calculations.

I think the misconception made is that there is a "last 9". There is no final 9--there is an infinite number of repeating 9s. Infinity plus one is still infinity.

It is easy to see how multiplying by 10 shifts the decimal place. Let's consider a less ambiguous repeating decimal: 1/3.

1/3 = 0.333...

1/3 * 10 = 10/3 = 3.333...

There isn't a different number of 3s after the decimal place in the decimal representation of 1/3 and 10/3 -- they are both infinite.

I hope this example helps clear up the steps in the proof and why they are valid. If this proof still troubles you, perhaps the example offered it #3 is easier to understand.

Some related reading that may help clear things up as well:

http://en.wikipedia.org/wiki/Decimal_representation

http://en.wikipedia.org/wiki/Repeating_decimal

http://en.wikipedia.org/wiki/0.999...#Analytic_proofs

The proof using infinite series may give more insight into "why" it is true than the algebraic approach I offered.

I think I understand your approach. And the 1/3 example comes closer to clarifying that 1/3 and 10(1/3) both have an infinite number of 3's after the decimal point.

My concern is that I don't believe infinity and infinity plus one are the same number. In this regard, and more related to these discussions, I don't think 0.99 is the same as 0.999. 0.99 is equal to 0.990, which is different from 0.999. So what if we had one number with an infinite number of 3's after a decimal, and another number with an infinite number of 3's PLUS a zero. If you look at the last three digits of these numbers, one would end in 333, and the other would end in 330. Clearly, they aren't the same number. Doesn't this make sense?

The idea infinity and infinitely plus one being the same quantity is very counterintuitive. This idea is actually known as "Hilbert's paradox of the Grand Hotel." The page on Wikipedia does a good job of explaining it: http://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel

(Note: I find this topic to be really interesting. I can try to come up with some more examples illustrating this if you're interested.)

When we talk about a repeating decimal, such as 0.99..., the nines repeat forever. The fallacy is that there is a last digit. In fact there is no last digit--they go on forever. So even when we multiply by 10 (thus removing one of the nines), there is still nines repeating forever, without a final zero.

Your reasoning makes perfect sense, and illustrates that we could not claim that a nine with any finite number of zeros after the decimal place is equal to one--we can say they are equal *only* when there are infinite number of nines, as in 0.999....

I disagree with 0.999... being 1 It will be "almost" identical, but, really, even just by looking at it, it does not make sense (at least when i first looked at it, I had no idea why it would be true). However, by reading the responses, I was mind-blown of how depth it went in. But I thought about this for "a while", because the topic was preventing me from doing anything else. And I tried to prove it with grade 11 brain...

So, let x represent 0.999...

x=0.999...

10x = 10(0.999...)

10x-x = 10(0.999...) - 0.999...

9x = 9(0.999...)

x = 0.999...

Therefore, x is not 1.

Here, I did not change the value of the infinity. Only 1 value of infinity exists, unlike 0.999... and 9.99... used in previous responses.

And this could be random thing, but if there were infinite number like 0.999...., and if you multiply it by 10, it would be,

0.999... X 10 = 9.99...0 (this is totally random but if you treat the value of infinity as "...", it makes sense.).

Maybe if you're not a big fan of decimals, you might be a fan of fractions!

I think we can all agree about the results of infinite geometric series. We can rewrite 0.99... as the following:

0.9999... = 9*(1/10 + 1/100 + 1/1000 + ...)

We can now add -1 and 1 to the inside of the parentheses:

= 9*(-1 + 1 + 1/10 + 1/100 + 1/1000 + ...)

Now, we can apply the formula of an infinite geometry series, that given a fraction, 0<r<1, the following holds true:

1 + r + r^2 + r^3 + ... = 1/(1-r)

We can apply the same reasoning to the 1 + 1/10 + 1/100 + 1/1000... on the inside of the parentheses (here r = 1/10):

= 9 * (-1 + 1/(1-1/10)) = 9*(-1 + 10/9) = 9*(1/9) = 1

I like that we have some dissenters in the group. I'll throw in a couple observations, which are similar to previous points but may help someone nonetheless:

1/9 = 0.1111...

2/9 = 0.2222...

3/9 = 0.3333...

It's not that 0.1111... is ALMOST 1/9 - they're just two different representations of the same quantity. Multiply both quantities by 9 and, similarly, 0.9999... is just another representation of the quantity 1.

Here's another quick idea: if 0.9999... and 1 were not equal, then we would be able to locate a number in between them (their mean, for example), but this is not the case.

To extend this a bit: 0.249999... = 0.25, etc.

@nathanshields

Assuming that this is true is also assuming that a function that approaches a number at infinity equals that number well before infinity.

Mathematically this is not true because (x-1)/x will never equal 1, but it will get really really REALLY close.