Is 0.9999... = 1?

The problem with this issue is that we find it difficult to get out heads around limits of series. Since these can be pretty weird, that is forgiveable!.

Let us assume that the number $a = 0.\dot{9}$ exists. Since $0.999\cdots9$ (for any finite number of $9$s) is less than $1$, we can reasonably deduce that $a \le 1$.

Suppose now that $0 < b < 1$. If we consider the decimal expansion of $b$, we will eventually find a number after the decimal point which is less than $9$. In other words, we will be able to write $b < 0.999\cdots9$, where the number on the right of this inequality has a finite number of $9$s after its decimal point. Thus $b < a$.

In other words the number $a$ is less than or equal to $1$, but is greater than any number less than $1$. This means that $a$ must be equal to $1$. In other words, if $0.\dot{9}$ exists, it must be equal to $1$.

The only problem left is that of accepting that the number $a$ exists in the first place. This is a nontrivial issue, and many mathematicians, starting with the ancient Greeks, have had problems with infinite limits. Consider such "paradoxes" as Zeno's paradox or the Achilles and the tortoise paradox.

If we are prepared to accept a mathematical system within which limits exist, then $0.\dot{9}$ will both exist and be equal to $1$, and $0.\dot{3}$ will exist and be equal to $\tfrac13$. The existence of such limits is based upon what is called the Completeness Axiom of the Reals; while mathematicians might try to see what can be deduced in the absence of this Axiom, I would think that few disbelieve it.

It is interesting that students find the existence of $0.\dot{9} = 1$ harder to accept than they do the existence of $0.\dot{3} = \tfrac13$. Accepting that either of these numbers exists involves coping with the idea of an infinite number of entries in a decimal expansion; there is something about the fact that the identity $0.\dot{9} = 1$ increases the digit before the decimal point that gets people agitated, while students at a very early age have few problems with $0.\dot{3} = \tfrac13$. From a teaching point of view, therefore, observing that $3 \times 0.\dot{3} \,=\, 0.\dot{9}$, and therefore that $0.\dot{9} = 3\times\tfrac13 = 1$ is an argument which can content most students, at least until they can get to grips with the proper definition of the limit of a series.

I am aware of more elegant proofs for this than the one that follows.

However, I find this version to be easier to grasp for people struggling with the concept.

1) $\frac{1}{3} = 0.333... = 0.\overline{3}$

2) $3 \times \frac{1}{3} = 3 \times 0.333... = 0.999... = 0.\overline{9}$

However,

3) $3 \times \frac{1}{3} = \frac{3}{3} = 1$

From 2) we get: 4) $3 \times \frac{1}{3} = 0.\overline{9}$

From 3) we get: 5) $3 \times \frac{1}{3} = 1$

If we combine 4) and 5) we get: 6) $1 = 3 \times \frac{1}{3} = 0.\overline{9}$

And finally, 6) reduces down to: 7) $1 = 0.\overline{9}$

The more elegant proofs are... well... more elegant in the way that they convey the general pattern going on beneath the surface that can then be abstracted to other scenarios, such as the ever-popular $0.\overline{12}$. However, those can be hard for people to grasp.

I find that the $3 \times \frac{1}{3}$ proof is easier for most people to grasp, in that it opens the door enough for them to realize that there's something really weird going on here.

Which is important. Because, and let's be honest about this, this $1 = 0.\overline{9}$ business is really freaking weird. Anything that helps crack the door open to let in the light is a good thing, in my book. :)

$x=0.\overline { 9 } \quad \Rightarrow \quad 10x=9.\overline { 9 } \\ \\ \Rightarrow \quad 10x-x=9.\overline { 9 } -0.\overline { 9 } \\ \\ \Rightarrow \quad 9x=9\quad \Rightarrow \quad x=1$

In one solution I wrote, there is an argument about whether 0.99999... equals to 1, and many of this community's respected member point out that it is, when I don't personally believe it is equal. The evaluation of limit in the infinite sum of $\sum_{n=1}^{\infty}9(0.1)^n$

shows that it tends to reach 1, but never exact 1. I will keep an eye on respected members for this

More proofs here than I have the patience to read

Yes it is. there are 3 ways:

1/3 = 0.3333333333333333... so multiplying by 3 on both sides we get 1= 0.9999999999999999999.... OR 0.9 can be thought as $\frac{9}{10^{1}} +\frac {9}{10^{2}} +\frac{9}{10^{3}} ....$ (infinite GP.) OR x = 0.999999999... 10 x = 9.99999999999.... 9 x = 9.9999999999...-0.999999999999.... = 9 x = 1

There is a wiki about this topic here: is 0.999... = 1?

What you need to ask yourself regarding this topic is which set of numbers are you using? On the real line .9 repeating equals 1. The only way .9 < 1, as your answer to this question would have us believe, is if we were dealing in rational numbers only. The problem with asking this question of the rational numbers is that .9 repeating doesn't exist. Because of the analytical definition of the real line the limit of the sum that is .9 repeating exists and it is one.

Interesting how many proofs rely on the concept that 0.3recurring = 1/3. This is only true is 0.9recurring =1. In other words you are proving nothing. 0.3recurring is an approximation to 1/3.

bing bang theory depents on thant 1 is infinetely larger the any lim-- 0.9999999999999....