How can 0.2999... be exactly 0.3

Question

Maslanka asked a bizarre Pyrgic question on the 4th of May (see below (I had to type it in because I'm not allowed to post images yet)). What I couldn’t get my head around was that, as I understand it, 0.2999 recurring is very nearly (infinitely close to) 0.3 but it is not actually 0.3 but Chris Maslanka shows that they are equal. (Where he arrives at 90X = 27 he should have got X = 3/10 of course)

Where am I going wrong?

Question:

Find two distinct ways of writing 1/3 as a non-terminating decimal. (sic)

Answer:

0.333... and 0.2999... in which the 3s and 9s go on without termination. If 0.2999... = X then 100X = 29.999... and 10X = 2.999... Subtracting we get 90X = 27, whence X = 1/3 (sic²). A similar process shows 0.333... = 1/3.

Explanation:

Clearly 0.2999... does not equal 1/3 but 0.3. My disruptive gaslighting fooled very few. Most stopped at "it does not compute"; a few imaginative ones thought further.

You have to know what is meant by an infinite decimal expansion like $0.2999\ldots$. For one, it's a converging infinite series. — Randall, Jun 02 '24 at 14:08
Also, take a look at this question for a thorough discussion. Your question seems very, very like the very frequently asked “Is $.999…$ exactly equal to $1$?” — Paul Tanenbaum, Jun 02 '24 at 14:11
The quotations you shared in `Answer' seem to imply that $0.333...=0.2999...$. Have I got that right? If so, that is not correct. What is true is that $0.2999...=0.3$ as you say in your question. — Joshua Tilley, Jun 02 '24 at 14:12
Well, $0.2999999.....$ does indeed equal $0.3$ exactly and actually and *NOT* "very nearly (infinitely close to)". (read https://math.stackexchange.com/questions/11/is-it-true-that-0-999999999-ldots-1... that is the final word on the subject. Nothing more can, nor will be said on the matter). But $0.3$ is not equal to $\frac 13$. And $0.2999.....$ is not equal to $0.33333.....$. There is only one way to write $\frac 13$ as a decimal (in base $10$) and that is $0.333333......$ — fleablood, Jun 02 '24 at 15:58
In quoting the Question, Answer, and Explanation you are not at all clear of who is asking the question and who is giving the answer and the explanation. You ask were are you going wrong but we don't actually know if you are the one making the answer, or the person rejecting the (wrong) answer or what. — fleablood, Jun 02 '24 at 16:01
The argument that $X=0.2999....$ so $100x = 29.9999...$ and $10x=2.999....$ is true. And so $90x = 27$ but $x =\frac {27}{90}\ne \frac 13$ and I kind of have to wonder why in the world he thought it would fool anyone. $\frac {27}{90} = \frac {3\times 9}{10\times 9} =\frac 3{10} = 0.3$ and absolutely true but in no way does $\frac {27}{90} = \frac 13$. $3\times 27=81$... not $90$. — fleablood, Jun 02 '24 at 16:05
Could you give a reference to the article where this question was asked. I had never heard of Chris Maslanka or of his Guardian Column "Prygic Puzzles" and could not google this particular column. — fleablood, Jun 02 '24 at 16:18
Hmm... I thought there was a way to open a discussion page. I have 3 comments for the OP as to why $0.29999....$ should equal $0.3$ exactly and not be "infinitely near" but different from $0.3$ but they aren't appropriate as an answer. I'll add them as comments but they should be discussions. 1) We need to overcome the intuitive concept of "infinitely close". There is no such thing. Things either equal something exactly or they are some measurable (finite) distance away. If $0.299...\ne 0.3$ it must be some distance from $0.3$. What distance is that? — fleablood, Jun 02 '24 at 17:01

score 0 · Answer 1 · answered Jun 02 '24 at 14:10

0

If you agree that $0.99999... = 1$, just subtract $0.7$ from both sides

answered Jun 02 '24 at 14:10

Davide Masi

1,677

This is an intuitively appealing way to look at it, but it leaves unanswered the question, “What does it mean to subtract a rational number from a non-terminating decimal string?” For that you need to understand how we choose to define things like $0.999…$, which is as the limits of infinite series. – Paul Tanenbaum Jun 02 '24 at 14:17
Alternatively, multiply by 10 and subtract 2. – Semiclassical Jun 02 '24 at 14:51
Again, @Semiclassical, this requires accepting or deciding what it means to do arithmetic (here multiplication by a natural number) on what I should have described above as limits of infinite sequences of finite series. – Paul Tanenbaum Jun 02 '24 at 15:11
@PaulTanenbaum if the op doesn't know that $0.afd99999999..... = 0.(afd+1)$ already then the subtlety of objections will be utterly loss. Yes, the old argument that $x=0.99999...\implies 10x = 9.9999.... \implies 9x = 10x -x = 9\implies x=1$ makes a lot of epistimological assumptions that would need to be verified in an rigorous formal analysis class but the OP is no-where ready for such rigor and subtlety. For better or worse, this is the cost of teaching decimal at the grade school level before real analysis. Bottom line: OP must take on faith $0.299...=0.3$ exactly because we say so. – fleablood Jun 02 '24 at 16:29
Oh, @fleablood, I disagree with your condemning people to (admittedly temporary) ignorance in this matter. I believe that even those with a fairly unsophisticated understanding can benefit from intuitive demonstrations that, for instance, $\sum_{i=1}^N \frac{1}{2^i} = 1 - \frac{1}{2^N}$ gets arbitrarily close to $1$ and never exceeds it, so we define the infinite series to be equal to $1$. And that absent some definition, such infinite-length strings of digits have no meaning. – Paul Tanenbaum Jun 02 '24 at 16:38
I'm not condemning and I agree with you that an understanding of the definitions definitely help. But the intuition of $0.299999.... +0.7 = 0.99999.... = 1$ so $0.2999.... = 1- 0.7=0.3$ although unrigorous and riddled with logical inconsistencies, is useful in getting the student to a level of maybe beginning to get that intuition that $0.29999 < 0.3$ is not the absolute he thought it had to be. I think if a student is saying "I just don't get it" and one gives a naive explanation the student can get, I feel we are just compounding if we point out the naive explanation is... well, naive. – fleablood Jun 02 '24 at 17:32
The idea that any finite approximation is arbitrarily close means we can *define* an infinite sum to be exactly is subtle and very confusing. I think a student thinks "well, if a mathematician just makes up a definition then in some way it's not really true, is it". We have to somehow teach that even thought it is a constructed definition, it is a definition of a very "real" thing that does actually exist and makes sense and is "real". There is a very precise real upper limit of $0.29, 0.299,...etc$ and if $0.29999.....$ equals anything it has to equal that upper limit. – fleablood Jun 02 '24 at 17:39

How can 0.2999... be exactly 0.3

1 Answers1