When proving that the number .999... exactly equals to 1, using basic algebra, is it an uncountably-infinite number of 9s?

Question

When proving that the number .999...999... equals 1, using nothing more than high school algebra, how should I think of the number of 9s being added to each slot that was previously a 0? Is it an uncountably-infinite number of 9s or is it countable? How could I justify this?

I want to show this idea to my students this week, but I want to prepare for them asking me the above questions.

(It's for an elementary calculus course, not analysis.)

Thanks,

Answer 1

To the best of my recollection, this fact is simply a given in 'high school algebra', so there's nothing to prove. At that level, all you can do is to give motivations.

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Anyways, as to the context to answer your question, the places of a decimal numeral are indexed by integers. The most convenient indexing is:

• ...
• Place -1 is the tenth's place
• Place 0 is the one's place
• Place 1 is the ten's place
• Place 2 is the hundred's place
• ...

The reason this is convenient is that if we have decimal numeral that has the digit $a_n$ in place $n$, then that numeral represents the number

$$ \sum_{n=-\infty}^{\infty} a_n 10^n $$

One can take this formula as a definition of which number is represented by a decimal numeral. (pedantic note: I'm using $a_n$ both to refer to a digit and to the number that digit corresponds to. Depending on the fine details of how you lay the foundations, these may or may not be different types of objects)

Incidentally, the value of the numeral $0.\overline{9}$ is an infinite geometric series, so you could use the geometric series sum formula to prove it's equal to 1. Assuming, of course, that the audience is comfortable with infinite geometric series.

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So, to answer your questions

how should I think of the number of 9s being added to each slot that was previously a 0?

You shouldn't think that way at all. That description does not resemble one of the numeral $0.\overline{9}$ — it sounds like you are trying to describe the infinite sequence $s$ whose terms are

• $s_0 = 0$
• $s_1 = 0.9$
• $s_2 = 0.99$
• ...
• $s_n = 0.\underbrace{99\ldots9}_{n \text{ nines}}$
• ...

The numeral $0.\overline{9}$ is not a "process" or an "algorithm" or even a "sequence of terminating decimals". It is simply the decimal that has a 9 in every negative place and a 0 in every nonnegative place.

And I should make it explicit that the symbols $0.\overline{9}$ consisting of a zero, a dot, and a nine with a line over it is shorthand for the decimal described in the previous paragraph.

Is it an uncountably-infinite number of 9s or is it countable?

The places are indexed by integers, so there are only countably many digits.

How could I justify this?

You can't "prove" any of the above since they're simply givens; it's simply the way we've defined decimal numerals. The justification you can give, however, are that decimals so defined are useful, and furthermore we have theorems

• Every non-negative real number is the value of a decimal numeral
• Every non-negative real number has at most two representations as the value of a decimal numeral.
• Every left-bounded decimal numeral has a real value.

Left-bounded means that once you go far enough to the left, everything else to the left is zeroes. Formally, if $a_n$ is the sequence of digits, there is some $N$ such that $a_n = 0$ for all $n > N$.

The exceptions that have two decimal representations are, of course, the cases like $1.\overline{0} = 0.\overline{9}$.