Do asymptotes disprove 0.9 repeating equal 1?

Question

I am in 9th grade and taking geometry. Several of my friends taking pre-calc say that 0.9999... does not equal 1, but is just an asymptote. I have not taken that subject yet and they don't give any proofs for why it is an asymptote. I couldn't find any definate answers on the internet that dealt with asymptotes. It would be much appreciated if someone could explain this to me.

Answer 1

Suppose, that $0,99... \ne 1$. In that case, $0,999... < 1$. If $a < b, \exists c$, such that $a < c < b$.

In our case, how this $c$ would look like? It would have the form $0,999...$, but somewhere it can't have $9$. If somewhere we don't write $9$, then $c < a$. Since we can't find such $c$, it is a contradiction. Therefore $0,999...=1$.

Second reason: Consider $\frac13$. It is easy to see, that $\frac13=0,333...$. Multiply both sides with $3$, and you get $1=0,999...$(This is not a proof, just to make it easier to see).

Answer 2

The number 0.99999... is a fixed number, and not the asymptote of anything, even though you might happen to read the number as a succession of values, each corresponding to 1/10th the value of the previous component. I presume you're familiar with the elementary proof that this number equals 1.0, that is, it is just another way of writing the value 1.0:

$10 I = 9.99999...$

$I = 0.99999...$

subtract to find:

$9 I = 9$,

therefore $I = 1.0$.

Answer 3

The concept of an asymptote is when some behavior is reached after an infinite amount of steps, which for you would be exactly the idea: any finite number of 9's in the sequence $0.999\ldots$ is indeed strictly less than $1$, but taking an infinite amount of these 9's will make it equal exactly.

It's not easy to define or argue these without formal notions of limits, but here is an intuitive argument. Notice that $$ 0.999\ldots = 9 \times (1/10)^1 + 9 \times (1/10)^2 + 9 \times (1/10)^3 \ldots = 9a (1+a+a^2+\ldots) $$ for $a = 1/10$. If the infinite sum in brackets exists, say it is evaluates to $L$. Then $L*a = a + a^2 + \ldots = L-1$, so $aL = L-1$ which means that $$ L = \frac{1}{1-a} = \frac{1}{1-1/10} = \frac{10}{9}, $$ and our sum would be $9aL = 9 \times (1/10) \times \frac{10}{9} = 1$, as your intuition suggested.