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I fully understand that $0.9999...$ mathematically will equal 1.0 exactly as the repeating decimal continues infinitely. This would be contrary to conventional logic where such a fraction is considered approximately 1.0 and not exactly 1.0. Here, of course, the conventional logic fails to understand what it mathematically means when the decimal is repeated in infinity.

However can this principle be translated to an infinite discount as well?

Say a number is reduced by 10% constantly (n-10%-10%-10%...). Will this too mathematically equal zero (as 0.0), or will it only ever get near zero?

Again, conventional logic will say that this will never actually be zero, it will only ever get closer to zero. Is conventional logic mathematically correct in such a case?

Thank you.

Jon
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  • If you're asking about $\lim\limits_{n\to\infty}c\cdot(\frac{9}{10})^n$ then the answer is yes, it is equal to $0$ for any constant value $c$ that you choose. – barak manos Jan 29 '15 at 10:55
  • Hey Jon, please let me know if my edits to your question preserve your original question's meaning. Thanks! – daOnlyBG Jan 29 '15 at 10:55

3 Answers3

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The phrase "is reduced by e.g. 10% in infinity" is perhaps ambiguous, but let me try to make it more precise:

If you reduce a quantity $A$ by $10\%$ (i.e., a proportion of $\frac{1}{10}$, then the remaining value is $$\left(1 - \frac{1}{10}\right) A = \frac{9}{10} A,$$ and, inductively, if you do so $n$ times, then the remaining values is $$\frac{9}{10}\left( \cdots\frac{9}{10}\left(\frac{9}{10} A\right) \cdots \right) = \left(\frac{9}{10}\right)^n A.$$ But for any small number $\epsilon > 0$, we can make this quantity small than $\epsilon$. (In fact, a little algebra shows that taking $n \geq \left\lceil \frac{\log \epsilon - \log A}{\log r}\right\rceil$ will do.) On the other hand, for any $n$, we have $\left(\frac{9}{10}\right)^n A \geq 0$, and so the as $n \to \infty$, the remaining quantity $\left(\frac{9}{10}\right)^n A$ goes to zero.

Travis Willse
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  • Thank you so much for this explanation. Just to be sure that I read it correct: "A goes to zero", means A equals zero when n is ∞. yes? – Jon Jan 29 '15 at 12:10
  • No, strictly speaking it doesn't make sense to substitute $n = \infty$ into the expression $\left(\frac{9}{10}\right)^n$. "Goes to zero" here means that the sequence $\left{\left(\frac{9}{10}\right)^n A\right}$ converges to $0$, or just as well (in this case) that $\lim_{n \to \infty} \left(\frac{9}{10}\right)^n A = 0$. – Travis Willse Jan 30 '15 at 00:39
  • I see. So a discount in infinity will not strictly equal zero in the same way that 0.999...in infinity equals 1.0? – Jon Feb 02 '15 at 13:26
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    The discount $\lim\limits_{n\to\infty}\left(\frac{9}{10}\right)^nA = 0$ and the $0.999... = \lim\limits_{n\to\infty}\sum\limits_{k=1}^n \frac{9}{10^k} = 1$, and they are equal in exactly the same sense. However, we usually don't say "when $n$ is infinity", we look at the limit. – Eff Feb 02 '15 at 13:34
  • 3/3=0.999...=1.0, and according to those equations, a constant reduction of 10% of n will also equal 0.0 in infinity (and only in infinity, like the 0.999... example only equal 1.0 in infinity). If the principle can be translated to both cases, I then understand these two examples are equivalent, but I am not exactly sure if this is true. The 0.999... example is an infinity within a natural finite number [0-1], and so works logically. The discount example is not within a finite number (I think...) and so is more difficult to grasp. – Jon Feb 03 '15 at 13:13
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I respect your curiosity for asking these kinds of questions; they're important questions, and I think the downvoters were too harsh.

The basic distinction that needs to be understood is the distinction between a number and sequence of numbers. A number is basically a distinct point on the real line. Hence, a sequence of numbers is basically an infinite sequence of points on the real line.

For example, $0.9999\ldots$ is a number; it happens to be the same number as $1$, because it is represented by the same point on the real line. On the other hand, the sequence:

$$(0,0.9,0.99,\ldots)$$

should be thought of as an infinite sequence of points on the real line. The first point in this sequence is $0$, the second is $0.9$, the third is $0.99$ etc.

So, what is the connection between the number $0.9999\ldots$ and the above sequence? To explain this, we need the concepts of "convergent sequence" and of "limit." Unfortunately, these have fairly technical definitions, but let me try to explain them intuitively.

Some sequences have the property that the points are getting closer and closer to some fixed value. The sequence $(0,0.9,0.99,\ldots)$ falls into this category; it gets closer and closer to the number $1$. When this happens, we say that the sequence is convergent, or we might say that it "converges to some fixed value." Okay, what are we going to call that fixed value? Mathematical convention is that we call it the limit of the sequence. So we can say that the sequence $(0,0.9,0.99,\ldots)$ is convergent, with limit $1$.

On the other hand, some sequences don't become arbitrarily close to any point on the real line. For example, the sequence $(0,1,0,1,0\cdots)$ falls into this category. We therefore say that this sequence is not convergent, and that it does not have a limit. (You may object and say that it becomes arbitrarily close to two different points, namely $0$ and $1$. But that's not how the definition of a limit works; you'll have to dive into the technical details to see why. This stuff takes time, unfortunately, and it can be quite challenging.)

Okay, now that we understand the meaning of "convergent sequence" and "limit", lets get back to the main point. What is the relationship between the number $0.9999\ldots$, and the sequence $(0,0.9,0.99,\ldots)$? Well the very definition of $0.9999\ldots$ is that its the limit of the sequence $(0,0.9,0.99,\ldots).$ Hence the following statements (which are both true) are really expressing the same thing.

  • $0.9999\ldots$ equals $1$
  • the sequence $(0,0.9,0.99,\ldots)$ is convergent, with limit $1$.
goblin GONE
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In a sense, it's equal to zero in the same way your example is equal to 1. Both converge on to their target, but only after an infinite number of digits/reductions.

JTP - Apologise to Monica
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