Problem 56 - Calculus Single Variable

Question

At first glance, it may appear that the number \(y=0 . \overline{9}=0.999 \ldots\) is just a little smaller than \(1 .\) In fact, \(y=1 .\) Let \(x=1-y .\) Explain why it is true that \(x<10^{-n}\) for every positive \(n .\) Deduce that \(0 \leq x

Step-by-Step Solution

Answer

Since \( x = 0 \) in the equation \( x = 1 - y \), it confirms \( y = 1 \), proving \( 0.999... = 1 \).

Step 1: Express y and understand its decimal representation

We start with the understanding that \( y = 0.\overline{9} = 0.999\ldots \). This means \( y \) is a repeating decimal with an infinite sequence of nines after the decimal point.

Step 2: Define x as the difference from 1

Let \( x = 1 - y \). Given \( y = 0.999... \), substituting gives \( x = 1 - 0.999... \). The subtraction suggests that \( x \) represents the leftover from an infinite sequence approaching 1.

Step 3: Calculate x for nines approximation

Consider \( x = 1 - 0.999... \) as an infinite sum approach. As \( 0.999... \) gets infinitely closer to \( 1 \), it effectively means that \( x \) approaches \( 0 \). This implies \( x = 0.000...1 \), which reduces to \( 0 \) as the nines continue indefinitely.

Step 4: Examine x in terms of 10^{-n}

For any positive integer \( n \), we observe that the difference \( x \) is so small it cannot meet or exceed \( 10^{-n} \). Specifically, \( x \) can be thought of as \( 0.000...1 \), but since the sequence of nines is infinite, there is no actual "1" fitting in any decimal place to create a fixed, positive number smaller than \( 10^{-n} \).

Step 5: Deduce x in relation to any positive b

If \( x < 10^{-n} \) for any positive \( n \), it follows that \( x \leq 0 \). Since \( 10^{-n} \) can represent any fraction greater than 0, \( x \leq b \) for every positive \( b \). Consequently, \( x \), being 0, satisfies \( 0 \leq x < b \).

Step 6: Conclude y=1 effectively

Since \( x \) is 0 and \( x = 1 - y \), solving gives \( y = 1 \). Thus, the infinite sum \( 0.999... \) is precisely \( 1 \), confirming that \( y \) equals 1.

Key Concepts

Repeating Decimals

Repeating decimals are numbers where one or more digits repeat infinitely after the decimal point. A classic example is the number \(0.\overline{9} = 0.999\ldots\). In this number, the digit '9' repeats forever.
Such numbers are interesting because they have unique properties when it comes to their decimal and fractional representations.
For example, \(0.\overline{3} = 0.333\ldots\) translates directly to the fraction \(\frac{1}{3}\). This helps us understand that repeating decimals can often be expressed as an infinite series, which we can use calculus to analyze.
Repeating decimals are not just mathematical oddities. They play a role in understanding the nature of irrational numbers and infinite processes.

Infinite Series

An infinite series is a sum of infinitely many terms following a particular pattern. In calculus, this concept helps us understand repeating decimals more deeply.
When we look at \(0.\overline{9} = 0.999\ldots\), we think of it as an infinite geometric series. Each term in the series represents a contribution to the total value. Specifically, \(0.9 + 0.09 + 0.009 + \ldots\), where each term is ten times smaller than the previous.
The sum of this infinite series is simply \(1\), which can be represented by the formula for an infinite geometric series, \(S = \frac{a}{1-r}\), where \(a = 0.9\) and \(r = 0.1\).
By understanding it as an infinite series, we comprehend why \(0.999\ldots\) equals \(1\).

Limit Approach

The limit approach in calculus is a way of finding the value that a sequence or series approaches as the number of terms increases without bound. This is highly relevant to analyzing repeating decimals like \(0.\overline{9} = 0.999\ldots\).
Consider the expression \(y = 1 - 0.999\ldots\). As the series \(0.9 + 0.09 + 0.009 + \ldots\) gets closer and closer to \(1\), we find that \(y\) tends towards \(0\).
This is represented by the limit, \(\lim_{n \to \infty} (1 - (0.9 + 0.09 + \dots + 0.09\times10^{-n})) = 0\).
Using the limit approach, we prove that the difference between \(1\) and \(0.999\ldots\) is negligible, effectively making \(0.999\ldots = 1\). This concept highlights the power of limits to resolve counterintuitive scenarios.

Decimal Representation

Decimal representation is how we express numbers using the base-10 system, which is familiar to us in everyday arithmetic. The decimal system is composed of digits and a decimal point, with each digit holding a place value.
In the case of repeating decimals like \(0.\overline{9}\), the representation is not straightforwardly finite, yet it conveys a clear fractional value such as \(1\).
Understanding decimal representation helps us to realize the nuances of infinitely repeating decimals. Despite appearances suggesting that \(0.999\ldots\) might be less than \(1\), correct interpretation reveals their equivalence.
This equivalence is rooted in the understanding that the mechanics of decimal representation support this harmony between apparent endlessness and finiteness, enforced by the aforementioned concepts such as infinite series and limit approaches.