Problem 51 - Prealgebra

Question

Show that \(0.999 \ldots=1\)

Step-by-Step Solution

Answer

The repeating decimal \(0.999\ldots\) is equal to \(1\).

Step 1: Understanding the Problem

We are tasked with proving that the repeating decimal \(0.999\ldots\) is equal to \(1\). This is often counter-intuitive, but through a mathematical approach, we can show that they are indeed the same.

Step 1: Define the Repeating Decimal

Let's use a variable \(x\) to represent the repeating decimal: \[ x = 0.999\ldots \] This means essentially that \(x\) is a number with infinite 9's after the decimal point.

Step 2: Set up the Equation

Consider the equation for \(x\) and scale it up by a factor of 10: \[ 10x = 9.999\ldots \] This represents shifting the decimal of the repeating 9's one place to the right.

Step 3: Subtract the Original Equation

We subtract the original equation \(x = 0.999\ldots\) from the scaled equation \(10x = 9.999\ldots\):\[ 10x - x = 9.999\ldots - 0.999\ldots \]Simplifying both sides gives:\[ 9x = 9 \]

Step 4: Solve for x

Divide both sides of the equation \(9x = 9\) by 9:\[ x = 1 \] Thus, we've shown that the variable \(x\), representing \(0.999\ldots\), equals \(1\).

Step 6: Conclusion

The calculation has proven that \(0.999\ldots\) is indeed equal to \(1\) by using algebraic manipulation of the repeating decimal.

Key Concepts

Infinite Series

Infinite series are mathematical expressions that denote sums of an infinite number of terms. These come in handy when we want to represent repeating decimals as exact numbers. In the case of \(0.999\ldots\), it can be viewed as an infinite series where each term signifies a fractional power of 10: \(0.9 + 0.09 + 0.009 + \cdots\). This can be expressed in terms of a series:

• The first term is \(a = 0.9\).
• The common ratio is \(r = 0.1\).

For an infinite geometric series with \(|r|<1\), the sum is given by the formula:\[ S = \frac{a}{1 - r} \]Thus, for our repeating decimal:\[ S = \frac{0.9}{1 - 0.1} = 1 \]This provides a powerful perspective: endless repetitions converge towards a single precise value.

Algebraic Manipulation

Algebraic manipulation allows us to rearrange and simplify expressions to discover hidden truths in equations. In the case of proving \(0.999\ldots = 1\), we use algebra to solve for the variable \(x\), where \(x = 0.999\ldots\). By multiplying both sides of \(x\) by 10, we create an equation that can easily "cancel out" infinite decimals when subtracted:

• Initial equation is \(x = 0.999\ldots\).
• Scaling equation is \(10x = 9.999\ldots\).
• Subtract to find: \(9x = 9\).

Solving \(9x = 9\) leads to \(x = 1\). This neatly shows that algebra isn't just about solving rigid problems, but also about revealing how seemingly different expressions can actually be equivalent.

Decimal Representation

Decimal representation refers to the way we write numbers using digits and a decimal point. It’s the conventional way of expressing fractions and whole numbers in a linear sequence that is easily understandable. Repeating decimals, like \(0.999\ldots\), can often challenge our intuitive understanding of number representation. This is because the number appears different visually, yet it represents the same value as its whole number counterpart. When we perform operations like the ones used in the algebraic manipulation of repeating decimals, we are essentially uncovering the real numerical identity hidden within the infinite series of digits. This concept is key:

• Repeating decimals can appear different from fixed numbers, but can be equivalent.
• They challenge us to think beyond face value and understand deeper numerical concepts.

Understanding this helps demystify why, mathematically, \(0.999\ldots = 1\).