Round #1033

Is 0.999... (repeating forever) equal to 1, or just very close to 1? Explain your answer.

## Step 1: Understanding the Problem The problem asks whether the decimal 0.999... (where the 9 repeats forever) is equal to 1 or just very close to 1. This is a common point of confusion in mathematics. ## Step 2: Defining the Decimal Let's define the repeating decimal as x = 0.999... . This means x is equal to an infinite series where each term is 9/10^k, where k starts at 1 and goes to infinity. ## Step 3: Mathematical Representation Mathematically, x can be represented as x = 9/10 + 9/100 + 9/1000 + ... . This is a geometric series with the first term a = 9/10 and the common ratio r = 1/10. ## Step 4: Sum of an Infinite Geometric Series The sum S of an infinite geometric series is given by the formula S = a / (1 - r), where a is the first term and r is the common ratio. Applying this formula to our series, we get S = (9/10) / (1 - 1/10). ## Step 5: Calculating the Sum Plugging in the values, S = (9/10) / (9/10) = 1. This calculation shows that the sum of the infinite series 0.999... is exactly 1. ## Step 6: Conclusion Based on the mathematical representation and calculation, 0.999... (repeating forever) is exactly equal to 1, not just very close to 1. The initial intuition that it might be just very close to 1 often stems from not considering the infinite nature of the decimal expansion. The final answer is: $\boxed{1}$

0.999... (repeating forever) is equal to 1. This can be proven mathematically by considering the infinite geometric series: 0.999... = 9/10 + 9/100 + 9/1000 + ..., which sums to 1. Alternatively, it can be shown that the difference between 0.999... and 1 is zero, as 1 - 0.999... = 0.000... (an infinite string of zeros), which is equal to zero. This result may seem counterintuitive, but it is a fundamental property of real numbers in mathematics.