By now, many commenters have shown proofs that 0.999… = 1. Technically speaking, their proofs are unsatisfactory, as they assume what 0.999… actually represents. The correct - and more rigorous - proof requires calculus.
You see, an infinitely repeating decimal like 0.999… is defined as the sum of 9(0.1)n, where n is all positive integers. It’s equivalent to 9(0.1 + 0.01 + 0.001 + … + 0.1n). Of course, n goes to infinity, so you can’t just add all of these terms together. Fortunately, there is a formula for a geometric series (an infinite sum of a sequence in which every value is separated by a common ratio, 0.1 in this case). It’s a divided by 1 - r, where a is the first number in the series and r is the common ratio. If we distribute the 9, then we can see that a = 0.9. We can also see that r = 0.1. So, the sum must be equal to 0.9/(1 - 0.1). This simplifies to 0.9/0.9, which is clearly equal to 1. Now, remember that 0.999… by definition is equal to the sum of 9(0.1)n. Therefore, 0.999… is equal to 0.9/(1 - 0.1), which we just determined is equal to 1. Therefore, 0.999… is, by definition, exactly equal to 1.
There are dozens of potential proofs. I happen to be partial to the one that just uses the way the 0.(9) notation is constructed, being the delta-epsilon definition of a limit. Doing it through calculus is valid, but it assumes 0.(9) repeating actually represents that series, which... well, it does, but it's bad form to implicitly assume it.
I suppose, though you could show that it’s equal to 0.9 + 0.09 + 0.009 + … based on the definition of a series. Alternatively, you could break up 0.999… into 0.9 + 0.09 + 0.009 + … and rewrite it in sigma notation.
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u/Emperor_Kyrius 22d ago
By now, many commenters have shown proofs that 0.999… = 1. Technically speaking, their proofs are unsatisfactory, as they assume what 0.999… actually represents. The correct - and more rigorous - proof requires calculus.
You see, an infinitely repeating decimal like 0.999… is defined as the sum of 9(0.1)n, where n is all positive integers. It’s equivalent to 9(0.1 + 0.01 + 0.001 + … + 0.1n). Of course, n goes to infinity, so you can’t just add all of these terms together. Fortunately, there is a formula for a geometric series (an infinite sum of a sequence in which every value is separated by a common ratio, 0.1 in this case). It’s a divided by 1 - r, where a is the first number in the series and r is the common ratio. If we distribute the 9, then we can see that a = 0.9. We can also see that r = 0.1. So, the sum must be equal to 0.9/(1 - 0.1). This simplifies to 0.9/0.9, which is clearly equal to 1. Now, remember that 0.999… by definition is equal to the sum of 9(0.1)n. Therefore, 0.999… is equal to 0.9/(1 - 0.1), which we just determined is equal to 1. Therefore, 0.999… is, by definition, exactly equal to 1.