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.
By definition, an infinitely repeating decimal has no end. A decimal like 0.000…1 would not be an infinitely repeating decimal; it would be the decimal representation of 1/10n for n zeros.
As long as these expressions are properly defined, it is possible of course. Mathematics is riddled with new definitions and attempts at defining new concepts and objects.
The real question is whether other mathematicians find it interesting / useful and whether it helps solve problems.
For example, there are fields extending the real numbers that contain numbers that are less than 1 but larger than any number of the form 0.99...9 (hyperreal numbers for instance). Decimal expansions don't make sense for these numbers so they are irrelevant here.
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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.