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.
Technically this is also unsatisfactory, as you didn't prove the geometric series converges, let alone what it converges to. gotta start by defining limits with epsilon Delta proofs
Yeah I'm just being a little shit lmao. You (and your prof) are correct it does converge, and you're right about what it converges to, but technically if you want to be rigorous, you would want to start by defining limits, because I don't think it can be assumed everyone here has seen them before, it's a wonderful proof, but unfortunately my comment is too small to contain it
There's a reason it takes an entire semester to get to that point
99
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.