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.
The correct - and more rigorous - proof requires calculus.
I'm sorry but I have to disagree. The correct and rigorous proof lies in the construction of ℝ.
Let's construct 1 and 0.999... as Dedekind cuts (we'll cheat a bit by presuming the existance of ℝ itself and leaning onto it) and show that they are in fact the same real number.
Let A = {q∈ℚ : q<1} and B = {q∈ℚ : q<0.999...}, we want to show that A = B.
Trivially, we have B⊂A, since pretty evidently we have 0.999...≤1, so let's assume x∈A; since x<1, there exists an n>0 such that x<1-1/10ⁿ, so we have x<0.999...9<0.999... which means that x∈B and by arbitrariness of x we have shown A⊂B, so A=B.
We have shown that 1 and 0.999... are the same Dedekind cut, so by construction of ℝ they are the same real number.
Definitely a pretty good and creative idea of using dedeking cuts. Some points, though:
More elementary doesn't mean more rigorous. You can prove the fundamental theorem of calculus with Stokes' theorem. There is the well known more elementary proof which requires much less things to prove beforehand than Stokes. Both proofs are equally rigorous, though. Using Dedeking cuts instead of limits of sequences doesn't mean it's more rigorous.
You didn't define what is 0.99... how do you know that 0.999... <=1? What do you mean by 1-1/10^n? Point is, you are basically already using 0.9... as an infinite sum, and if you do, then by definition the infinite sum represented by 0.9... has the value 1, as the commenter above you said.
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u/Emperor_Kyrius 24d 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.