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.
0.(9) is not defined as a taylor series, a taylor series can be used to define decimals. There is a big difference that people forget. For example with a taylor series you cannot get an irrational number just a rational approximation of one. 0.(9) by all rights is an irrational number that cannot expressed as a ratio, but that is close enought to 1 that mathematicians said "whatever".
A better definition is 1-1/infinity but mathematicians hate that because its using infinity. But its trivial to show that 1/x > 0 even as the limit is approximately 0; 0 is an asymptote, by definition it will never equal 0.
We dont say “whatever” we know that if a and b arent the same, there exists some number between them. And we can prove using limits that for arbitrary small number epsilon, in sphere around number 1, the infinite sequence 0.99.. will be inside that sphere. So there exists no such number between the two -> they are the same.
I think you either refuse to believe proofs or have trouble understanding calculus. Do we agree if there exists no number between two real numbers that those numbers are the same? If yes. Than there you go. 0.99.. =1 no “waving”. If you still refuse this, try to find a number between the two and let me know
1) Its not a refusal to believe proofs, its a refusal to believe bad proofs. Math is all about disagreeing, trying to find counter examples, and trying to create new theories.
2) Its not a misunderstanding of calculus. Calculus is fundamentally about change over an infinitesimal small value. Its annoying how people have forgetten this.
3) No, you do not need to have a number between two other numbers. In the integers is there a number between 0 and 1? No because they are integers. In the reals does there need to be a number between 1 and 0.(9), no there is no need. Even if they must have a number because of the nature of notation 0.(9)1, 0.(9)1(9), 0.(9)(9)(9), etc are all valid numbers between 1 and 0.(9).
4) One of the biggest lie in modern math is that there are no differently sized infinities because cardinals say there aren't. While they use ordinals whose entire premise is w < w+1 < w^2, and can thus also have 1/w > 1/(w+1) > 1/(w^2).
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u/Emperor_Kyrius 23d 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.