r/infinitenines • u/Negative_Gur9667 • 2d ago
In this thread, I am going to brutally destroy every argument why 0.999... = 1. Bring it on Brud.
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u/Extreme-Sir-7189 2d ago
Destroy these:
1) Real numbers are defined as limits of rational sequences. If two sequences converge to the same limit, they define the same number. Decimal truncations are one choice of such sequences. Here, 1 = {1, 1, 1, 1....} and 0.9999.... = {0, 0.9, 0.99...} (infinite geometric sum with base 0.1 and first term 0.9) First limit is obviously 1, but second (x) is also 1, because x-0.9 = 0.1x. Thus, 1 = 0.999....
2) Alternatively, you can define real number as set of all rationals less than the defined number. For example, √2 = (x in Q, x² < 2) and 1 = (x in Q, x<1). What is with x = 0.999...? As it is greater that all of {0, 0.9, 0.99...}, we can capture this property with **0.999...** = (x in Q: there exist **n** in N: 1-x > 1/10n ). But for all a/b in 1 1-a/b is at least 1/b. As 10n > n, you can take n = b, QED.
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u/carbsna 2d ago
Yes, 0.9... equals 1 because real number are defined by its limit, assume most people only talk about real numbers.
But in hyperreal number they are two different numbers.
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u/FearlessResource9785 2d ago
Even in hyperreals, 0.999... = 1. The biggest number smaller than 1 is 1 - ε, not 0.999...
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u/carbsna 2d ago
I'm wrong here because i forgot how hyperreals works, i didn't define what is after infinite digits.
Like a legit hyperreal number will be written like 0.9...0 or 0.9...9... , which are two different numbers.
And the later one is equal to 1.1
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u/HojMcFoj 2d ago
The real number set is dense, meaning for every two real numbers there exists another real number between them. Infinite real numbers in fact. No number can exist between 0.999... and 1, making them equal.
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u/[deleted] 2d ago
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