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I am about to destroy this board. First we need a trivial proposition
If a and b are two real numbers and R(x) is a rounding function then
[math]R(a)\neq R(b) \implies a \neq b[/math]
Hopefully this result is obvious for everyone. By the way I am speaking of rounding as in 'rounding to the next x'.
Now consider the numbers 1.00 and 0.9999999...
If you round them to the second decimal place you get
R(1.00) = [1.00] = 1.00 = 1
R(0.999...) = [0.99]99... = 0.99
1 is not equal to 0.99
Therefore 1 is not equal to 0.99999..
EASY
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Wow your shitty rounding function doesn't even round properly congrats
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>>8241722
no rounding function is injective. the "trivial" proposition is a lie
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>>8241722
That's truncating, not rounding, you fucking idiot
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>>8241722
that's not a function
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>>8241729
Then consider it a 'decimal cutoff' function.
For example.
1.5 = 1.5 because no matter where you cut, the equality holds.
1 = 1
1.5 = 1.5
Another example
1.8473737273
1 = 1
1.8 = 1.8
1.84 = 1.84
etc.
How come this decimal cutoff method does not work for 0.99... and 1?
Clearly this implies that the real numbers are incomplete and ought to be completed with infinitesimals.
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>>8241722
1/3 = 0.333...
1/ = 0.33
1 = 0.3
I don't get it.
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>>8241722
>R(0.999...) = [0.99]99... = 0.99
This is not rounding. This is truncating or rounding down.
You can certainly define an equivalence relationship based on rounding/truncating. You need to better define your equivalence relationship (eg round up/down, truncate, number of decimal places).
What you have proven in your example is that:
1 is not equivalent to 0.99999... under an equivalence defined as truncating the decimal representation to the 1/100's decimal place. Your equivalence relationship basically groups everything from 0 - 0.9999... as equivalent. Also all decimals between 1 - 1.9999... are equivalent.
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>>8241742
Your truncating function, while it corrects one problem, now does not hold the property that R(x)!=R(y) implies x!=y.
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>oh look another 1≠.999... thread
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>>8241722
Wow, you just destroyed professional mathematicians.
>Totally not sarcasm
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>>8241722
you are not rigorous enough
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>>8241742
>How come this decimal cutoff method does not work for 0.99... and 1?
Because real numbers do not necessarily have a unique decimal expansion, which makes your function poorly defined.
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Well anon. One of two things is definately true. Either important properties of real numbers are false, or your shitty bait proposition is false.
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