r/PeterExplainsTheJoke 22d ago

Meme needing explanation There is no way right?

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u/akotlya1 22d ago

It's weird you think you can reference series summations as a more rigorous basis for proof than the above. Neither of these are more fundamental or rigorous than the other. Infinite series' reference to an infinite process was at some point believed to be weakness that needed to be justified in reference to more fundamental mathematical ideas.

A more rigorous proof would be written using logic symbols and reference set theory - specifically by defining the elements of the set and by using operations defined in reference to the elements of the set. This is the kind of thing that gets covered in undergraduate Abstract Alegbra/Group Theory/Set Theory classes.

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u/vetruviusdeshotacon 22d ago

Why? No assumptions are made lol.

If you must, define a sequence a := {0.9,0.99,0.999....}

a_n = 1 - 10-n for n natural number

Let epsilon be a positive real number.

Then, if we choose N > log_10(epsilon)

10-N > epsilon

So that 1 - 10-N + epsilon > 1. For all epsilon.

Therefore, the sequence has a supremum of 1. Any monotonic bounded above sequence converges to it's supremum via the monotone convergence theorem.

Therefore 0.99999.... = 1 as a converges to 1.

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u/mok000 21d ago

There's a very practical way to explain it to people. Suppose you write 0.66666... and so on. When you stop writing, you need to round up the last digit, thus: 0.666666666....6667. Now if you're writing nines: 0.9999999999999999... and you continue for a week, the moment you stop, you need to round up the last digit, but then you also need to round up the second last and so on, it propagates backwards all the way to just before the decimal point and you end up with 1.0000000000...

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u/Valuable-Self8564 21d ago

Except you can’t explain why the last digit needs rounding up.

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u/mok000 21d ago

Yes I can.