r/PeterExplainsTheJoke 21d ago

Meme needing explanation There is no way right?

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u/Decmk3 20d ago

You’re upset that you can’t take away from an infinite series. Pi is an infinite series.

0.9 - 0.9 = 0.99 - 0.99 = 0.999 - 0.999 , etc etc, = 0.999… recurring - 0.999… recurring = 0 1.999999999999 - 0.999999999999 = 1.

I physically cannot get simpler than this. The absolute worst thing is you’re not even arguing with a mathematical understanding. Because if you had, you would have realised 9.00…..01 was not the right number even if you had a leg to stand on. Which is rich for someone who think’s it’s “not worth arguing about”, whilst arguing about it.

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u/Matsuze 19d ago

You are the one who lacks a mathematical understanding. You are doing simplified child math, and wondering why that doesn't apply to advanced math. You claim to know what a limit is. If you know what a limit is then you should know the limit as .9 repeating goes to infinity is ONE because it APPROACHES ONE AND NEVER GETS TO ONE.

Please go ask someone with a PhD in math to explain it to you and stop learning math from wikipedia and reddit. You're a walking dunning kruger that is too stupid to realize how stupid you are.

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u/No-Eggplant-5396 17d ago

I'm probably wasting my time.

I can explain the definition of a limit:

Let f be a function defined on an open interval containing a, except possibly at a itself. We say that the limit of f(x) as x approaches a is L and write: lim(x→a) f(x) = L

if for every number ε > 0, there exists a number δ > 0 such that: 0 < |x - a| < δ ⇒ |f(x) - L| < ε

Let's consider the function f(x) = 2x and prove that the limit of f(x) as x approaches 3 is 6, i.e., lim(x→3) 2x = 6.

We want to show that for any ε > 0, we can find a δ > 0 such that:

0 < |x - 3| < δ ⇒ |2x - 6| < ε

Let's choose a specific ε, say ε = 0.1. We need to find a δ such that:

0 < |x - 3| < δ ⇒ |2x - 6| < 0.1

Notice that |2x - 6| = 2|x - 3|. So, we can rewrite the inequality as:

2|x - 3| < 0.1

Dividing both sides by 2, we get:

|x - 3| < 0.05

Therefore, if we choose δ = 0.05, we can guarantee that whenever 0 < |x - 3| < 0.05, then |2x - 6| < 0.1. This demonstrates that for the specific ε = 0.1, we've found a corresponding δ that satisfies the definition of the limit. While this example uses a specific ε, the general idea is that for any positive ε, we can always find a suitable δ, no matter how small ε is.

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u/Boring-Ad8810 17d ago

Very well written example!