Talk:Conservapedia insights

I'm not sure which is funnier - that The Scotsman thought the Harvard student's column was serious or that Conservapedia is shouting at the rooftops about it. Read some of his other columns, like "Harvard Doesn't Hate Parties, They Just Hate Us" and the like, then reread the column's parts that are quoted. Then you can come back here and admit how embarrassed you are.--ThomasE 14:35, 11 May 2008 (EDT)

The point is made with good humor - as in your second example about the lack of parties at Harvard. Or ... perhaps you think your example is trying to say that Harvard is a party school??? I guess you don't think much of political cartoons either.--Aschlafly 16:08, 11 May 2008 (EDT)

Umm .... what? --CharlesN 16:22, 11 May 2008 (EDT)

Hahah never mind, you're right, it does make a good point. Please leave it unchanged.--ThomasE 18:27, 11 May 2008 (EDT)

Two equally valid insights.

1. Conservapedia observed that "Son of Man" is a liberal mistranslation to downplay the divinity of Jesus, and that a stronger, more precise translation would be "The Son as Man" or "The Son, a Man."
2. Conservapedia observed that pi contains pi and every other number, despite liberal denial of the breadth and power of infinity.

At least both insights are equally valid. --AugustO (talk) 13:01, July 25, 2021 (EDT)

Proof that π does not contain π:

1) Let π be represented by the sequence of digits x₀.x₁x₂x₃x₄x₅...... Therefore x₀ = 3, x₁=1, x₂=4, x₁₀₀₀=9, and so on.

2) Assume that π contains π: Then there is an index n such that x₀ = x_n = 3, x₁ = x_n+1 = 1, x₁₀₀₀ = x_1000+n = 9, or in general x_k = x_{n + k} for all natural numbers k.

3) But then, π's representation is periodic: x₀ = x_n = x_2n = x_3n = x_{w n} for all natural numbers w.

4) Every number with a periodic representation is a rational number. In this case, we could present π as x₀x₁x₂x₃...x_n-1 / (10^n-1 -1)

5) It is well known that π is not a rational number, so we have a contradiction. Ergo, π cannot contain itself.

--AugustO (talk) 13:18, July 25, 2021 (EDT)

Point (3) does not follow from points (1) and (2). A more fundamental flaw is that the above argument incorrectly denies the possibility of an infinite series embedded within an infinite series, which pi likely contains.--Andy Schlafly (talk) 15:21, July 26, 2021 (EDT)

Thanks for your input! That's how a proof should work: a sequence of steps, each one acceptable for the reader. So, I'll try to do it better:

1. Do we agree that π can be represented as a digital expansion x₀.x₁x₂x₃x₄x₅...... (as any other real number)? I mean, that's the reason that people learn the first hundreds or even thousands of digits of π, and that super-computer race to find ever more digits of the expansion - it is meaningful to say that x_1,000=9 and x_1,000,000=1.
2. Can we agree that "finding π in &pi" means that there is an index n such that π starts all over again at that place?

If we can agree on the above, then the following should be straight forward:

1. If π starts again at n, then x₀ = x_n = 3, x₁ = x_n+1 = 1, x₂ = x_n+2 = 4, x₃ = x_n+3 = 1 and so on. "And so on" means that x_k = x_k+n for each natural number k, i.e.,. k ∈ N₀. Andy, you seemed to have no qualms with this step.
2. What happens if k equals n? What is the value of x_{n + n}? According to the step above, x_{n + n} =x_n. But x_n=x₀=3 ! So, x_{n + n} = 3. That's true for all multiples of n: if w ∈ N₀, then x_{w n} = x_{(w-1) n} = .... = x_{2 n} = x_n = x₀ = 3.
3. What about x_{w n +k} ( 0 ≤ k < n )? Again, x_{w n +k} = x_{(w-1) n +k} = x_{(w-2) n +k} = x_{2 n +k} x_{n +k} = x_k
4. So, for every natural number w and every number k with 0 ≤ k < n we have: x_{w n +k} = x_k. That makes π periodic, the length of the period is n.
5. Therefore, π can be written as x₀x₁x₂x₃...x_n-1 / (10^n-1 -1) . But this is a rational number, and we know that π is irrational.
6. We have a classical contradiction: As our result is wrong, our assumption that π is contained in π has to be wrong.

I hope this proof is acceptable, if necessary, additional steps can be entered. --AugustO (talk) 03:55, July 27, 2021 (EDT)

The possibility of an infinite series embedded within an infinite series

I do not deny the possibility of an infinite series embedded within π: in fact, there is a countable number of infinite and different sub-series in π each one corresponding to the position of the digit it starts with.

Do I miss something? --AugustO (talk) 08:48, July 28, 2021 (EDT)

See my proof that pi contains pi at Talk:Pi#Pi_contains_every_other_number.3F.--Andy Schlafly (talk) 13:10, July 29, 2021 (EDT)