Pi x 10 ^ 62,800,000,000 = Pi with no decimals.

Orion

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#2
Pi x 10^infinity, you mean. Because that would move the decimal an infinite number of places to the right.
 

Lemniscate Biscuit

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#3
3.14159265358979323846263... × 10 = 31.4159265358979323846263...

31.4159265358979323846263... ^ 62.8 = 4.5447957777 × 10⁹³

Therefore, (10π)⁶²⁸⁰⁰⁰⁰⁰⁰⁰⁰ is 10 times larger than 4.5447957777 × 10⁹³

Because π is infinite, you can never have π with no decimals since infinity is not a number but rather an amount. Therefore, π without decimals cannot exist.
 

Orion

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#4
3.14159265358979323846263... × 10 = 31.4159265358979323846263...

31.4159265358979323846263... ^ 62.8 = 4.5447957777 × 10⁹³

Therefore, (10π)⁶²⁸⁰⁰⁰⁰⁰⁰⁰⁰ is 10 times larger than 4.5447957777 × 10⁹³

Because π is infinite, you can never have π with no decimals since infinity is not a number but rather an amount. Therefore, π without decimals cannot exist.
On a slightly more interesting note, who here thinks pi contains itself?
 

Orion

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#6
Justify your position.
 

J3 Mângo

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#8
Pi is one of the oldest mathematical symbols made (250 BC)
 

Altaïr

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#9
π = 355/113
At least it's very close.
 

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#11
Here's another good one (two):
Screenshot_20231020_184829_DuckDuckGo.jpg
 

Altaïr

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#12
I started nicely, but you guys hit really strong with the Ramanujan formulas or the Riemann zeta function :p

Fair, I'll answer with the Gaussian integral:
Screenshot_20231021_131855_Chrome.jpg


There's also that one I like from Ramanujan:
Screenshot_20231021_131723_Chrome.jpg


Even when limiting yourself to the "n=0" term you get π = 9801×√(2)/4412 = 3.14159273... This is already a very good approximation.
 

Orion

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#15
Screenshot_20231021_071528_DuckDuckGo.jpg

Meanwhile the equations:
Screenshot_20231021_071601_DuckDuckGo.jpg
 

TheeZackMichael

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#21
Each digit from right to left doubles each time. In the binary system, if the last digit is 1, it's an odd number. 2 is obviously an even number, so the last digit is 0. The second to last digit is 2, the third to last is 4, the fourth to last is 8, and so on. If I want 10 in binary, I need to add 8 and 2 together so that means 10 in binary is 1010. There's already a 2 in binary so we need that 2 only so 10 is 2 in binary.
 

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#22
Yes,
Its like that,

1+1=2+0
=3-1=1+1
=3-1=4-2
=6-4=4-2
=6̶³4̶¹-=-4̶¹2̶¹ or 6/2 - 4/4
1= 3×1×1×1
= 3
So, 1+1=3
Each digit from right to left doubles each time. In the binary system, if the last digit is 1, it's an odd number. 2 is obviously an even number, so the last digit is 0. The second to last digit is 2, the third to last is 4, the fourth to last is 8, and so on. If I want 10 in binary, I need to add 8 and 2 together so that means 10 in binary is 1010. There's already a 2 in binary so we need that 2 only so 10 is 2 in binary.
ITS THREEEEEEE!!101
 

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#24
3 IS 11, NOT 101!1!1!1!1!1!
You're actually right. But Base 6 is a lot better.
0 dec = 0 sen
1 dec = 1 sen
2 dec = 2 sen
3 dec = 3 sen
4 dec = 4 sen
5 dec = 5 sen
6 dec = 10 sen
7 dec = 11 sen
8 dec = 12 sen
9 dec = 13 sen
10 dec = 14 sen
11 dec = 15 sen
12 dec = 20 sen
Becuase senary has six digits (0,1,2,3,4,5) each hand can represent 6 dec (10 sen) which allows for finger counting up to 30 dec. (I think. I haven't actually done this though)

You see, one hand can represent the 6's place (one's place if you're a decimal freak) and the other hand can represent the 12's place (tens place if you still use decimal)

one hand can count up to 6 dec (10 sen) before needing the other hand to act as a carryover.
 
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#25
I actually wanted to post something, but at some point during the process I noticed that my imagination couldn't be packaged into a graphic or words that humanoid beings or pseudo-humanoid entities could understand...:p