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| 10 Oct 2013 06:47 AM |
Ok, so, yeah, probably one of the stupidest things ever, but meh, listen to this for a while.
[Rules:] If a uses [op] by itself, it would be equal to a + (# of times you [op]'d it to itself.) It must have a counterpart.(+/-, ^/sqrt)
[Self-check:] 2 [op] 2 = 4(Any operation with 2, to itself, results in 4.)
[Disproving:] Every operation more complex than the most basic operation(addition), results in x operated by 1, is equal to x. However, if we make a more basic operation, than addition, it would break such pattern, for x ~= x+1. |
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Oysi
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| 10 Oct 2013 07:20 AM |
| ^ Exactly, I don't really understand how it works, but it'd be pretty cool to find a math more basic than addition. |
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Oysi
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| 10 Oct 2013 07:23 AM |
Oysi, think of it as this. 2 + 2 + 2 + 2 + 2 = 2*5.
For the 1st rule: This operation we're finding, which we'll nickname as "op" would do 2 [op] 2 [op] 2 [op] 2 [op] 2 = 2 + 5. |
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| 10 Oct 2013 07:40 AM |
function op(numbers) return 7 end
Stupid Americans...
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Oysi
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| 10 Oct 2013 08:28 AM |
| Well, op must be used just as how addition/multiplication/exponentiation is used. |
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jobro13
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| 10 Oct 2013 01:03 PM |
Aha!
Reminds me of the game 24.
You get a card with 4 numbers. You have to make 24 from those four numbers. Division, multiplication, addition and substraction is allowed and they are all calculated every step (so 2 + 4 + 6 / 3 = 12/3 and not 2 + 4 + 2)
Interesting idea to make a "24 solver". |
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jode6543
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| 10 Oct 2013 02:51 PM |
| I think what he is trying to say is this: if exponents are repeated multiplication, and multiplication is repeated addition, can there be an operation that addition would be a repeat of? |
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rabbidog
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| 10 Oct 2013 02:54 PM |
It's x + 1
Idc if you don't get it. |
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| 10 Oct 2013 03:38 PM |
| Addition is repeated successor. As I just said. |
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| 10 Oct 2013 04:32 PM |
| It's all fine, UNTIL!!! A and B are not the same numbers. |
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rabbidog
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| 11 Oct 2013 12:03 AM |
Or, if you are not working with just integers, it is:
1/infinity + 1/infinity |
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noliCAIKS
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| 11 Oct 2013 10:12 AM |
No, it's not, because 0 * infinity = 0, not 1. PRO TIP: don't try to simplify arithmetic operations on infinity; otherwise you'll get stuff like 2 * infinity = infinity so 2 = 1. |
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| 12 Oct 2013 03:41 AM |
@Oysi
I tried to understand what he means and this is what I could understand:
Basically, he thought of the fact that 4 * 3 can be decomposed in 4 + 4 + 4. Here, you take a complex operation (multiplication, 4 * 3) and decompose it into a sequence of simpler operations (additions, 4 + 4 + 4). His idea is to take 4 + 3 and to decompose it into another simpler operation which we have no name for so far. The idea is that this would be decomposed into 4 [op] 4 [op] 4.
The logic behind this is very, very, very doubtful, but I think that's the idea he had in mind. |
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| 12 Oct 2013 03:49 AM |
| Colorful, that's pretty much, but I'd also like trying to operate 2 different numbers, which is currently out of my knowledge. |
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noliCAIKS
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| 12 Oct 2013 03:59 AM |
| You could compute the integral ∫(1)dx from -A to B? |
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| 12 Oct 2013 04:01 AM |
| How'd we reach to calculus? ;-; |
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