1200-year-old problem

Reply #16

JADELottery — Sat, 16 Dec 2006 22:29:39 GMT

Also, I did a mthematical work out for the X_LOW and X_HIGH equations with 0 and treated 0 as if it were a variable and was able to get the right answer.

Not by adding 1, but just working the equation like 0 is a variable.

02 ¹ 0 and 3 * 0 ¹ 0

02 0 and 3 * 0 0 are computational operations not mathematical operations.

02 and 3 * 0 in an equation, the 0 is a variable and needs to be handled through mathematical equation solving operations not computational operations.

It's also... [ More ]

Reply #15

JADELottery — Sat, 16 Dec 2006 22:26:16 GMT

This problem actually came up in a topic a while back --> The combinatorics of a permutation

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.

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Now, the tricky math part.

Using the values of MIN, MAX, SUM, and PRODUCT we can find the values of X_LOW and X_HIGH.

Taking the SUM of W, X, Y, and Z, if we subtract the values of MIN and MAX we are left with the sum of the two middle numbers X_LOW and X_HIGH.

Sum of X_LOW and X_HIGH: n = SUM - MIN - MAX

Taking the PRODUCT of W, X, Y, and Z, if we div... [ More ]

Reply #14

ayenowitall — Sat, 16 Dec 2006 14:22:06 GMT

Mathematics Forum:

The place to discuss the strong ties between lotteries and mathematics.

Excuse me, but I think I'm missing the strong tie between lotteries and mathematics in this thread.

Reply #13

JAP69 — Sat, 16 Dec 2006 13:56:24 GMT

WHEW

I know how to add, subtract and multiply.

I think.

Reply #12

dartiger — Sat, 16 Dec 2006 13:45:56 GMT

doug I agree

Reply #11

JADELottery — Mon, 11 Dec 2006 22:10:52 GMT

The mathematical operation of X / X = 1 must be continuous for all Real values of X; including {- , 0, }.

1 - If X / X = 1, then - / - = 1

- / - = 1

- = 1 * -

- = - , true \ - / - = 1

2 - If X / X ¹ 1, then - / - ¹ 1

- / - ¹ 1

- ¹ 1 * -

- ¹ - , false \ - / - = 1

3 - If X / X = 1, then 0 / 0 = 1

0 / 0 = 1

0 = 1 * 0

0 = 0, true \ 0 / 0 = 1

4 - If X / X ¹ 1, then 0 / 0 ¹ 1

0 / 0 ¹ 1

0 ¹ 1... [ More ]

Reply #10

JADELottery — Mon, 11 Dec 2006 15:39:23 GMT

Some more:

n =

( n ) / = /

n ( / ) = /

n ( 1 ) = 1

n = 1, \ n = true when n = 1

n ¹

( n ) / ¹ /

n ( / ) ¹ /

n ( 1 ) ¹ 1

n ¹ 1, \ n ¹ true when n ¹ 1

n = -

( n ) / = - /

n ( / ) = - /

n ( 1 ) = -1

n = -1, \ n = - true when n = -1 <--- correction from previous post

n ¹ -

( n ) / ¹ - /

n... [ More ]

Reply #9

JADELottery — Mon, 11 Dec 2006 15:24:12 GMT

And just a bit more:

0 * = n

( 0 * ) / = n /

0 ( / ) = n /

0 ( 1 ) = n /

0 = n / , if and only if n = 0 *

0 * = n

( 0 * ) / 0 = n / 0

( 0 / 0 ) = n / 0

( 1 ) = n / 0

= n / 0

= n / 0 , if and only if n = 0

Reply #8

JADELottery — Mon, 11 Dec 2006 15:04:30 GMT

Some more:

n =

( n ) / = /

n ( / ) = /

n ( 1 ) = 1

n = 1, \ n = true when n = 1

n ¹

( n ) / ¹ /

n ( / ) ¹ /

n ( 1 ) ¹ 1

n ¹ 1, \ n ¹ true when n ¹ 1

n = -

( n ) / = - /

n ( / ) = - /

n ( 1 ) = -1

n = -1, \ n = true when n = -1

n ¹ -

( n ) / ¹ - /

n ( / ) ¹ - /

n ( 1 ) ¹ -1... [ More ]

Reply #7

JADELottery — Mon, 11 Dec 2006 10:50:28 GMT

In addition, -1 is the infinitesimal unit and is exactly and only exactly the first number after 0.

Reply #6

JADELottery — Mon, 11 Dec 2006 10:39:09 GMT

Also, 0 * is a non-reduceable number much like 2i is a non-reduceable number, it is what it is, a number all by itself.

Reply #5

JADELottery — Mon, 11 Dec 2006 10:15:00 GMT

Something of a proof why -1 ¹ 0 and -1 > 0

1 - If -1 = 0 , then 0 + -1 = 0

0 + -1 = 0

0 + ( 1 / ) = 0

0 ( / ) + ( 1 / ) = 0

( ( 0 * ) / ) + ( 1 / ) = 0

( 0 * + 1 ) / = 0

0 * + 1 = 0 *

( 0 * ) - ( 0 * ) + 1 = ( 0 * ) - ( 0 * )

0 + 1 = 0

1 = 0

1 = 0 , is false \ -1 ¹ 0

2 - If -1 > 0 , then 0 + -1 > 0

0 + -1 > 0

0... [ More ]

Reply #4

JADELottery — Mon, 11 Dec 2006 04:03:23 GMT

0 -1 = (1 / 0) = and -1 = (1 / ) = 0 are incorrect and 0 -1 is a new number I, the Hyper-infinite number.

1 - Reciprocal Operation with Inequality

a > b

taking the reciprocal of both side causes the > to change to <.

reciprocal of a > b is a-1 < b-1

a > ba-1 < b-1

aba-1b-1

10.251.000004.00000

10.51.000002.00000

210.500001.00000

21.50.500000.66667

1050.100000.20000

1090.100000.11111

10005000.001000.00200

10009000.001000.0... [ More ]

Reply #3

JADELottery — Mon, 11 Dec 2006 03:17:55 GMT

I see two incorrect statements:

1. 0-1 = (1 / 0) = , is not correct

2. -1 = (1 / ) = 0 , is not correct

Reply #2

for99c — Sat, 09 Dec 2006 03:22:42 GMT

http://www.bookofparagon.com

Reply #1

MADDOG10 — Sat, 09 Dec 2006 02:40:20 GMT

That was actually solved 1199 years ago, by Dr.goodfornuttin...

1200-year-old problem

for99c — Sat, 09 Dec 2006 02:20:50 GMT

Dr James Anderson, from the University of Reading's computer science department, says his new theorem solves an extremely important problem - the problem of nothing.

4/4 = 1 3/3 = 1 2/2 = 1 1/1 = 1 0/0 = 1