Results 1 to 8 of 8

- December 16th 2009, 04:48 AM #1

- Joined
- Dec 2009
- Posts
- 79

- December 16th 2009, 04:54 AM #2

- Joined
- Jun 2009
- From
- Africa
- Posts
- 641

- December 16th 2009, 04:59 AM #3

- Joined
- Dec 2009
- Posts
- 79

- December 16th 2009, 05:41 AM #4

- December 16th 2009, 06:14 AM #5

- Joined
- Jun 2009
- From
- Africa
- Posts
- 641

- December 16th 2009, 06:18 AM #6

- Joined
- Jun 2009
- From
- Africa
- Posts
- 641

- December 16th 2009, 03:47 PM #7

- Joined
- Dec 2009
- Posts
- 79

- December 18th 2009, 07:15 AM #8

- Joined
- Dec 2009
- Posts
- 79

This my way of proof

1 - can not be any of the X, Y number is negative because in this case it would output the left end in a fraction equations absolute value smaller than 1

2 - can not be any of the X, Y equal to zero because in this case would be the left end result = either 1 or zero or undefined

3 - can not be any of the X, Y equal to 1 because it has achieved one of the equations can not be achieved while the other

And therefore there is no option for X, Y only be a Moajabin greater than 1

4 - can not be greater than Q Ooisawi 3 because when you is no more achieved the first equation and therefore the only value for Q is 2

5 - can not be greater than p. 4 Ooisawi because it is no more realized when the second equation and therefore the only value of r is 3

What do you think