A Craps Tutorial

Other Topics Section    --    Bracket Function Properties

	Theorem   6
		Given any real number x, let   c1   =   -   0.5 - x     and   let   c2   =     x + 0.5   .   If x is an integer,   then c1 and c2 will both equal x.   But if x is not an integer then x must lie somewhere between two consecutive integers   x     and     x + 1.   If x happens to lie exactly half way between   x     and     x   + 1,   then   c1 will equal   x   and c2 will equal   x   + 1. Otherwise, c1 and c2 will both equal the nearest integer to x.

First we will show that when x is an integer c1 and c2 both equal x.
Assume that x is an integer.   Then by theorem 1,
  0.5 - x     =   -x   +     1 / 2   →   c1   =   - ( -x )   =   x.

Also by theorem 1,   c2   =     x + 0.5     =   x   +     1 / 2     =   x.

Now assume that x is any real number.

Regardless of whether or not x is an integer,
there must exist a real number   α   such that
x   =     x   +   α      where 0 ≤ α < 1.    See Fig 1.

	Let n   =   the nearest integer to x
	Note that
		α   <   1 / 2   ↔   n   =     x
		α   >   1 / 2   ↔   n   =     x   + 1
		α   =   1 / 2   ↔   n   =   ambiguous

	Let's start with	x	=	x   +   α   where 0 ≤ α < 1	statement 1
	Add 0.5 to both sides	x + 0.5	=	x   +   α   +   0.5	statement 2
	By thm 1,	x + 0.5	=	x   +   α + 0.5	statement 3
	From statement 1,	-x	=	- x   -   α	statement 4
	Add 0.5	0.5 -x	=	- x   + ( 0.5   -   α )	statement 5
	By theorem 1,	0.5 - x	=	- x   +   0.5 - α	statement 6
	Multiply by -1,	- 0.5 - x	=	x   -   0.5 - α	statement 7
	Split into 3 cases according to the value of α.
	Case 1    α < 0.5
	Then n   =   closest integer to x   =   x
	From the inequality in statement 1,			0   ≤   α   < 0.5	statement 8
	Add 0.5 throughout			0.5 ≤ α + 0.5 < 1	statement 9
	From statement 3,	x + 0.5	=	x   →   c2 = n	statement 10
	Thus   c2 is the closest integer to x.
	From statement 1,	0.5 -x	=	- x   +   ( 0.5 - α )	statement 11
	By theorem 1,	0.5 -x	=	- x   +   0.5 - α	statement 12
	From the inequality in statement 8,			0   ≥   -α   > -0.5	statement 13
	which implies			0.5   ≥   0.5 -α   > 0	statement 14
	And so, by stmt 12 and definition of c1,			0.5 - α is 0.	statement 15
	Therefore,   c1 = x - 0 = n;     i.e.   c1   =   the closest integer to x.				statement 16
	Case 2    α > 0.5
	Then	n	=	closest integer to x = 1 + x
	By thm 1 & stmt 2,	c2	=	x + 0.5
			=	x   +   α + 0.5	statement 17
	But   0.5   <   α   <   1   →   1   <   α + 0.5   <   1.5   →   α + 0.5   =   1				statement 18
	So,	c2	=	x   +   1	statement 19
	By thm 1 & stmt 7,	-c1	=	0.5 - x
			=	- x   +   0.5 - α	statement 20
	Thus,	c1	=	x   -   0.5 - α	statement 21
	But   0.5   <   α   <   1   →   -0.5   >   -α   >   -1   →   0   >   0.5 - α   >   -0.5				statement 22
	Therefore,	0.5 - α	=	-1.	statement 23
	And so,   by stmts 21 and 23,        c1		=	x   -   ( -1 )
			=	1   +   x
	Case 3    α =   0.5
	Then	n	=	closest int to x     is ambiguous
	From statement 5,	0.5 - x	=	- x	statement 24
	By theorem 1,	0.5 - x	=	- x   →   c1   =   x	statement 25
	From statement 2,	x + 0.5	=	x + 1	statement 26
	Therefore,	c2	=	x + 0.5	statement 27
			=	1 + x