Tetrits, just checking FTIA...

[Dan Zuras Intervals <intervals08@xxxxxxxxxxxxxx>]

	Nate's recent assertion that {domainFalse} would serve just
	as well as {} caused me to double check if the fundamental
	theorem holds for either decoration assignment.

	I find that it holds for {} but not for {domainFalse}.

	Recall, for divide we will define:

		domainFalse = {There exists x in xx & y in yy such
			that x/y is not in the domain of divide}
		(i.e. an x & y such that y = 0)

		domainTrue = {There exists x in xx & y in yy such
			that x/y is in the domain of divide}
		(i.e. an x & y such that y != 0)

	Here are two towers of ever decreasing intervals.  Each has
	the property that the operands are contained in the operands
	above & the results are contained in the results above,
	including decorations:

		[1,2]/[entire] = {[entire],{domainFalse,domainTrue}}
		[1,2]/[-2,2] = {[entire],{domainFalse,domainTrue}}
		[1,2]/[0,2] = {[1/2,oo],{domainFalse,domainTrue}}
		[1,2]/[0,0] = {[empty],{domainFalse}}
		[1,2]/[empty] = {[empty],{}}
		[empty]/[empty] = {[empty],{}}

		[1,2]/[-2,2] = {[entire],{domainFalse,domainTrue}}
		[1,2]/[1,2] = {[1/2,2],{domainTrue}}
		[1,2]/[1,1] = {[1,2],{domainTrue}}
		[1,2]/[empty] = {[empty],{}}
		[empty]/[empty] = {[empty],{}}

	Notice that in the second tower a decoration of {domainFalse}
	for an [empty] result would violate the fundamental theorem.

	Here is a similar pair of towers for sqrt():

		sqrt([entire]) = ([0,oo],{domainFalse,domainTrue}}
		sqrt([-2,4]) = ([0,2],{domainFalse,domainTrue}}
		sqrt([-2,-1]) = ([empty],{domainFalse}}
		sqrt([empty]) = ([empty],{}}

		sqrt([entire]) = ([0,oo],{domainFalse,domainTrue}}
		sqrt([0,4]) = ([0,2],{domainTrue}}
		sqrt([1,1]) = ([1,1],{domainTrue}}
		sqrt([empty]) = ([empty],{}}

	I just had to check...


				   Dan