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 Metamath The Collision In The Octal Equating the field GF(4)'s parts to 0 = (the octal) 1 = (unity) or "hot" x = "cold" x + 1 = "lukewarm" We may relabel this set of four as [0,a,b,c] where we will treat a = 1 as hot and {b,c} as cold and lukewarm, respectively. In the octal we would have (0,a,b,c) acted upon by multiplication, say (a,b,d,c,f,g,e). This would hold static the subgroup (0,b,d,f). Clearly then if "a" is "hot" and "b" "cold" we note that the cycle (a,b,d,c,f,g,e) sends unity to "b" so this seven cycle is actually equal to the element "b". Now this "b" holds static the subgroup [b,d,f] whereas the sum of a and b, the "lukewarm" is the element "c" which upon repeatedly applying the multiplication to find c, we obtain c = (a,c,e,d,g,b,f) which holds {c,e,g} static. This triple is an "antichrist" "bow" which we have seen elsewhere - it will become as "lukewarm". The cycle also holds fixed [b,d,f] since the inverse of the cycle squared sends b=>d=>f. Yet "c" is positioned within the "bow" itself and the elements {c,e,g} correspond to the subgroups that contain unity. There is in this sense a "rest" but only in the coset of {a,c,e,g}, not in the K4 form of the ultrafilter. Then "hot" is to be found in Christ and the Father, having obained rest as "a" in both! The "cold" is to be one of the other two either "x" or "y = x+1". In the father with frobenius this equates to a = 1 = (b,d,f) b = (c,d,g) c = (a,f,g) and the subgroups of the right do not form a group in K4 addition. (Christ is "sat at the right hand" in the form of (b,d,f) in this example, rather than (a,b,c)) In this manner (a,b,c) is clearly opposed to Christ - It is impossible for (a,b,c) to be static and contain unity. To be "cold" or "lukewarm" is then to be as one of the non-unity elements that hold fixed "a" and (0,b,d,f) - to be as "b" is to not have a group of subgroups on the right. (yet "b" only is cold and not "c" for holding fixed (b,d,f) and not being contained in {c,e,g}) Lukewarm then, which is to hold fixed {c,e,g} is then to have a = 1 = {c,e,g} b = {a,e,f} c = {b,d,e} from which we can deduce that "e" corresponds to [a,b,c] And lukewarm is to have a subgroup of both the right and of the groups of the left handed octal(s), but is spiritually to be completely found in the left handed octal, and not in the K4 form of the filter where a = 1 = [b,d,f] = {[a,b,c],[a,d,e],[a,f,g]} which is at the right hand. The individual is found in the earthy elements {a,c,e,g} only. Spiritually then Jesus would rather that a man were "hot" (at rest in Him and the Father) or "cold" (not at rest in the left hand of God, but dwelling in or towards the right) rather than "lukewarm" which is the position where the "hail and fire as mingled with blood." of a falsy rest in the left hand (and the earth) rather than to be found in Christ in the right hand, and also therefore in the K4 form of the ultrafilter. Finally then we should state that "cold" is to be opposed to the K4 form of the ultrafilter - with the same association of a = 1 = [b,d,f] that is the fixed position of the gospel at hand. The lukewarm appears to be in agreement in the octal, but is actually a different fixed position; our example showed that [a,b,c] = "e" rather than a = [b,d,f]. At least the cold has the correct fixed position of "a = 1" even if they are in opposition to the K4 form. The difference is of the static subgroup; and the cold and lukewarm are both "two parts cut off" - One represents the cold in the elements that may not exist in GF(4) intersecting the octal, and the other elements exist in the octal but take a "wider path"than the K4 form, being in the left hand only rather than at rest in Christ - cold and lukewarm respectively. Continue To Next Page Return To Section Start Return To Previous Page