Written In Black And White
It is vogue amongst many students of the book of revelation to assume that the 144,000 sealed saints are "super christians" and there is a third and broader way for most believers who must go under the heel of the beast whilst these sealed saints escape the tribulation somehow - that they are present on earth during the wrath of God but are immune to the beast through some great protection.
That, however is a nonsense, because God has no preference for one of His elect over another - and it is sensible to point out that if you are not a "sealed" christian - you must somehow be considered inferior by God - but if that is the case how are you saved? Logic would point out that men are created equal - and there are no classes amongst the saved, and the "unsaved" are not "inferior". That is then worse than what would be racism. It would make us all hypocrites as we were all sinners. Rather the saved are saved and the non-sealed are not the elect and there is preservation of the election by faith, and faith alone.
The calculation then, that states whether you are sealed or not is as follows. There are 7 choices for a floating unity in the octal, and for each choice of unity there may apply eight groups of seven cycles. Alternatively, in seven symbols there are 35 ways to choose three elements. Then we immediately have the ways to choose all possible K4 subgroups of the generalised octal under those seven cycles for multiplication. Now under multiplication the static subgroup may not contain the unity of the finite field of eight elements (under frobenius) so we may state as follows that there are...
35 * 4 = 140 such possible arrangements of static K4 subgroups The factor of four comes from one of the remaining non-zero elements chosen as the unity element. Now, each one of these 140 excludes just one of the others as implausible, it belongs in the finite field as a static triple, reffered to as an antichrist "bow" elsewhere on the site. (In the complement octal or "left hand".) So we actually have...
(35 * 4) - 1 = 140 - 1 = 139.
Lastly we note that there are six possible octals (under twelve seven cycles, each octal sharing two such seven cycles) containing each one of those K4 subgroups, giving us 840 - 6 = 139 * 6 = 834 possible octal containers for a generalised static K4 subgroup in seven symbols under it's various multiplications.
However if you are a sealed Christian then you have an identification or association with the unity element in the octal, and we have 24 multiplicative automorphisms (three each for each valid C7 group preserving the subgroups under multiplication) over 7 choices of (a floating) unity. (Note that every possible antichrist bow is present in the 28 possible left handed triples (35 - 7, the seven in the right hand) , so we only need reference the one right handed "Sun octal")
Now, simply stated, the total number of sealed is not 144, but we must include the 24 elders that surround the throne of God referencing the primacy of the Lord as the reference, the "unity" or origin of the "Sun octal" itself. Then the 144 refer to the secondary elements of the floating unity's other six possible positions.
So we have 144 + 24 = 7 * 24 = 168.
Without further ado, we simply state that 834 - 168 = 666.
Again we have some sleight of hand, because we should rather state that automorphisms of a field do not provide additional octals. We have to provide otherwise for 168 /3 or include a factor of three to cancel out the effect of the one field under three automorphisms.
We actually have so far, 7 * (24/3) = 56. We would wish to recover the factor of three to get 834 - (3 * 56) = 666.
For any particular octal we have those same 56 elements. Rewinding above for a while, we could state that before we counted 840 - 6 we in fact had 6 * (140 - 1). In fact we may confidently state we have 6 * (140 - 1 - (56 / 2)). We halve the 56 because we did count static triples rather than the pair of seven cycles on each octal permitting each of those triples.
So, in order to commensurate properly the 144 with the 139 we actually have to remove down to four pairs of seven cycles on 7 choices of unity per octal group. (giving us 28 rather than 168)
So we again find 6 octals per static (subgroup/bow) pair (leaving the unity element uniquely determined.) So our calculation is correctly
6*(139 - 28) = 6*111 = 666.
Thus the number of sealed correctly align to the uniqueness of unity rather than a set of three found in either the elements of the static K4 subgroup or antichrist bows. Then "In all the land I cut off two parts, and the thrid I bring through the fire." as Zechariah related. You either find yourself opposing Christ, or placing yourself in His place without reason - or you are in a unique position of rest. (Being sealed.)
So, if you are not sealed, you have the number of the beast. There is no "middle way" and if you doubt you are sealed (especially if you are part of a church that profess you have no such witness for example) you need Jesus Christ, and you need Him now. If you simply state that you must be sealed already because you are a Christian but were previously unsure or unaware, you are in deadly peril.. you need effective prayer alone with God, not as in a church full of the methods used in the beast and its image and those with the mark amongst you.
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