An odd question re Satan's power


askandanswer

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His power I would think would only be decreased per person by an increase in the number of righteous individuals, if that is an accurate way to view it. Of course, once he wins people to his side I think that would increase his influence or perhaps magnifies it. 

D&C 19:5 Pray always, that you may come off conqueror; yea, that you may conquer Satan, and that you may escape the hands of the servants of Satan that do uphold his work.

Edited by laronius
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2 hours ago, askandanswer said:

Since Satan's power is finite and limited, does it then follow that as human population increases, Satan's power, per person, decreases?

Satan's power is not a finite quantity, like a gallon of milk. Satan is a liar. The more people there are who listen to his lies, the more people there are who are deceived. The only way to dilute the influence of Satan is to cast him out.

I have wondered before if the reality of human nature means that for any large random population of human beings, some fixed percentage--a third, perhaps?--will be especially vulnerable to seduction by Satan's lies. I'm not proposing this as a valid model, just musing.

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While trying to ascertain where I had heard the idea that Satan will be bound during the Millennium because of the righteousness of the Saints, I came across this page with some interesting thoughts:

https://www.ldsscriptureteachings.org/2017/04/01/satan-bound-during-the-millennium/

Edit: And I probably first heard it in The Book of Mormon.

Quote

And because of the righteousness of his people, Satan has no power; wherefore, he cannot be loosed for the space of many years; for he hath no power over the hearts of the people, for they dwell in righteousness, and the Holy One of Israel reigneth.

1 Nephi 22:26

 

Edited by SilentOne
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On 10/7/2022 at 8:46 AM, askandanswer said:

Since Satan's power is finite and limited, does it then follow that as human population increases, Satan's power, per person, decreases?

Well, it seems you haven't taken into account the killoquad redundant molochortz ramostadt differential.  Then you'd need to calculate the bilateral kellolakterals of the differential elements of the sinewave.  If you include the isophilavial constant in the complex heisenberg formula for the pheromantral function denominator, then you'd realize that your premise is a bit off.

See if you can take those into account and get back to me.

Edited by Carborendum
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35 minutes ago, Carborendum said:

Well, it seems you haven't taken into account the killoquad redundant molochortz ramostadt differential.  Then you'd need to calculate the bilateral kellolakterals of the differential elements of the sinewave.  If you include the isophilavial constant in the complex heisenberg formula for the pheromantral function denominator, then you'd realize that your premise is a bit off.

See if you can take those into account and get back to me.

Lol, of course I took these into account, these are fairly basic and fundamental components of the overall calculation. Equally fundamental is the need NOT to include the isdphilavial constant in the Heisenberg formula for the pheromantral function denominator. That actually comes in much later in the process. (Don’t worry, it’s a common rookie mistake to include it prematurely in calculations of this kind and perfectly understandable.) And when we exclude it, as we properly should, we see that my premise is actually right on target. ;)

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4 minutes ago, askandanswer said:

Lol, of course I took these into account, these are fairly basic and fundamental components of the overall calculation. Equally fundamental is the need NOT to include the isdphilavial constant in the Heisenberg formula for the pheromantral function denominator. That actually comes in much later in the process. (Don’t worry, it’s a common rookie mistake to include it prematurely in calculations of this kind and perfectly understandable.) And when we exclude it, as we properly should, we see that my premise is actually right on target. ;)

So, if it is premature to include it in the function's denominator, at what point would be the appropriate time?  After the function is completed, the resultant is solved.  So, we would never include it at all in the calculations... unless that is what you're driving at.  If so, why did you say it was premature instead of unnecessary?

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