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This word is slipping my mind and it is driving me crazy. Not sure if it is a math term or computer science term, but I use it a lot in development (when I can remember it!)

Basically, you have the system in some state S. You perform action A on the state putting it in S'. Now you apply the opposite action, ~A. The system is now back to state S because this set of actions acting on this system are _____.

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    .... Reversible.
    – Edwin Ashworth
    Commented Jun 1, 2015 at 19:00
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    Well, I'm almost certain the word you're looking for is idempotent, but it doesn't have the meaning hot describe. The state of idempotent systems do not change in response to external stimuli. So applying any external stimulus is, in a sense, "safe".
    – Dan Bron
    Commented Jun 1, 2015 at 19:02
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    If Action P sets a state and Action Q unsets it, P and Q might be said to be complementary.
    – Jim Mack
    Commented Jun 1, 2015 at 19:36
  • @DanBron Yes, sorry, the description does not fit the meaning exactly and I probably use it somewhat inappropriately since I do not know of a word that makes more sense. In the case I am describing, somebody is toggling a button in some software (think: click on, click off, click on), which should not change the overall state. Thanks, that was the one.
    – Ryan
    Commented Jun 1, 2015 at 19:39
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    Reciprocal? Inverse?
    – Lefty
    Commented Jun 1, 2015 at 22:30

1 Answer 1

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It sounds to me like an Inverse Function, which is beautifully explained in this Youtube video. He takes a number, plugs it into an equation with an inverse function, and the answer is the number he started out with.

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    @JJJ - Because it's the inverse function of A.
    – Bread
    Commented Mar 18, 2018 at 5:51
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    @JJJ - You have explained that in your own answer which is right here on the same page. So why do you feel you must argue your case in the comments of my answer?
    – Bread
    Commented Mar 18, 2018 at 6:40
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    Mathematician here. @Bread is correct: the two actions are inverses of each other.
    – Brendan W. Sullivan
    Commented Mar 28, 2018 at 16:25
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    @JJJ: No, you are very much mistaken. The Schroder-Bernstein Theorem says that if you have an injective function f:A->B and an injective function g:B->A then there must exist some bijective function h:A->B. However, this does not necessarily imply that f and g must have been inverses of each other to begin with. (Indeed, if that were the case, then f would already necessarily be a bijection from A to B, and we would not even require the result of the theorem.)
    – Brendan W. Sullivan
    Commented Mar 28, 2018 at 18:06
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    @JJJ: Always glad to put my expertise to use :-)
    – Brendan W. Sullivan
    Commented Mar 28, 2018 at 18:13

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