Within an infinitesimally wee portion of the real number line the following 5 simple “equalities” hold for [z], whose definition is to be deduced.  This puzzle strays from strict mathematics as [z] denotes a function of  z , while  =  means “infinitesimally close enough in value to be considered equal.”  All values are real.

 1.           [ [a] ]  =  [u]        for any a but a constant value of u

 2.   [a] + [b] − [c]  =  [a+b-c]    for any a, b, c

 3.         [a] × [b]  =  [a+b]      for any a, b

 4.         [a] / [b]  =  [a-b]      for any a, b

 5.           [a]^[b]  =  [a]        for any a, b

           ( “[a]^[b]” means “[a]  raised to the power of  [b]” ) 
  • What is  [a] − [b]  in terms of  [a-b]without an additional  −  minus sign?

Solutions that correctly identify  u  or define [z] deserve votes of appreciation. A check mark goes to the solution which most cleanly shows its deductive path and most clearly proves equations 1 through 5 (as always along the subjective-yet-persuadable impressions of puzzle’s poser).

  • 1
    $\begingroup$ I'm not quite sure what equation 1 means here - does that mean for all a, there exists exactly one u satisfying the equation? The "but" doesn't quite make sense, because that implies that the quantifiers are independent... $\endgroup$
    – Deusovi
    Nov 1, 2021 at 22:08
  • $\begingroup$ Wait, no, you want us to "identify u" - does that mean [[a]] is a(n infinitesimally-close-to-) constant function? $\endgroup$
    – Deusovi
    Nov 1, 2021 at 22:10
  • $\begingroup$ u is the same for all a and i will try to make that clearer in an edit and yes, @Deusovi, [[a]] is infinitesimally close to [u] $\endgroup$
    – humn
    Nov 1, 2021 at 22:10
  • $\begingroup$ Are all the expressions that occur as arguments of the "[]" function (in particular, u,a+b,a-b,a+b-c and [a]) understood to be part of that "infinitesimally wee portion of the real line" or is this only guaranteed for a,b,c or something different still? $\endgroup$
    – loopy walt
    Nov 1, 2021 at 23:56
  • $\begingroup$ Yes, @loopy walt, everything in [...] square brackets is in the wee portion. I should make that clearer in the puzzle statement. $\endgroup$
    – humn
    Nov 2, 2021 at 3:02

2 Answers 2


I'm not 100% sure about how to answer this question but hopefully somebody else would be able to formalise this into something meaningful.

To answer the main question, I think that

[a] - [b] = [a-b] - 1

The function [z] is something along the lines of

[z] = 1 + εz with ε an infinitesimal

Specifically for the given equations

[ [a] ] = 1 + ε[a] = 1 + ε(1 + εa) = 1 + ε + ε^2 a = 1 + ε = [1]
ignoring the square of the infinitesimal ε (note I will use this throughout).
[a] + [b] - [c] = 1 + εa + 1 + εb - 1 - εc = 1 + ε(a+b-c) = [a+b-c]
[a] x [b] = (1 + εa)(1 + εb) = 1 + ε(a+b) + ε^2 ab = 1 + ε(a+b) = [a+b]
Note how, considering [a] x [-a], this implies that 1/[a] = [-a] so that [a]/[b] = [a] x [-b] = [a-b]
[a]^[b] = (1 + εa)^(1 + εb) = 1 + εa + ε^2(...) = [a]

Note: I haven't completely pulled this out of thin air.

The given equations reminded me of the theory of Lie algebras and how the derivative of the Lie group action at the identity induces a Lie algebra homomorphism. There is a way to formalise this properly but I'm not an expert and it might be too technical for the question at hand anyway.

  • $\begingroup$ You understood how to stray from strict mathematics alright, @hexomino, and got the checkmark even though you added a minus sign to [a-b]. No doubt you can eliminate that minus sign if you'd care to. Ps. Hurray for understanding and mentioning how 1/[a] works. At one stage of pre-flight it was a 6th equation. $\endgroup$
    – humn
    Nov 2, 2021 at 16:26

You don't need to understand what [z] represents.

by 4: 1 = [a]/[a] = [a-a] = [0]
by 2: [0] + [a] − [b] = [0+a-b] = [a-b]
therefore [a-b] = [a] − [b] + [0] = [a] - [b] + 1


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