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I was looking at this question (writing a program such that 2 + 2 = 5) and tried to write something in Mathematica, but I couldn't get something to work without completely replacing Plus. Neither of these worked:

Unprotect[Plus]
Plus[2, 2] := 5
Unprotect[Integer]
2 + 2 ^= 5

This book suggests that Plus and Times have special rules. Is this really impossible to do?

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    $\begingroup$ Something like Block[{Plus = Times}, 2 + 3] $\endgroup$ – Jacob Akkerboom Jun 2 '14 at 11:25
  • $\begingroup$ @JacobAkkerboom that's what I meant with "completely replacing Plus" $\endgroup$ – Artefacto Jun 2 '14 at 12:13
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Another solution for this weird exercise is to make a combination from using $Pre and defining a new plus function. You use $Pre to replace every occurrence of Plus by your own definition which only act special at the input plus[2,2] and calls the normal Plus otherwise:

SetAttributes[plus, Attributes[Plus]];
Unprotect[plus];
plus[2, 2] = 5;
plus[args___] := Plus[args];
$Pre = Function[Null,
   ReleaseHold[Hold[#] /. Plus :> plus], HoldAllComplete];

Now, the obvious input 2+2 works as well as e.g. 4/2 + (6 - 4) but I wouldn't bet that this works in all circumstances.

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  • $\begingroup$ A simpler use of \$Pre would be just \$Pre = (# /. Plus[2, 2] :> 5) & $\endgroup$ – Bob Hanlon Jun 2 '14 at 21:57
  • $\begingroup$ Presumably, because Function has the attribute HoldAll. $\endgroup$ – Bob Hanlon Jun 3 '14 at 14:36
  • $\begingroup$ @BobHanlon That solution won't work unless you use HoldPattern and set the attribute HoldAll on the function, as in $Pre = Function[x, Unevaluated[x] /. HoldPattern@Plus[2, 2] :> 5, HoldAll]. Otherwise it will only replace the input 4 with 5, not replace 2+2 with 5. Consider the input 3(2+2) or the input 4. $\endgroup$ – Szabolcs Jun 4 '14 at 14:44
  • $\begingroup$ Disregard my original comment. $\endgroup$ – Bob Hanlon Jun 4 '14 at 17:05
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How about this ;-)

\!\(\*InterpretationBox[2,3]\)+2
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  • $\begingroup$ This one made me laugh. Very creative! $\endgroup$ – Oleksandr R. Jun 2 '14 at 22:38
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This works:

mp[2, 2] = 5
Block[{Plus = mp}, 2 + 2]
(*5*)

Now of course the interesting question is why Unprotect does not work as expected? I have not yet figured it out. Nothing special is seen in the Attributes. Actually doing 

Unprotect[Plus]
Plus[2, 2] = 5
??Plus

shows the new rule.

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  • $\begingroup$ Plus is handled specially and is subject to all sorts of evaluation shortcuts, at least for numeric arguments. In this case it would probably be better to think of the definitions as showing you how Plus behaves, rather than that the behavior actually comes from those definitions as in the usual case. $\endgroup$ – Oleksandr R. Jun 2 '14 at 22:42

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