
Re: Is it possible that 3+4=8?
Posted:
Jan 23, 2011 6:44 PM


On 20110123, Craig Feinstein <cafeinst@msn.com> wrote: > Let's pretend that we live in a world where there are invisible pixies > running around messing everything up. Let's say you have 3 balls in > your left hand and 4 balls in your right hand. Whenever you combine > the balls together, one of these invisible pixies secretly throws in > another ball, so that whenever you count the balls, you come up with 8 > balls. So in this pretend world, it is a known fact that 3+4=8. Anyone > who disputes this fact in our pretend world is considered either crazy > or stupid. > > In our real world, it is a known fact that 3+4=7. But how can we be > sure that there are no invisible pixies running around taking balls > away from us causing us to think that 3+4=7 when really 3+4=8? > > My point is that mathematics is considered a deductive science, in > which everything is absolutely certain. But how can mathematics prove > that the above scenario cannot be true?
I'm going to ignore your point and just consider whether it's possible to have a selfconsistent "addition" where 3+4=8. Or, rather, since I want to also talk about normal addition without getting the two mixed up, I'm going to reserve the sign "+" for that, and use "&" for the other "addition", so that 3+4=7, but 3 & 4 = 8. Or, more generally,
x & y = x + y + 1
for any positive integers x and y.
I'll asume that this equation holds in both directions, so that if you split your pile of 8 balls into two piles, the same invisible pixies who added the extra ball will take it back so that you're left with piles of 3 and 4 (or 2 and 5, or 1 and 6) balls. Otherwise you could obtain an arbitrary number of balls (or make an arbitrary number of them vanish) just by splitting and merging piles repeatedly.
I'll also assume that
x & 0 = 0 & x = x
for all x, since the adding no balls to a pile (or taking no balls out of it) shouldn't change its size. (Otherwise I could again obtain an arbitrarily large pile of balls just by repeatedly adding 0 balls to it, forcing the pixies to keep throwing in extra balls.)
So far so good; our "addition" satisfies all the usual laws we'd expect it to, such as
x & y = y & x (commutativity) and x & (y & z) = (x & y) & z (associativity).
We also have the equivalence
a & x = a & y <=> x = y,
so we can solve equations in the usual way by introducing and removing common terms from both sides. However, we can't solve all equations that way; for example, the equation
8 & x = 3
isn't satisfied by any natural number x. If it was, though, we could apply the laws given above to show that x would also have to satisfy
4 & x = 0.
The usual way of handling this situation, of course, is to introduce for each natural number x an additive inverse ~x, defined such that
x & ~x = ~x & x = 0.
Of course, ~0 = 0. We can also show just from the definition that
~(~x) = x (cancellation) and ~x & ~y = ~(x & y) (distributivity of inverse).
However, whereas with ordinary addition we could then show that any equation using only addition and one unknown term could be solved by natural numbers or their inverses, this is not true with the otherworldly addition. For example, the equation
1 & x = 2
still doesn't have a solution. In particular, we can rewrite this equation as
x = 2 & ~1,
which still has no solution  that is, there is no number which equals 2 & ~1!
What does this mean in terms of the pixies? It means that the pixies can't let us divide a pile of 2 balls into two piles. Why not? Because if we could do that and the pixies took one of the balls away, we'd be left with just 1 & 0 = 1 ball, and thus could make any number of balls vanish by repeatedly splitting two balls out and one ball in to a large pile. Whereas, if the pixies _didn't_ take a ball away when we split a pile of two balls, we'd be left with 1 & 1 = 3 balls and could thus use the same process to obtain an unlimited number of balls.
But we'd still like to have a solution to that equation, even if it wasn't a proper number, if only to preserve the arithmetic laws I gave above. Can we do the formal solution trick again? Let's introduce a special number "o" which satisfies 1 & o = 2. We could think of o as a "ghost" pile which contains no balls but still somehow isn't quite empty. By applying the arithmetic laws above, we can then show that
x & o = x + 1
for all positive integers x: when we add the ghost pile to a real pile of balls, the pixies throw in an extra ball just as if we'd added a real pile of balls to it.
What about adding two of these ghost piles together? We know that
x & (o & o) = (x & o) & o = x + 2 = x & 1,
so it would seem to make sense to define
o & o = 1.
From here, a bunch of other equations naturally follow:
o & 0 = 0 & o = o, o & ~o = 0, 1 & ~o = o, o & ~1 = ~o, x & ~o = x  1, and o & ~x = x + 1
for all x > 1.
In fact, it turns out that this system is now closed in the sense that all equations with addition and one unknown term have a solution; in particular, all numbers have inverses and any two numbers can be added to yield a third.
It also turns out that the otherworldly addition as we have defined it corresponds in a simple way to ordinary addition. In particular,
x & y = g( f(x) + f(y) ),
where
f(0) = 0, f(o) = 1, f(~o) = 1, f(x) = x+1 and f(~x) = (x+1)
for all positive integers x, and g is the inverse of f, i.e.
g(0) = 0, g(1) = o, g(1) = ~o, g(x) = x1 and g(x) = ~(1x)
for all integers x > 1.
Using this correspondence, we could also define a multiplication operator "#" for our pixieinfested world, e.g. as
x # y = g( f(x) f(y) ).
However, multiplication defined like this looks a bit weird, even by the standards of people used to invisible pixies: for example, the multiplicative identity becomes o, not 1. It also fails to satisfy the basic rule that
x # y = x & x & ... & x
where x occurs y times on the right hand side. (Instead, the equation holds if x occurs f(y) times on the right.)
A more sensible way to define multiplication might be
x * y = g( f(x) y ),
where x is an otherworldly number but y is an ordinary integer like in our world. (In particular, x * o is not defined.) This version of multiplication isn't commutative, so in general x * y =/= y * x even if both are defined. However, it does seem to make sense in the weird and wacky world of ballstealing (and donating) pixies: after all, with the pixies involved, you get more balls by merging five piles of two balls than two piles of five balls.
This also implies that, while an inhabitant of the pixie world would use the otherworldly numbers and "&" to count and add together balls, they'd still use our ordinary numbers and arithmetic to count piles of them. Which does make sense  after all, the pixies only toss in and steal away single balls, not entire piles.
 Ilmari Karonen To reply by email, please replace ".invalid" with ".net" in address.

