The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Is it possible that 3+4=8?
Replies: 21   Last Post: Jan 29, 2011 9:06 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Ilmari Karonen

Posts: 311
Registered: 2/1/07
Re: Is it possible that 3+4=8?
Posted: Jan 23, 2011 6:44 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On 2011-01-23, Craig Feinstein <> 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 self-consistent "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

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) ),


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) = x-1 and g(-x) = ~(1-x)

for all integers x > 1.

Using this correspondence, we could also define a multiplication
operator "#" for our pixie-infested 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 ball-stealing (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 e-mail, please replace ".invalid" with ".net" in address.

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.