# TobyBartels comments on Intellectual Hipsters and Meta-Contrarianism - Less Wrong

128 13 September 2010 09:36PM

You are viewing a comment permalink. View the original post to see all comments and the full post content.

Sort By: Best

You are viewing a single comment's thread.

Comment author: 17 September 2010 11:02:26PM 5 points [-]

As a mathematician, I offer my services for anybody who wants arguments (mathematical arguments, not philosophical ones) that 1+1 = 3. But beware: as a meta-contrarian mathematician, I will also explain why these arguments, though valid in their own way, are silly.

Comment author: 28 February 2012 11:50:05PM 4 points [-]

1.3 + 1.4 = 2.7, which when reported to one significant figure...

Comment author: 29 February 2012 12:30:09AM 1 point [-]

As the "old" computer science joke goes, 2 + 2 = 5 (for extremely large values of 2).

Comment author: 29 February 2012 01:08:21AM *  0 points [-]

The physicist-typical version is that 3=4, if you take lim(3->4).

Comment author: 01 March 2012 05:15:34AM 1 point [-]

This reminds me that the difference between a physicist and astronomer is that a physicist uses π ≈ 3 while an astronomer uses π ≈ 1.

Comment author: 01 March 2012 10:52:55AM *  3 points [-]

I remember someone in a newsgroup saying the average person is about one metre tall and weighs about 100 kilos, and when asked whether maybe they were approximately a bit too roughly, they answered “I'm an astronomer, not a jeweller.”

(And physicists sometimes use π ≈ 1 too -- that's called dimensional analysis. :-) The problem is when the constant factor dimensional analysis can't tell you turns out to be 1/(2π)^4 ≈ 6.4e-4 or stuff like that.)

Comment author: 20 January 2011 07:37:59PM 1 point [-]

You know, I am seized with a sudden curiosity. You have arguments such that 1 is still the successor of 0 and 3 is still the successor of the successor of 1, where 0 is the additive identity?

Comment author: 25 January 2011 06:03:51PM *  2 points [-]

Ah, now I have to remember what I was thinking of back in September! Well, let's see what I can come up with now.

One thing that I could do is to redefine every term in the expression. You tried to forestall this by insisting

1 is still the successor of 0 and 3 is still the successor of the successor of 1, where 0 is the additive identity

[Note: I originally interpreted this as "3 is still the successor of 2" for some dumb reason.] But you never insisted that 2 is the successor of 1, so I'll redefine 2 to be 1 and redefine 3 to be 2, and your conditions are met, while my theorem holds. (I could also, or instead, redefine equality.)

But this is silly; nobody uses the terms in this way.

For another method, I'll be a little more precise. Since you mentioned the successor of 0, let's work in Peano Arithmetic (first-order, classical logic, starting at zero), supplemented with the axiom that 0 = 1. Then 1 + 1 = 3 can be proved as follows:

• 1 + 1 = 1 + S(0) by definition of 1;
• 1 + S(0) = 1 + S(1) by substitution of equality;
• 1 + S(1) = 3 by any ordinary proof in PA;
• 1 + 1 = 3 by transitivity of equality (twice).

Of course, this is also silly, since PA with my new axiom is inconsistent. Anything in the language can be proved (by going through the axiom that 0 = S(n) is always false, combining this with my new axiom, and using ex contradictione quodlibet).

Here is a slightly less silly way: Modular arithmetic is very useful, not silly at all, and in arithmetic modulo 1, 1 + 1 = 3 is true.

But however useful modular arithmetic in general may be, arithmetic modulo 1 is silly (for roughly the same reasons that an inconsistent set of axioms is silly); everything is equal to everything else, so any equation at all is true. In other words, arithmetic modulo 1 is trivial.

You can get arithmetic modulo b by replacing the Peano axiom that 0 = S(n) is always false with the axiom that 0 = b and b − 1 additional axioms stating (altogether) that a = b is false whenever (in ordinary arithemetic) 0 < a < b. But you could instead add an arbitrary axiom of the form b = c (and another finite set of inequalities between smaller numbers). So let us use the arithmetic given by the axiom that 2 = 3. Then 1 + 1 = 3 is easy to prove (since the proof that 1 + 1 = 2 doesn't rely on the axiom that we've removed, and we still have transitivity of equality). And yet this system is not trivial; it is basically (0, 1, 2, 2, 2, …).

Actually, this example is minimal; let's go for a little overkill and instead use the axiom that 1 = 2. Of course, we can still prove that 1 + 1 = 3 (this time leaving the formal proof entirely to the reader). This system is a bit more trivial than the last one, but not quite trivial; it is basically (0, 1, 1, 1, …).

Now, although these systems of arithmetic are nontrivial, I really ought to give some mathematical reasons why anybody would be interested in them at all. I can give several profound reasons for the last one, which I will skip on the grounds that you can find them elsewhere; it boils down to this: this system is the system of truth values in classical logic (Boolean algebra). I don't even have to tell you how to interpret 0, 1, and + (much less =) in this system; I'm simply using them with their standard meanings in this context.

So now that you see that 1 + 1 = 3 in Boolean algebra, I need to turn around (as meta-contrarian) and explain why it is still silly. One reason is that nobody doing Boolean algebra (or even the slightly less trivial system of arithmetic based on 2 = 3) should ever want to write "3"; they should just write "1" (or "2" in the other system) instead. Another reason is that you shouldn't just throw "1 + 1 = 3" or even "1 + 1 = 1" out without explanation; the default meanings of those terms are in Peano arithmetic (or an extension thereof), not Boolean arithmetic. So in a general context, you really ought to say "1 + 1 = 3 in Boolean arithmetic", or something like that, instead. Just saying "1 + 1 = 3" and expecting people to know what the heck you're talking about is, well, silly.

I have no idea if that's what I was thinking in September, but that's what I thought of now. I hope that you like it.

Comment author: 25 January 2011 06:12:05PM 2 points [-]

1 is still the successor of 0 and 3 is still the successor of the successor of 2 [you wrote 1 here, but I understand that this was a typo], where 0 is the additive identity

I wrote "successor of the successor of" - 3 is the successor of 2, which is the successor of 1. But I understand that this was a typo. :P

But yes, I enjoyed that. Thank you.

Comment author: 25 January 2011 06:14:40PM 1 point [-]

Ha, that would be a reado!

But seriously, I should have read that again. I got it in my head that you had done this while I spent time planning my response and forgot to verify.