Wikipedia:Reference desk/Archives/Mathematics/2009 June 1

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June 1

Could set theory (ZFC) be wrong?

I've wondered this for a while. Most people these days believe ZFC is consistent on intuitive and other grounds, so let's assume that it really is consistent. What reason do we have to believe that it is arithmetically sound? Could ZFC be consistent but false in the sense that it proves arithmetic statements that are false about the standard integers? Nik Weaver wrote an article suggesting this is actually the case, that ZFC has no models whose ω is the standard integers and that it is $\Sigma _{1}$-unsound, or anyway suggesting that we don't have much evidence to the contrary. But, I get the impression that his writings are considered a bit out of the mainstream. If we assume PA is sound, then since ZFC proves PA, any arithmetically false theorem of ZFC would have to be independent of PA, already putting such theorems in a pretty rarefied realm. Are there some numerical experiments that might give evidence one way or the other? 67.122.209.126 (talk) 00:40, 1 June 2009 (UTC)

Wouldn't any statement about arithmetic independent of PA automatically involve infinite sets and consequentially be inaccessible to experiments? —Preceding unsigned comment added by 84.187.84.107 (talk) 02:54, 1 June 2009 (UTC)

Certainly not (e.g. PA's own consistency statement). Per Gödel's incompleteness theorem there will always be independent statements expressible within the system for any sufficiently powerful axiom system (such as PA). Goodstein's theorem is a well known theorem that can't be proved in PA, but I think it can be proved if you explicitly include an assumption of PA's consistency (i.e. it is provable in PA+CON(PA)). Kruskal's tree theorem is a well known theorem that can be stated in finitary terms but can't be proved without considerably stronger axioms than PA. But is Kruskal's theorem true? Possibly not, if the axioms used to prove it aren't true. 67.122.209.126 (talk) 05:27, 1 June 2009 (UTC)

... that ZFC has no models whose ω is the standard integers...: right. By definiton, the standard integers validate only true arithmetical statements.

... and that it is $\Sigma _{1}$-unsound...: wrong. It's perfectly possible that ZFC is arithmetically unsound (or even $\Sigma _{2}$-unsound) but $\Sigma _{1}$-sound.

... already putting such theorems in a pretty rarefied realm...: this "rarefied" realm contains a wealth of sentences of complexity as low as $\Sigma _{1}$. ZFC positively proves plenty of arithmetical sentences independent of PA (and considered true), so it's in no way rare, unusual, suspect, or paradoxical. — Emil J. 12:12, 1 June 2009 (UTC)

About Sigma-1, I was stating Weaver's conjecture that ZFC is Sigma-1 unsound, which is of course an even stronger statement than ZFC being unsound at some higher quantifier depth. It turns out this topic (of ZFC unsoundness) was discussed extensively on the FOM mailing list earlier this month: [1], see many posts from Weaver and others on "arithmetical soundness of ZFC" in that thread. I just came across this while googling.

Of course ZFC in the abstract proves many arithmetical statements that PA does not. By "rarefied", I mean there aren't many such theorems that actually turn up in mathematics as practiced by humans, e.g. Kruskal's theorem. All the examples I can think of are provable in second-order arithmetic, but I'm no expert. I do know about Harvey Friedman's artificial examples that use ZFC, large cardinals, etc. 67.122.209.126 (talk) 14:15, 1 June 2009 (UTC)

There are extremely few "natural" mathematical statements that can be stated in second order arithmetic but are not provable in second order arithmetic. However there are many "set theoretic" statements, such as $V\cap \mathbb {R} =L\cap \mathbb {R}$, that can be stated (with effort) in second order arithmetic but are not provable even in ZFC. The best example of a theorem that is provable in ZFC but not second-order arithmetic is probably Borel determinacy. — Carl (CBM · talk) 23:15, 2 June 2009 (UTC)

Regarding Weaver's claims, I do not see them as very strong, and I did not see any support for them on the FOM list. If ZFC is not arithmetically sound then second-order ZFC would be inconsistent, which would be remarkable. — Carl (CBM · talk) 23:19, 2 June 2009 (UTC)

Hmm, what does it mean for second-order ZFC to be inconsistent? If ZFC is arithmetically unsound, I can see how that would imply that second order ZFC has no models. But the completeness theorem doesn't apply to second order logic, so (I could very well be missing something) I don't see how the lack of models means that the axioms prove a contradiction like in first order logic. Is there a necessarily a way to reach an actual contradiction in ZFC-2, if ZFC-1 is unsound? If not, is ZFC-2's lack of models in the event of ZFC-1's unsoundness still remarkable? Thanks. 207.241.239.70 (talk) 03:01, 4 June 2009 (UTC)

I was just referring to semantical inconsistency. I do think that the inconsistency (= lack of a model) of second order ZFC would be very surprising. For example, there is a classical result of Zermelo that models of second order ZFC are exactly of the form V_κ for κ an inaccessible cardinal. So an inconsistency in second-order ZFC would apparently have great ramifications for our understanding of the cumulative hierarchy. Of course one would have to see the proof of inconsistency before these ramifications could be worked out. — Carl (CBM · talk) 04:14, 4 June 2009 (UTC)

Just to check I understand: when Nik Weaver claims that ZFC is likely $\Sigma _{1}$-unsound, he means that probably either asserts the existence of functions $N\rightarrow N$ that are not functions over the naturals in the standard model, or denies the existence of such functions that are? And, since his complaint is that ZFC overgenerates, presumable that ZFC asserts the existence of such phantom functions? My gut feeling is that overgeneration here would lead to inconsistency. — Charles Stewart (talk) 08:37, 4 June 2009 (UTC)

Weaver means that ZFC might prove a $\Sigma _{1}^{0}$ sentence such that there is no actual natural number that satisfies the existential quantifier. For example, the negation of Goldbach's conjecture is of this form. — Carl (CBM · talk) 11:38, 4 June 2009 (UTC)

Sure, and for the purposes of soundness, statements proving the totality of functions are $\Sigma _{1}^{0}$, since the open formula one gets by dropping the universal quantifier is equivalent wrt. validity. — Charles Stewart (talk) 13:03, 4 June 2009 (UTC)

This only works for recursive functions. — Emil J. 13:43, 4 June 2009 (UTC)

Euler's identity

This whole $e^{i\pi }+1=0\,\!$ business, I'm really struggling to understand it, and this seems to be the right place to ask about the thing lol.

When I read the derivation given here I have the impression of having a really good card trick done on me, that it's impressive and you can't figure out why or how it does what it does, but you don't really believe that it's doing what it appears to be. My confusion is that deriving it requires measuring angles in radians, and that seems to be the only reason for bringing π into it in the first place. Wouldn't it work equally well as $e^{180i}+1=0\,\!$ if you were using degrees? Or any positive number in fact, depending on what you were measuring angles in?

Am I massively understanding the ideas here lol, all I have is AS level Maths & Further at the moment and my understanding is rather limited heh, I was just astonished by the beauty of this thing and I was wondering how it worked in greater detail. —Preceding unsigned comment added by Dan Hartas (talk • contribs) 20:47, 1 June 2009 (UTC)

The short answer is that the formula $e^{ix}=\cos x+i\sin x$, on which this derivation is based, is only correct when cos and sin are given their argument in radians.

The long answer depends on what you know. If you're comfortable with Taylor series, then one way to prove this so-called Euler's formula is to note that

$e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}$,

$\sin x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}$

and

$\cos x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{(2n)!}}$,

and the rest is trivial. These Taylor series won't work if sin and cos get their argument in degrees. -- Meni Rosenfeld (talk) 22:25, 1 June 2009 (UTC)

Angles are, strictly speaking, dimensionless. They don't have units. An angle is defined as the ratio of the arc length and radius of a sector of a circle, so it is a length divided by a length, so it doesn't have a unit. However, it is often convenient to arbitrarily multiply all angles by 180/π in order to get nice numbers, and we call those new numbers "angles in degrees". When necessary to avoid ambiguity we call regular angles "angles in radians", but mathematicians generally just say "angle" because in mathematics we rarely need the convenience of using degrees (which is actually inconvenient for most things mathematicians do since you have to add in factors of 180/π to all your equations). It is best not to think about degrees and radians are being units because they don't work anything like metres, feet, kilograms, pounds, etc. Angles in radians are the "real" size of the angles, angles in degrees are the real angle multiplied by something to make it a nicer number. --Tango (talk) 22:44, 1 June 2009 (UTC)

This is an excellent question whose answer is worthy of its own Wikipedia page. If we set a = e^π/180, then a¹⁸⁰ⁱ = −1. In a similar way, you can find a number b such that b¹⁵ⁱ = −1, or put any other number (except 0) in the role in which we see 15 in this example. If the number you want in that role is π, then what you need in the role of a or b is e. Recall something about exponential functions and about e. If we let ƒ(x) = c^x, then ƒ is an exponential function, satisfying the important law

$f(x+y)=f(x)f(y).\,$

Now suppose we let

$\operatorname {cis} (x)=\cos(x)+i\sin(x).\,$

Using the two trigonometric identities

$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y),\,$

$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y),\,$

and the definition of multiplication of complex numbers

$(a+bi)(c+di)=(ac-bd)+(ad+bc)i,\,$

we can see that

$\operatorname {cis} (x+y)=\operatorname {cis} (x)\operatorname {cis} (y),\,$

and this is that same "important identity" that ƒ satisfies. In that way, cis is seen to behave like an exponential function, as if we had this identitiy:

$\operatorname {cis} (x)=d^{x}.\,$

But what number would be the base, d? Now remember something about exponential functions: if g(x) = c^x, then the derivative is

$g'(x)={\text{constant}}\cdot g(x),\,$

and that is easy to derive from the definition of the derivative. The "constant" is equal to 1 precisely if the base c is equal to e, and one can say that's why e is so "natural". In other words, when x = 0, then g(x) is changing 1 times as fast as x is changing, precisely if the base c is equal to e. Now suppose x = 0, so that g(x) = 1, and x is changing at a rate of i. That means as x passes the number 0 while moving "upward", along the "imaginary" axis, g(x) passes the number 1 while moving to the "upward", in the "imaginary" direction, at the same rate. Now the point(cos(x), sin(x)) moves along the circle at the same rate as the rate at which x moves, precisely when x is in radians—that's how radians are defined. (So the answer to the question posed earlier, "what is d if cis(x) = d ^x, is d = e ⁱ.)

Bottom line: measuring the angles in radians is the same as letting e be the base of the exponential function. Michael Hardy (talk) 02:32, 2 June 2009 (UTC)

(@OP) In fact, with a very small effort of abstraction, you can define exp(z) independently from the trigonometric functions (e.g. exp(z):=lim_n→∞(1+z/n)ⁿ or by the exponential series), then prove all its elementary properties, and then define cos(x) and sin(x) as the real and the imaginary part of exp(ix), that is the true definition. As to Euler identity, if you understand geometrically the complex multiplication, you can easily prove that the length (=classical total variation) of the curve exp(ix) on the interval [0,t] is exactly t. --pma (talk) 06:45, 2 June 2009 (UTC)

You can think of the formula as the limit of (1+i/N)^Nπ. This iterates a tiny rotation so you go round π amount which is half a circle, so 1 goes to -1. Dmcq (talk) 10:19, 2 June 2009 (UTC)