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January 15

leibniz's idiot proof of the product rule

Another q. I don't get Leibniz's proof of the product rule. your article quotes it as: Let y=uv. Then $y+dy=(u+du)(v+dv)$ so expanding we get $y+dy=uv+udv+vdu+dvdu$ and we cancel y=uv and we assume dvdu is inconsequential, then dy=d(uv)=vdu+udv, QED. But how does he reason that $y+dy=(u+du)(v+dv)$ ? Or a better question is, is the proof "right" at all? — Preceding unsigned comment added by 200.60.11.20 (talk) 04:59, 15 January 2012 (UTC)[reply]

It's not a rigorous proof. But his reasoning for

y+dy=(u+du)(v+dv)

is, since

y=uv

, then an infinitesimal increase in u and an infinitesimal increase in v will yield an infinitesimal increase in y. Widener (talk) 05:57, 15 January 2012 (UTC)[reply]

In general terms: If

y=f(u,v)

then

y+dy=f(u+du,v+dv)

if f is continuous. Widener (talk) 06:01, 15 January 2012 (UTC)[reply]

The formal proof of the product rule is quite easy. If you don't know what it is, you should learn it first. Widener (talk) 06:13, 15 January 2012 (UTC)[reply]

Subfield of $\mathbb {R}$ isomorphic to $\mathbb {R}$ ?

This talk page seems to be pretty inactive, so I was wondering if anybody here happened to know for certain whether the claim that there are proper subfields of the reals isomorphic to the reals is true.

Of course, the statement would be much more appropriate in the page on Real numbers rather than the page on complex numbers regardless, but yeah. --COVIZAPIBETEFOKY (talk) 19:42, 15 January 2012 (UTC)[reply]

Unless I'm missing something, a back-and-forth argument shows that a real closed field is characterized by its transcendence degree (working in the language of fields, not the language of ordered fields). So take any real closed subfield of size continuum.--188.28.188.102 (talk) 21:23, 15 January 2012 (UTC)[reply]

Well, I'm not really knowledgable enough to know for sure whether that makes sense, but the important issue here is, can you find a reference for it? Because I searched as well as I could, and I know someone who actually knows what they're talking about will be able to do better. --COVIZAPIBETEFOKY (talk) 21:38, 15 January 2012 (UTC)[reply]

There can be no such field, because the only field endomorphism from R to R is the identity. This can be proved as follows. Let f be such an endomorphism. If x is any nonnegative number, write x = t². Then f(x) = f(t²) = f(t)² ≥ 0. Thus for all y we have f(y + x) = f(y) + f(x) ≥ f(y), which shows that the function f is increasing. Furthermore, it is straightforward to prove that f must induce the identity on Q. Combining these facts, we deduce that f is the identity. 82.124.238.180 (talk) 21:34, 19 January 2012 (UTC)[reply]

That seems pretty convincing to me. Sławomir Biały (talk) 21:46, 19 January 2012 (UTC)[reply]

188's arguments may all be correct; it may simply be the case that there is no real-closed proper subfield of size continuum. You can try to add continuum-many reals one at a time (in a transfinite recursion of length

2^{\aleph _{0}}

and closing under whatever you have to close under to get a real-closed field, each new real being transcendental over what went before, but how do you know you can do that without eventually picking up every real? --Trovatore (talk) 21:54, 19 January 2012 (UTC)[reply]

On rethinking it, I think someone here has to be at least a little wrong. I think it has to be at least consistent with ZFC that there's a real-closed proper subfield of R having the cardinality of the continuum. Proof: Add a single Cohen real r to V. Then R in the sense of V is such a subfield of R in the sense of V[r]. --Trovatore (talk) 08:14, 21 January 2012 (UTC)[reply]

So to summarize: we have two seemingly legit arguments, the first of which seems to indicate that there is such a proper subfield of the reals isomorphic to itself, and the second of which indicates that this is impossible (and I understand this one), and these two might not actually be in direct conflict if it turns out that the first description doesn't actually provide a proper subfield. However, it is consistent with ZFC that they are in direct conflict, so we still have a problem.

Why is it so difficult to find publications on this topic? This seems so basic. --COVIZAPIBETEFOKY (talk) 15:30, 22 January 2012 (UTC)[reply]

I would say the reason for the lack of publications is precisely because it is so basic. The argument suggesting that an injective field homomorphism f from $\mathbb {R}$ to $\mathbb {R}$ must actually be the identity is incomplete but the general approach is correct. From first principles you can indeed show that f must be the identity on Q. The next step is to show that f preserves the ordering; so if x<y then f(x) <f(y). Now we use the fact that real numbers are limits of convergent sequences of rationals. Assume that f(x)<>x and aim for a contradiction. Let's suppose that f(x)>x -- the argument is similar if f(x)<x. Since a rational sequence converges to x we can choose rationals a and b so that a<x<b<f(x), since f(x) is fixed and we can find rationals arbitrarily close to x on either side. But f preserves the ordering so we can apply it to the inequalities to get f(a)=a<f(x)<f(b)=b. But it is impossible for f(x) to be both larger and smaller than b. This contradiction shows that the original assumption that f is not the identity is incorrect. So any such f must be the identity and there is no proper subfield of the reals isomorphic to the reals. Puzl bustr (talk) 16:59, 22 January 2012 (UTC)[reply]

I should also say the reason why the more abstract and general argument in favour of there being a proper subset fails is probably because the reals is rather a special field due to the existence of an ordering. More general infinite fields lack orderings and there is plenty of room for proper subfields. Puzl bustr (talk) 17:32, 22 January 2012 (UTC)[reply]

188 was referring to real closed fields, which necessarily have an order. My meagre understanding is that he was suggesting that one could give a proper subfield of the reals that remains a real closed field and has the same transcendence degree (presumably by picking a generating set of size continuum, but to somehow ensure that it wouldn't generate all reals), and that it would somehow necessarily follow that the subfield was isomorphic.

The other argument makes sense to me, though. --COVIZAPIBETEFOKY (talk) 18:33, 22 January 2012 (UTC)[reply]

My bad. As you say real closed fields have an order, and are in fact generalisations of the reals. What I meant to say is that it is an odd way to go, to try and establish intricate properties of the reals by working back from generalisations. As Trovatore said, there might not be such a proper subfield, only the reals themselves, and my direct argument proves just that. Puzl bustr (talk) 11:17, 23 January 2012 (UTC)[reply]

To get a field isomorphic to a proper subfield you can look at the field of rational functions over some field F. So, the elements are quotients of polynomials in one variable x with coefficients in F, with the denominator non-zero. The map induced by sending x to x² is an injective field homomorphism whose image is the proper subfield where the numerator polynomial has no odd powers of x. Not sure if this appeal to transcendentals is the only way to do it.Puzl bustr (talk) 14:15, 23 January 2012 (UTC)[reply]

counterproof of four-color theorem rule

draw a circle and connect the center with each o'clock. now 12 (or 11 or possibly 13, I haven't actually done this) are touching each other at the center. so, since you have 4 colors, and 12 (or 11 or maybe 13) regions all touching, you're out of luck and have to start mixing your colors into shades if you don't want two regions with the same color to touch.

if you would argue that having them all touch at just one point is not enough, consider this image. As you can see, five colors all touch each other in not one, but two places. since they're all touching, any two are touching, and, therefore, if any two of the five are the same colors then the map is uncolorable. by the pigeonsh** principle, four colors are not enough to make five different colors from. Therefore, the 4-color principle has been disproved, both at the top and bottom of the graphic. 80.98.112.4 (talk) 23:19, 15 January 2012 (UTC)[reply]

This is why some care has to be taken when attempting to state the four color theorem as a theorem about maps, instead of about graphs. You'll note that our article specifies that points shared by more than 2 regions don't count for adjacency.--188.28.188.102 (talk) 23:27, 15 January 2012 (UTC)[reply]

I think it was Archimedes Plutonium who observed that every map can be colored with only two colors — white, for the countries, and black, for the borders. --Trovatore (talk) 23:53, 19 January 2012 (UTC)[reply]

I still get the feeling like in Harry Potter about He Who Must Not Be Named when his name is written down in case he suddenly appears ;-) Dmcq (talk) 00:43, 20 January 2012 (UTC)[reply]

January 15

leibniz's idiot proof of the product rule

Subfield of R {\displaystyle \mathbb {R} } isomorphic to R {\displaystyle \mathbb {R} } ?

counterproof of four-color theorem rule

Subfield of $\mathbb {R}$ isomorphic to $\mathbb {R}$ ?