Shown below is the second shortest published math paper; it’s the shortest published math paper, except for this one. This paper relates to an extension to Fermat’s last theorem. That’s well known math, though I think a few words of background would help the educated lay reader.
By way of background, Fermat’s last theorem states that there is no set of integers for which An + Bn = Qn, where n is an integer larger than 2. Thus, there is no set where A3 + B3 = Q3 or A5 + B5 = Q5, etc. This theorem was really a conjecture until recently though Fermat claimed to have proven it in 1695.
The Euler conjecture of the title here, is related to Fermat’s conjecture/theorum: Either conjectured that the smallest collections of A, B, C, D.., whose power to the n, summed, will equal some whole number to the power n, Qn , must have at least as many components (A,B,C,D,..) as the exponent value, n. Thus, while you might find a set of five numbers, A, B, C, D, E where A5 + B5 + C5 + D5 + E5 = Q5, you can’t find a set of four numbers where A5 + B5 + C5 + D5 = Q5. The paper above disproves this conjecture in a most clear way; it shows a counter-example where A5 + B5 + C5 + D5 = Q5.
This is, in a sense, the ideal math paper: clear, short, important, and true. For background to this conjecture, the authors merely reference a page of a math history book.
Shown below is the shortest math pater ever published. It appeared in the American Mathematical Monthly, 2004, and would have appeared in a more honored journal if the authors were willing to add more words as an editor requested. You’ll see that the paper itself has only pictures and one sentence with one English word: n2 + 2 can:, I thought I might as well try to explain it because as the editor commented, this is too few words for most readers.
The trick to understanding this at all is that most of the background is in the title, which is in the form of a question. The text of the article is in the form of an answer with the diagrams serving as proof. Even with this insight, you’ll likely need more background, but that’s the start.
Here’s the background: Most folks notice that you can make a big equilateral triangle of side-length n, out of n2 unit subtriangles, that is of subtrangles where the side lengths =1. For example, to make an equilateral triangle with length 10, requires 100 unit equilateral triangles, n2.
Now the question in the title involves what happens if all these n2 component triangles are made slightly larger, the sides of each becoming 1+ε/n, where ε is some very small amount. The side of the new big triangle is now n+ε. The question in the title now is can you cover this bigger, super triangle with n2+1 unit triangles. The authors provide two, half answers to this question by their drawings, suggesting two different ways that you can cover the bigger super-triangle with n2+2 unit triangles. That would be 102 for the case where you start with 100 unit triangles and expanded each by ε/n.
The first solution is the bottom of figure 1. This shows what happens if you add two more unit triangles to the bottom row of the old super triangle, and squish a bit from the sides so that the top of the new row matches the bottom of the old row. Doing this leaves you with a row that’s n+ε long at the bottom with wings at the top that expand the sides to n+ε as well. The drawing shows that this new row has effective height, 1+ε.
Now, take every other row and push them together slightly from top-down so that the height becomes (1-ε) but the length expands to n(1+ε). Adding rows like this, you’ll be able to cover the entirety of the bottom space of the new, larger super triangle. Notice that the thickness of each line now 1-ε as shown. Use these longer lines to cover the rest of the bigger super triangle. And that’s the end of the paper. Once again you needed n2+2 unit triangles to cover the bigger super-triangle.
An extension to the above paper was discovered since the original paper. It’s shown in the figure below. Here the original requirement of equilateral triangles is relaxed. For highly elongated triangles, you still find that a normal super-triangle requires n2 sub-triangles. But now, from this figure, you see that an expanded super-triangle (each side expanded by 1+ε/n say) can be covered using only n2+1 of the original size subtriangles.
The proof is clear enough that no words are needed. It’s conceivable that the authors could have published this as an even shorter paper, if it were ever published, but it was not. Instead, I saw this extension as a result from a math competition, here. These insights of geometry come from Princeton University, a top notch place where I was a grad student (in engineering). The school has gone somewhat to seed, IMHO, because of political correctness.
One way to look at dating and other life choices is to consider them as decision-time problems. Imagine, for example that have a number of candidates for a job, and all can be expected to say yes. You want a recipe that maximizes your chance to pick the best. This might apply to a fabulously wealthy individual picking a secretary or a husband (Mr Right) in a situation where there are 50 male choices. We’ll assume that you have the ability to recognize who is better than whom, but that your pool has enough ego that you can’t go back to anyone once you’ve rejected the person.
Under the above restrictions, I mentioned in this previous post that you maximize your chance of finding Mr Right by dating without intent to marry 36.8% of the fellows. After that, you marry the first fellow who is better than any of the previous. My previous post had a link to a solution using Riemann integrals, but I will now show how to do it with more prosaic math — a series. One reason for doing this by series is that it allows you to modify your strategy for a situation where you can not be guaranteed a yes, or where you’re OK with number 2, but you don’t like the high odds of the other method, 36.8%, that you’ll marry no one.
I present this, not only for the math interest, but because the above recipe is sometimes presented as good advice for real-life dating, e.g. in a recent Washington Post article. With the series solution, you’re in a position to modify the method for more realistic dating, and for another related situation, options cashing. Let’s assume you have stock options in a volatile stock company, if the options are good for 10 years, how do you pick when to cash in. This problem is similar to the fussy suitor, but the penalty for second best is small.
The solution to all of these problems is to pick a stopping point between the research phase and the decision phase. We will assume you can’t un-cash in an option, or continue dating after marriage. We will optimize for this fractional stopping point between phases, a point we will call x. This is the fraction of guys dated without intent of marriage, or the fraction of years you develop your formula before you look to cash in.
Let’s consider various ways you might find Mr Right given some fractional value X. One way this might work, perhaps the most likely way you’ll find Mr. Right, is if the #2 person is in the first, rejected group, and Mr. Right is in the group after the cut off, x. We’ll call chance of of finding Mr Right through this arrangement C1, where
C1 = x (1-x) = x – x2.
We could used derivatives to solve for the optimum value of x, but there are other ways of finding Mr Right. What if Guy #3 is in the first group and both Guys 1 and 2 are in the second group, and Guy #1 is earlier in the second line-up. You’d still marry Mr Right. We’ll call the chance of finding Mr Right this way C2. The odds of this are
C2 = x (1-x)2/2
= x/2 – x2 + x3/2
There is also a C3 and a C4 etc. Your C3 chance of Mr Right occurs when guy number 4 is in the first group, while #1, 2, and 3 are in the latter group, but guy number one is the first.
C3 = x (1-x)3/4 = x/4 – 3x2/4 + 3x3/4 – x4/4.
I could try to sum the series, but lets say I decide to truncate here. I’ll ignore C4, C5 etc, and I’ll further throw out any term bigger than x^2. Adding all smaller terms together, I get ∑C = C, where
C ~ 1.75 x – 2.75 x2.
To find the optimal x, take the derivative and set it to zero:
dC/dx = 0 ~ 1.75 -5.5 x
x ~ 1.75/5.5 = 31.8%.
That’s not an optimal answer, but it’s close. Based on this, C1 = 21.4%, C2 = 14.8%, C3 =10.2%, and C4= 7.0% C5= 4.8%Your chance of finding Mr Right using this stopping point is at least 33.4%. This may not be ideal, but you’re clearly going to very close to it.
The nice thing about this solution is that it makes it easy to modify your model. Let’s say you decide to add a negative value to not ever getting married. That’s easily done using the series method. Let’s say you choose to optimize your chance for either Mr 1 or 2 on the chance that both will be pretty similar and one of them may say no. You can modify your model for that too. You can also use series methods for the possibility that the house you seek is not at the last exit in Brooklyn. For the dating cases, you will find that it makes sense to stop your test-dating earlier, for the parking problem, you’l find that it’s Ok to wait til you’re less than 1 mile away before you settle on a spot. I’ll talk more about this latter, but wanted to note that the popular press seems overly impressed by math that they don’t understand, and that they have a willingness to accept assumptions that bear only the flimsiest relationship to relaity.
As of this month, the District of Columbia has joined 15 states in a pact to would end the electoral college choice of president. These 15 include New York, California, and a growing list of solid-blue (Democratic party) states. They claim the electoral must go as it robed them of the presidency perhaps five times: 2016, 2000, 1888, 1876, and perhaps 1824. They would like to replace the electoral college by plurality of popular vote, as in Mexico and much of South America.
All the big blue states and some small blue states have joined a compact to end the electoral college. As of 2019, they are 70% of the way to achieving this.
As it happens, I had to speak on this topic in High School in New York. I for the merits of the old system beyond the obvious: that it’s historical and works. One merit I found, somewhat historical, is that It was part of a great compromise that allowed the US to form. Smaller states would not have joined the union without it, fearing that the federal government would ignore or plunder them without it. Remove the vote advantage that the electoral college provides them, and the small states might have the right to leave. Federal abuse of the rural provinces is seen, in my opinion in Canada, where the large liberal provinces of Ontario and Quebec plunder and ignore the prairie provinces of oil and mineral wealth.
Several of the founding federalists (Jay, Hamilton, Washington, Madison) noted that this sort of federal republic election might bind “the people” to the president more tightly than a plurality election. The voter, it was noted, might never meet the president nor visit Washington, nor even know all the issues, but he could was represented by an elector who he trusted, he would have more faith in the result. Locals would certainly know who the elector favored, but they would accept a change if he could justify it because of some new information or circumstance, if a candidate died, for example, or if the country was otherwise deadlocked, as in 1800 or 1824.
Historically speaking, most electors vote their states and with their previously stated (or sworn) declaration, but sometimes they switch. In, 2016 ten electors switched from their state’s choice. Sven were Democrats who voted against Hillary Clinton, and three were Republicans. Electors who do this are called either “faithless electors” or “Hamilton electors,” depending on whether they voted for you or against you. Hamilton had argued for electors who would “vote their conscience” in Federalist PaperNo. 68. One might say these electors threw away their shot, as Hamilton did not. Still, they showed that elector voting is not just symbolic.
Federalist theory aside, it seems to me that the current system empowers both large and small states inordinately, and swing states, while disempowering Alabama and Massachussetts. Change the system and might change the outcome in unexpected ways.
That the current system favors Rhode Island is obvious. RI has barely enough population for 1 congressman, and gets three electors. Alabama, with 7 congressmen, gets 9 electors. Rhode Islanders thus get 2.4 times the vote power of Alabamans.
It’s less obvious that Alabama and Massachussetts are disfavored compared to New Yorkers and Californians. But Alabama is solid red, while New York and California are only sort of blue. They are majority Democrat, with enough Republicans to have had Republican governors occasionally in recent history. Because the electoral college awards all of New York’s votes to the winner, a small number Democrat advantage controls many electors.
In 2016, of those who voted for major party candidates in New York, 53% voted for the Democrat, and 47% Republican. This slight difference, 6%, swung all of NY’s 27 electors to Ms Clinton. If a popular vote are to replace the electoral college, New York would only have the net effect of the 6% difference; that’s about 1 million net votes. By contrast, Alabama is about 1/3 the population of New York, but 75% Republican. Currently its impact is only 1/3 of New York’s despite having a net of 2.5 million more R voters. Without the college, Alabama would have 2.5 times the impact of NY. This impact might be balanced by Massachusetts, but at the very least candidates would campaign in these states– states that are currently ignored. Given how red and blue these states are, it is quite possible that the Republican will be more conservative than current, and the Democrat more liberal, and third party candidates would have a field day as is common in Mexico and South America.
Proposed division of California into three states, all Democrat-leaning. Supposedly this will increase the voting power of the state by providing 4 more electors and 4 more senators.
California has petitioned for a different change to the electoral system — one that should empower the Democrats and Californians, or so the theory goes. On the ballot in 2016 was bill that would divide California into three sub-states. Between them, California would have six senators and four more electors. The proposer of the bill claims that he engineered the division, shown at right, so skillful that all three parts would stay Democrat controlled. Some people are worried, though. California is not totally blue. Once you split the state, there is more than three times the chance that one sub-state will go red. If so, the state’s effect would be reduced by 2/3 in a close election. At the last moment of 2016 the resolution was removed from the ballot.
Turning now to voter turnout, it seems to me that a change in the electoral college would change this as well. Currently, about half of all voters stay home, perhaps because their state’s effect on the presidential choice is fore-ordained. Also, a lot of fringe candidates don’t try as they don’t see themselves winning 50+% of the electoral college. If you change how we elect the president we are sure find a new assortment of voters and a much wider assortment of candidates at the final gate, as in Mexico. Democrats seem to believe that more Democrats will show up, and that they’ll vote mainstream D, but I suspect otherwise. I can not even claim the alternatives will be more fair.
In terms of fairness, Marie de Condorcet showed that the plurality system will not be fair if there are more than two candidates. It will be more interesting though. If changes to the electoral college system comes up in your state, be sure to tell your congressperson what you think.
