i-maginary World

We have seen in the previous post that anyone whose YOB is odd will only have either 1 or 0 birthday prime. It is natural to ask what about people whose YOB is even. It turns out that from the limited observation of the table in the previous post, anyone whose YOB is even will always have at least 1 birthday prime if they live long enough. This leads me to conjecture that

Given any positive even number, we can ALWAYS find a prime such that their sum is another prime.

(Remark: I’m unsure if this is already a known result. I just find out from Wikipedia about Polignac’s conjecture which looks similar to my question. But I’m not sure if my question here is equivalent to or just a special case of the Polignac’s conjecture.)

I find no counterexample so far for the first 1 million even numbers with the first 1000 primes. I understand that 1 million is nothing compared to the infinitude of primes, but it convinces me a little that this conjecture holds. I shall sketch out my idea here:

Let be the set of all odd prime numbers and be the set of odd prime numbers strictly less than . Let be the set of all positive even numbers and for an odd prime , let . For examples,

, .

The above conjecture is then equivalent to prove that

To be continued …

We define birthday primes as a pair (year, age) where your age on that year are both prime numbers. I have written a Matlab program to list out all the year of birth (YOB) of people born in  the 20th century who have some positive number (#) of birthday primes (BP). The only assumption made is life span of 80 years. The YOB that are not listed have 0 birthday prime.

YOB        # of BP

1900           5
1902           6
1904           3
1905           1
1906           4
1908           6
1910           3
1911           1
1912           4
1914           6
1916           2
1918           4
1920           9
1922           3
1924           3
1926           9
1928           5
1929           1
1930           5
1931           1
1932           8
1934           3
1936           5
1938           7
1940           5
1942           4
1944           8
1946           6
1947           1
1948           3
1949           1
1950           9
1952           3
1954           3
1956          10
1958           5
1960           6
1962           6
1964           4
1966           6
1968           9
1970           7
1971           1
1972           3
1974           9
1976           6
1977           1
1978           2
1980          10
1982           6
1984           5
1985           1
1986           9
1988           5
1990           6
1991           1
1992           8
1994           5
1995           1
1996           6
1997           1
1998           7

For my own case (1986), my birthday primes are as follows:

Year          Age

1993           7
1997          11
1999          13
2003          17
2017          31
2027          41
2029          43
2039          53
2053          67

Furthermore, people whose YOB is odd can either have only 1 birthday prime or 0 birthday prime, even assuming immortality, by a simple observation:

odd YOB +2 = odd (either prime or composite)

odd YOB + odd ‘prime age’ = even (composite)

A metric on a set X is a function . For any , the function satisfies the following conditions

(i) and if and only if ,

(ii) ,

(iii) .

The purpose of this post is to prove that is a continuous function. I’m unsure whether this proposition is so trivial that it is often omitted in standard books on analysis and topology. Anyway, here’s my proof.

Let and . We define a new metric on by

We can show that this definition satisfies all the three conditions above since itself is a metric. We also require the following inequality

which can be shown by using the triangle inequality

Since we are working on metric spaces, we shall use the classical definition for a function to be continuous. Given , we set . Then for , we have

Hence, is continuous.

Here’s a riddle: “How is a door different from a set?”

Here’s a mathematician’s answer: “A door can be either open or closed but not both. While a set can be either open, or closed, or both, or neither!”