Thursday, April 26, 2018 ... Français/Deutsch/Español/Česky/Japanese/Related posts from blogosphere

Why 1 isn't prime

Natalie Wolchover is among those who believe that one should be considered a prime integer:


Well, she's wrong. One isn't prime exactly because it's not a building block of natural numbers. More precisely, it's totally useless as a building block.

Or as a follower of hers wrote:


The point is that the factor "one" isn't needed to write the prime integer decomposition of an integer. You may always add "times one" whereover you want but it changes nothing.




Every positive integer \(N\) may be written as\[

N = 2^{k_2} 3^{k_3} 5^{k_5} \dots

\] where \(k_2,k_3,k_5,\dots\) are non-negative integers (except for a finite number of exceptions, all of them are zero – not to get a divergent product) and the product goes over all integers.




To see that this is really linear algebra, let us take the logarithm of the equation above:\[

\log N = k_2 \log 2 + k_3 \log 3 + k_5 \log 5 + \dots

\] So this equation says that the logarithm of any positive integer may be written as a linear combination of the logarithm of primes – where the coefficients are non-negative integers. Similarly, if we allow all the coefficients \(k_2,k_3,k_5,\dots\) to be any integers, possibly negative ones, we get a unique decomposition of the most general rational number!

Because we had the phrase "linear combination" in the previous paragraph, we deal with some kind of linear algebra. Except that the coefficients are integer-valued, not real, and we're used to real or complex or quaternionic coefficients ("numbers") in vector spaces.

But the numbers \(\log 2,\log 3, \log 5, \dots\) may be interpreted as the basis vectors. Is there also \(\log 1\) over there? Well, \(\log 1 = 0\) which is a special number. It's zero. It doesn't add anything. If we added the additional factor \(1^{k_1}\) to \(N\), it wouldn't change anything, either: the value of \(k_1\) would be immaterial. In the logarithmic form, \(\log 1\) is a zero vector and a zero vector cannot be an element of any basis. Again, the addition of \(k_1 \log 1\) wouldn't add anything. So we eliminate one from the list of primes for the analogous reason why the vectors equal to zero aren't included in any bases. In fact, because of the arguments above, the claim about primes is a special case of the claim about the general bases!

Wolchover is confused about the primarily of two, too:


Well, there is absolutely nothing "awkward" about the number two as a prime. The number two is the only even prime. But in the same way, three is the only prime that is a multiple of three, and 101 is the only prime that is a multiple of 101. The only difference between 2 and 101 is that we have a special word for a "multiple of 2", namely "even", while a concise word for a "multiple of 101" doesn't exist at least in most languages on Earth. ;-)

If you ignore the irrelevant social science about the special adjectives that human languages want to create to simplify a phrase, there is no qualitative difference between the primes 2 and 101.

Add to del.icio.us Digg this Add to reddit

snail feedback (0) :