The Fundamental Theorem of Arithmetic says that all integers greater than one can be written uniquely as the product of prime numbers. Another way of stating this is that, if \(P = (p_1, p_2, p_3, …)\) is the (infinite) set of all primes, in order from least to greatest, and \(K = (k_1, k_2, k_3, …)\) is a set of non-negative integers with the same number of elements, then all positive integers can be expressed in the form \(p_1^{k_1} p_2^{k_2} p_3^{k_3} …\). The key here is “non-negative”: Since \(6 = 2 \cdot 3\), for instance, \(k_1 = k_2 = 1\) while all other elements of \(K\) are 0. This restatement allows us to include the value of 1 in the Theorem: It is expressed as the set \(K\) containing all elements of 0.

Because most elements of \(K\) will be zero, particularly for small numbers, we can use a short-hand. We currently do this for standard place-value: \(517 = (5 \cdot 100) + (1 \cdot 10) + (7 \cdot 1)\). We don’t bother with the initial zeroes because they’re irrelevant, but also because there would an infinite number of them.

In the new system, each place represents the corresponding element of \(K\). Since \(P = (2, 3, 5, 7, 11, 13, …)\), the rightmost digit of a short-hand number is the power of 2, the next digit to the left is the power of 3, and so on as high as needed.

Here are the first ten positive integers: \(0, 1, 10, 2, 100, 11, 1000, 3, 20, 101\). This is short hand for, respectively, \(2^0, 2^1, 3^12^0, 2^2, 5^13^02^0, 3^12^1, 7^15^03^02^0, 2^3, 3^2, 5^13^02^1\).

We can modify this system for all positive rational numbers by underlining negative exponents. For instance, \(\frac{1}{2} = \underline{1}\), while \(\frac{2}{3} = \underline{1}1\). Note that this eliminates the idea of equivalent fractions: Each rational number would have a unique representation.

We could also modify this system for all non-zero rational numbers by placing a negative sign in front for negative values. We could even modify it for all non-zero complex numbers with rational coefficients.

What we can’t do is represent a value of zero without creating a special symbol. Because of the empty product, \(\emptyset\) is a representation of one.

Adding values written in this form as if they were in standard place-value notation has the effect of multiplying them. For instance, \(2 + 100 = 102\) in prime place notation corresponds to \(4 \cdot 5 = 20\) in decimal notation.

The first number whose prime place notation matches standard decimal is 12. Are there any others?

A challenge to this system comes at \(2^{10} = 1024\). We could work with this by extending the set of digits (as done in standard hexadecimal representations), but eventually we’ll run out of symbols. For instance, using all upper and lower case letters as well as the ten digits, we run out of characters at \(2^{62} \approx 4.6 \times 10^{18}\).