I'm having a bit of trouble understanding why this incorrect definition of primality only defines one prime number.

"p is prime if it is divisible by only itself and 1."

My understood definition of a prime number is a positive integer that has exactly 2 positive integer factors: 1 and itself. Is the problem with the definition related to the fact that it doesn't specify sign, so then -1 would be the only "prime" number?

Sorry if my question is related to others, I am quite new to number theory and could not find any other resources that would help me with this specific problem.


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    $\begingroup$ This related question may be helpful. $\endgroup$ – Brian Mar 24 at 17:28
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    $\begingroup$ What do you mean “defines only one prime number”?? $\endgroup$ – symplectomorphic Mar 24 at 17:39
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    $\begingroup$ In my opinion your professor is wrong. $\endgroup$ – amsmath Mar 24 at 17:43
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    $\begingroup$ That definition would be "wrong" because $1$ is divisible by $1$ and itself, but $1$ is not prime. $\endgroup$ – B.Swan Mar 24 at 17:50
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    $\begingroup$ You should ask your professor what he or she meant. You are most likely interpreting what he or she said incorrectly. $\endgroup$ – symplectomorphic Mar 24 at 17:54

Usually, in beginning number theory, we work only in the natural numbers (usually including $0$) so there are no negative numbers in this context. Assuming that, the definition of "divisor" would include the fact that a divisor is a natural number, than therefore positive. Then the definition:

A (natural) number $p$ greater than $1$ is prime iff it is divisible only by itself and $1$.

The "greater than $1$" part is important, because we don't want $1$ to be prime. (Although, historically it was considered a prime.) The implicit positiveness of the potential divisors is also important.

Later in the course, you can extend these definitions of "prime" and "divisor" to include the negative integers and then you'll have to adjust your definitions a bit.


Is the problem with the definition related to the fact that it doesn't specify sign, so then -1 would be the only "prime" number?

Actually is worse. There would be no prime numbers as for any $x \in \mathbb R$ then $x = n*\frac xn$ for any integer $n$ so $x$ has infinitely many divisors and can't be prime. $3$ is not prime because $\frac 32, \frac 34, \frac 35$, etc all divide $3$. (The definition of $a$ divides $b$ is that there exists an integer $k$ so that $b = ka$. There is nothing in the definition that says $a$ and $b$ are integers)


The definition of "Having only $1$ and itself as divisor" assumes that we are only considering natural numbers (i.e. positive integers) and even then we are considering $1$ and "itself" to be different.

This definition is not even considering negative numbers to be considered. Nor is it considering that that rational non integers can be divisors.

Which is okay. We just have to be specific. The following would be a perfectly acceptable definition:

A natural number is prime if it has exactly two natural divisors; itself and $1$.

We can extend it extend this to all integers by saying a prime in any integer other than $\pm 1$ with only two positive integer divisors; $1$ and the absolute value of itself. ... or a prime number is an integer which can not be expressed as a product of two or more non-unit ($\pm 1$) integers.

With more abstract algebraic concepts we can define primes more precisely as Flowers' answer and various comment links explain. These also allow for primes in systems other than just the integers.


The precise way to define prime and irreducible elements in a commutative ring are given here and here. For the integers, the definitions are equivalent, so you can use either. Rephrasing the second link for integers:

A unit in a ring is an element that has a multiplicative inverse. In the integers, the only units are $1$ and $-1$. So a nonzero non-unit element of the integers is irreducible (prime) if it cannot be written as the product of two integers which are both non-units (not 1 or -1).

Note that this implies that negative primes exist. Unique factorization is defined up to multiplication by units.

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    $\begingroup$ Beware that the 2nd link only defines "irreducible" for integral domains. In more general rings there may be zero divisors, which forces common notions of "associate" to bifurcate into a few inequivalent notions. This variation percolates into related notions, e.g. "irreducib;e" - see this answer $\endgroup$ – Bill Dubuque Mar 24 at 18:10
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    $\begingroup$ It's always a good choice confusing newbies with very general theory. So, well done. $\endgroup$ – amsmath Mar 25 at 0:12
  • $\begingroup$ @amsmath I think I would have appreciated this explanation a few years ago. $\endgroup$ – Flowers Mar 25 at 8:25

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