# Why is_prime(6/3) results as False?

```
sage: (6/3).is_integer()
True
sage: (6/3).is_prime()
False
```

Why is_prime(6/3) results as False?

```
sage: (6/3).is_integer()
True
sage: (6/3).is_prime()
False
```

add a comment

2

Note that the parent of `6/3`

is the `Rational Field`

:

```
sage: a = 6/3
sage: a.parent()
Rational Field
```

So, when you write `(6/3).is_prime()`

, you ask whether `6/3`

is prime *as a rational number,* not as an integer, see the documentation:

```
sage: a.is_prime?
A *prime* element is a non-zero, non-unit element p such that,
whenever p divides ab for some a and b, then p divides a or p
divides b.
```

So, since `6/3`

is a unit in `QQ`

, the answer should be `False`

```
sage: a.is_unit()
True
sage: a.is_prime()
False
```

To see if `6/3`

is prime as an integer, just do:

```
sage: ZZ(6/3).is_prime()
True
```

So, it is very important in mathematics and in Sage to know where your elements are living. For example the polynomial `x^2-2`

can not be factorized in $\mathbb{Q}[x]$, but it does in $\overline{\mathbb{Q}}[x]$:

```
sage: x = polygen(QQ)
sage: (x^2-2).is_irreducible()
True
sage: x = polygen(QQbar)
sage: (x^2-2).is_irreducible()
False
```

Asked: **
2015-03-06 12:46:49 -0600
**

Seen: **426 times**

Last updated: **Mar 06 '15**

partial fraction decomposition function for multivariate rational expressions

How to Rationalize the Denominator of a Fraction ?

how to get output in a mixed fraction?

eliminating fractions and roots from equations

Get a matrix to display answers as decimals/floats, not fraction?

Can this fraction be simplified ?

latex(-(x-1)/(x+1)) still broken

how to run fraction elenment in Multivariate Polynomial Ring in over Finite Field ring

Copyright Sage, 2010. Some rights reserved under creative commons license. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license.