# 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 -0500
**

Seen: **473 times**

Last updated: **Mar 06 '15**

eliminating fractions and roots from equations

how to get output in a mixed fraction?

Can this fraction be simplified ?

divide numerator and denominator by certain value

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

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

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.