# Linearly Independent over the Rational Numbers iff Linearly Independent over the Integers

## Theorem

Let $z_1, z_2, \ldots, z_n$ be complex numbers.

Then:

- $z_1, z_2, \ldots, z_n$ are linearly independent over the rational numbers $\Q$

- $z_1, z_2, \ldots, z_n$ are linearly independent over the integers $\Z$.

## Proof

### Forward implication

Let $z_1, z_2, \ldots, z_n$ be linearly independent over the rational numbers $\Q$.

That is, if $q_1, q_2, \ldots, q_n$ are rational numbers such that:

- $q_1 z_1 + q_2 z_2 + \cdots + q_n z_n = 0$

then:

- $q_1 = q_2 = \cdots = q_n = 0$

Let $a_1, a_2, \ldots, a_n$ be integers such that:

- $a_1 z_1 + a_2 z_2 + \cdots + a_n z_n = 0$

Then, by Integers form Subdomain of Rationals and the assumption, it follows that:

- $a_1 = a_2 = \cdots = a_n = 0$

Hence it follows by definition that $z_1, z_2, \ldots, z_n$ are linearly independent over the integers $\Z$.

$\Box$

### Backward implication

Let $z_1, z_2, \ldots, z_n$ be linearly independent over the integers $\Z$.

That is, if $a_1, a_2, \ldots, a_n$ are integers such that:

- $a_1 z_1 + \cdots + a_n z_n = 0$

then:

- $a_1 = \cdots = a_n = 0$

Let $q_1, q_2, \ldots, q_n$ be rational numbers such that:

- $q_1 z_1 + q_2 z_2 + \cdots + q_n z_n = 0$

Let the lowest common multiple of the denominators of $q_1, q_2, \ldots, q_n$ be $m$.

Then, by the definition of lowest common multiple:

- $m q_1, m q_2, \ldots, m q_n \in \Z$

Thus:

- $m q_1 z_1 + m q_2 z_2 + \cdots + m q_n z_n = 0$

where the coefficients of $z_1, z_2, \ldots, z_n$ are all integers.

Hence, by the assumption:

- $m q_1 = m q_2 = \cdots = m q_n = 0$

Since $m$ is non-zero, it follows that:

- $q_1 = q_2 = \cdots = q_n = 0$

It follows by definition that $z_1, z_2, \ldots, z_n$ are linearly independent over the rational numbers $\Q$.

$\blacksquare$