Integral Multiple Distributes over Ring Addition

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Theorem

Let $\left({R, +, \times}\right)$ be a ring, or a field.

Let $a, b \in R$ and $m, n \in \Z$.


Then:

$(1): \quad \left({m + n}\right) \cdot a = \left({m \cdot a}\right) + \left({n \cdot a}\right)$
$(2): \quad m \cdot \left({a + b}\right) = \left({m \cdot a}\right) + \left({m \cdot b}\right)$

where $m \cdot a$ is as defined in integral multiple.


Proof

We have that the additive group $\left({R, +}\right)$ is an abelian group.


$(1): \quad \left({m + n}\right) \cdot a = \left({m \cdot a}\right) + \left({n \cdot a}\right)$:

This is an instance of Powers of Group Elements: Sum of Indices when expressed in additive notation:

$\forall n, m \in \Z: \forall a \in R: m a + n a = \left({m + n}\right) a$

$\Box$


$(2): \quad m \cdot \left({a + b}\right) = \left({m \cdot a}\right) + \left({m \cdot b}\right)$:

This is an instance of Power of Product in Abelian Group when expressed in additive notation:

$\forall n \in \Z: \forall a, b \in R: n \left({a + b}\right) = n a + n b$

$\blacksquare$


Sources