Additive Group of Reals is Subgroup of Complex

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Theorem

Let $\struct {\R, +}$ be the additive group of real numbers.

Let $\struct {\C, +}$ be the additive group of complex numbers.


Then $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$.


Proof

Let $x, y \in \C$ such that $x = x_1 + 0 i, y = y_1 + 0 i$.

As $x$ and $y$ are wholly real, we have that $x, y \in \R$.

Then $x + y = \paren {x_1 + y_1} + \paren {0 + 0} i$ which is also wholly real.

Also, the inverse of $x$ is $-x = -x_1 + 0 i$ which is also wholly real.

Thus by the Two-Step Subgroup Test, $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$.

$\blacksquare$


Sources