# Proof

Let $$(\mathbb R,\oplus,\odot)$$ be the field of real numbers and let $$(\mathbb C, + ,\cdot)$$ be the field of complex numbers. We want to show that the function. $f:=\cases{\mathbb R\to\mathbb C,\\ x\to (x,0),}$ is an bijective field homomorphism. Obviously, if the real numbers $$a,b$$ are unequal $$a\neq 0$$, then $f(a)=(a,0)\neq (b,0)=f(b).$

Thus, $$f$$ is injective. Furthermore, and trivially, for all $$f(a)=(a,0)\in\mathbb C$$, there exists $$a\in\mathbb R$$. Thus $$f$$ is surjective. Because $$f$$ is injective as well as surjective, $$f$$ is also bijective.

Moreover, for all real numbers $$a,b$$ it follows from the existence of real zero and the definition of complex addition:

$\begin{array}{rcll} f(a\oplus b)&=&f(a\oplus b,0)&\text{by definition of }f\\ &=&(a\oplus b,0\oplus 0)&\text{because real zero is neutral with respect to the addition or real numbers}\\ &=&(a,0)+(b,0)&\text{by definition of complex addition}\\ &=&f(a)+f(b)&\text{by definition of }f \end{array}$

We also have for all real numbers $$a,b$$, by virtue of the multiplication of real numbers by real zero and the definition of complex multiplication. $\begin{array}{rcll} f(a\odot b)&=&f(a\odot b,0)&\text{by definition of }f\\ &=&(a\odot b\oplus 0,0\oplus 0)&\text{because real zero is neutral with respect to the addition or real numbers}\\ &=&(a\odot b\oplus 0\odot 0,a\odot 0\oplus 0\odot b)&\text{multiplication of real numbers by real zero}\\ &=&(a,0)\cdot(b,0)&\text{by definition of complex multiplication}\\ &=&f(a)\cdot f(b)&\text{by definition of }f \end{array}$

Thus, $$f$$ is field homomorphism.

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### References

#### Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013