Angenommen, es existiert ein Ringhomomorphismus f : R → Q . {\displaystyle f:\ \mathbb {R} \ \rightarrow \ \mathbb {Q} .} Sei q ∈ Q {\displaystyle q\in \mathbb {Q} } mit f ( 2 ) = q . {\displaystyle f\left({\sqrt {2}}\right)=q.} Es folgt
2 = 1 + 1 = f ( 1 ) + f ( 1 ) = f ( 1 + 1 ) = f ( 2 ) = f ( 2 ⋅ 2 ) = f ( 2 ) ⋅ f ( 2 ) = q 2 , {\displaystyle 2=1+1=f\left(1\right)+f\left(1\right)=f\left(1+1\right)=f\left(2\right)=f\left({\sqrt {2}}\cdot {\sqrt {2}}\right)=f\left({\sqrt {2}}\right)\cdot f\left({\sqrt {2}}\right)=q^{2},}
ein Widerspruch.
