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My book (Finite Dimensional Vector Spaces by Halmos) says to redefine (if possible) addition and/or multiplication in order to make $\mathbb{Z}$ a field. The only field proprety that $\mathbb{Z}$ lacks is the existence of a multiplicative inverse. I defined multiplication as normal, with the exception that $\forall ( a \not = 0) a \cdot 2 = 1)$. Since $a \cdot 2 = a \cdot (1+1) = a+a$, I also redefined addition as normal, except in the case of $a+a$, which is always equal to $1$. But I noticed that this gives me a contradiction because we cannot have $0+0=1$.

What is a better way to solve this problem, and more importantly, how can I know that it won't give me a contradiction somewhere?


marked as duplicate by Watson, egreg linear-algebra Aug 31 '16 at 16:25

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    $\begingroup$ Use a bijection $f\colon \mathbb{Q}\to\mathbb{Z}$ and transfer the field operations. $\endgroup$ – egreg Aug 31 '16 at 16:20
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    $\begingroup$ Suppose $a_1 \neq a_2$, then $a_1 \cdot 2 = a_2 \cdot 2 \implies a_1 \cdot 2 \cdot 2 = a_2 \cdot 2 \cdot 2 \implies a_1 \cdot 1 = a_2 \cdot 1 \implies a_1 = a_2$. $\endgroup$ – Sloan Aug 31 '16 at 16:23