# Redefining multiplication so that $\mathbb{Z}$ is a field? [duplicate]

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 $0+0=1$.
## marked as duplicate by Watson, egreg linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 31 '16 at 16:25
• Use a bijection $f\colon \mathbb{Q}\to\mathbb{Z}$ and transfer the field operations. – egreg Aug 31 '16 at 16:20
• Suppose $a_1 \neq a_2$, then $a_1 \cdot 2 = a_2 \cdot 2 \implies a_1 \cdot 2 \cdot 2 = a_2 \cdot 2 \cdot 2 \implies a_1 \cdot 1 = a_2 \cdot 1 \implies a_1 = a_2$. – Sloan Aug 31 '16 at 16:23