[MATH](E, +, \cdot)[/MATH] is a vector space [MATH]\in \mathbb{R}[/MATH] with [MATH]u+v = u-v[/MATH] and [MATH]a\cdot u= -a \cdot u[/MATH], for any [MATH]u,v \in E[/MATH] and any [MATH]a,b \in \mathbb{R}.[/MATH]
I tried to see if the multiplication axioms are satisfied by this definitions:
-Ax1: [MATH]a(u+v) = au + av \Leftrightarrow a(u+v) = -a(u+v) \Leftrightarrow au+av = -au-av \Leftrightarrow -2au = 2av \Leftrightarrow -u= v.[/MATH] So axiom 1 isn't satisfied by the definition.
-Ax2: [MATH](a+b)u = au + bu \Leftrightarrow (a+b)u = -au-bu \Leftrightarrow au+bu = -au-bu \Leftrightarrow 2au = -2bu \Leftrightarrow au = -bu \Leftrightarrow a = -b [/MATH] So axiom 2 isn't satisfied by the definition.
-Ax3: [MATH](ab)u = a(bu) \Leftrightarrow -(ab)u = -a(bu) \Leftrightarrow -abu = -abu [/MATH] So axiom 3 is satisfied by the definition.
-Ax4: [MATH] 1u = u \Leftrightarrow 1u= -1u \Leftrightarrow u = -u [/MATH] So axiom 4 isn't satisfied by the definition.
Is this correct?
I tried to see if the multiplication axioms are satisfied by this definitions:
-Ax1: [MATH]a(u+v) = au + av \Leftrightarrow a(u+v) = -a(u+v) \Leftrightarrow au+av = -au-av \Leftrightarrow -2au = 2av \Leftrightarrow -u= v.[/MATH] So axiom 1 isn't satisfied by the definition.
-Ax2: [MATH](a+b)u = au + bu \Leftrightarrow (a+b)u = -au-bu \Leftrightarrow au+bu = -au-bu \Leftrightarrow 2au = -2bu \Leftrightarrow au = -bu \Leftrightarrow a = -b [/MATH] So axiom 2 isn't satisfied by the definition.
-Ax3: [MATH](ab)u = a(bu) \Leftrightarrow -(ab)u = -a(bu) \Leftrightarrow -abu = -abu [/MATH] So axiom 3 is satisfied by the definition.
-Ax4: [MATH] 1u = u \Leftrightarrow 1u= -1u \Leftrightarrow u = -u [/MATH] So axiom 4 isn't satisfied by the definition.
Is this correct?
Last edited: