# Prove a Proposition

• Feb 16th 2008, 12:14 PM
heathrowjohnny
Prove a Proposition
1 (a) If $x \neq 0$, and $xy = xz$, then $y = z$.
(b) If $x \neq 0$, and $xy = x$, then $y = 1$.
(c) If $x \neq 0$, and $xy = 1$, then $y = 1/x$.
(d) If $x \neq 0$, then $1/(1/x) = x$.

Axioms
(1) If $x \in F$ and $y \in F$, then their product $xy$ is in $F$.
(2) $xy = yx$, for all $x,y \in F$.
(3) $(xy)z = x(yz)$ for all $x,y,z \in F$.
(4) $F$ contains an element $1 \neq 0$, such that $1x = x$ for every $x \in F$.
(5) If $x \in F$ and $x \neq 0$, then there exists an element $1/x \in F$ such that $x \cdot (1/x) = 1$.

So (c) and (d) follows from Axiom (5), and (b) follows from Axiom (2). How about (a)?
• Feb 16th 2008, 02:30 PM
xifentoozlerix
Quote:

Originally Posted by heathrowjohnny
1 (a) If $x \neq 0$, and $xy = xz$, then $y = z$.
Axioms
(1) If $x \in F$ and $y \in F$, then their product $xy$ is in $F$.
(2) $xy = yx$, for all $x,y \in F$.
(3) $(xy)z = x(yz)$ for all $x,y,z \in F$.
(4) $F$ contains an element $1 \neq 0$, such that $1x = x$ for every $x \in F$.
(5) If $x \in F$ and $x \neq 0$, then there exists an element $1/x \in F$ such that $x \cdot (1/x) = 1$

(a) follows from axioms (2), (5), (3), and (4), in that order. since xy=xz, you can say that yx=zx using (2) on both sides. you then can show that yx(1/x)=zx(1/x) using (5). this becomes y(1)=z(1) using (3). your equation then becomes y(1)=z(1). using axiom (4) will reduce your equation to the desired y=z.