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Thread: Prove a Proposition

  1. #1
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    Prove a Proposition

    1 (a) If $\displaystyle x \neq 0$, and $\displaystyle xy = xz$, then $\displaystyle y = z$.
    (b) If $\displaystyle x \neq 0$, and $\displaystyle xy = x$, then $\displaystyle y = 1$.
    (c) If $\displaystyle x \neq 0$, and $\displaystyle xy = 1$, then $\displaystyle y = 1/x$.
    (d) If $\displaystyle x \neq 0$, then $\displaystyle 1/(1/x) = x$.


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

    So (c) and (d) follows from Axiom (5), and (b) follows from Axiom (2). How about (a)?
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  2. #2
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    Quote Originally Posted by heathrowjohnny View Post
    1 (a) If $\displaystyle x \neq 0$, and $\displaystyle xy = xz$, then $\displaystyle y = z$.
    Axioms
    (1) If $\displaystyle x \in F$ and $\displaystyle y \in F$, then their product $\displaystyle xy$ is in $\displaystyle F$.
    (2) $\displaystyle xy = yx$, for all $\displaystyle x,y \in F$.
    (3) $\displaystyle (xy)z = x(yz)$ for all $\displaystyle x,y,z \in F$.
    (4) $\displaystyle F$ contains an element $\displaystyle 1 \neq 0$, such that $\displaystyle 1x = x$ for every $\displaystyle x \in F$.
    (5) If $\displaystyle x \in F$ and $\displaystyle x \neq 0$, then there exists an element $\displaystyle 1/x \in F$ such that $\displaystyle x \cdot (1/x) = 1$
    (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.
    Last edited by xifentoozlerix; Feb 16th 2008 at 02:37 PM. Reason: forgot a step
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