Exercise 1.2.9 (Simplification)

Question

Show that if $ac\mid bc$, then $a\mid b$.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

\begin{proof} By Definition 1.1, if $ac \mid bc$, then $ac 
e 0$, so $c 
e 0$. Then is some integer $k$ such that $bc = kac$. Since $c 
e 0$, $b = ka$, so $a \mid b$.
\end{proof}

\bigskip
Note: in the definition 1.1, $a \mid b$ presuppose $a 
e 0$. This restriction is not universal. If we define, for all $a,b \in \Z$, $a \mid b \iff \exists k \in \Z,\ b = ka$, then $0 \mid 0$, so $2 	imes 0 \mid 3 	imes 0$, but $2 
mid 3$. We must then reformulate Exercise 1.2.9:

``If $ac \mid bc$ and $c 
e 0$, then $a\mid c$.''

Exercise 1.2.9 (Simplification)

Answers

Comments

Exercise 1.2.9 (Simplification)

Answers

Comments

Add answer