For all integers a, b, c, if a | b and b | c then a | c.
Solution 1
HINT: This is just a matter of applying the definitions, which is always the first thing that you should try. What does $a\mid b$ mean? It means that there is an integer $m$ such that $b=am$. Similarly, $b\mid c$ means that there is an integer $n$ such that $c=bn$. You now have the equations
$$\left\{\begin{align*} &b=am\\ &c=bn\;, \end{align*}\right.\tag{1}$$
and you want to show that $a\mid c$. Once again you should turn to the definition: this means that there is an integer $k$ such that $c=ak$. Can you combine the equations in $(1)$ to find such an integer $k$?
Solution 2
$$ a \mid b \Rightarrow b= na , n \in \Bbb Z$$
$$ b \mid c \Rightarrow c = kb = (kn)a \Rightarrow a \mid c , k \in \Bbb Z $$
Related videos on Youtube
user103010
Updated on October 25, 2020Comments
-
user103010 about 3 years
Is this T or F? and most importantly, why?
I'll be using any answers for a basis or completely my other questions, since my understanding is still a little poor.
-
dfeuer about 10 years@TheChaz2.0, that looks very different.
-
-
user103010 about 10 yearsI really like your answer and hopefully I didn't take it too literally, but then would be c=amn? "mn" would be k, so then the statement would be true?
-
Brian M. Scott about 10 years@user1030101: That’s exactly the right argument. The product of two integers is an integer, so setting $k=mn$ works perfectly.
-
user103010 about 10 yearsFantastic! Thank you so much! =D
-
Brian M. Scott about 10 years@user103010: You’re welcome!