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Just wanted some feed back on the following proof

"if $a$ divides $b$ and $b$ divides $c$ then $a$ divides $c$"

I came up with this:

If $a|b$ then there exist some $x$ that $a * x = b$ and if $b|c$ there exist some integer $y$ that $b * y =c$ therefore if $a|c=z$ and $z=xy$ then $(a|b)(b|c)=a|c )$.

Please let me know if I got this correct and I also wanted to know if there are more than one way to prove this statement, Thank you.

Ng Chung Tak
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Mr Hons
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1 Answers1

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$b=ka$ and $c=k'b$ implies $c=kk'a$ and thus $a\mid c$.

Surb
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  • can you explain what ' on k'b represents, haven't learned that yet in my class. – Mr Hons Apr 04 '16 at 14:30
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    $k$ and $k'$ are just simply non-zero integers. – Ng Chung Tak Apr 04 '16 at 14:31
  • It's just some variable. When two variables are related in usage or meaning, we often write them as __, __' to hi-light that fact, but it's e same as if we had used an entirely new symbol. Here, it highlights the fact that $k,k'$ are the numbers introduced by the definition of divides, as opposed to one of the numbers we start with in the problem. It's read "kay-prime" – Stella Biderman Apr 04 '16 at 14:33
  • @NgChungTak Does $k$ and $k'$ have to be non-zero integers? It seems the proof still works if they are zero, too. – john Oct 09 '19 at 09:26
  • @john that's OK for $k'=0, k\ne 0$ which yields $a|b \land b|0 \implies a|0$. But for $k=0$, only $a|0 \land 0|0 \implies a|0$ is valid (though somehow redundant in some sense) which means $k=0 \implies k'=0$. Special care will be needed for $a|0$, etc. – Ng Chung Tak Oct 09 '19 at 11:31