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This is part of homework.

Let $a \in \mathbb Z $. I know I can express $ 2|a $ as $\exists d\in\mathbb Z : 2d=a $ and $ 9|a $ as $\exists e\in\mathbb Z : 9e=a$ but don't know how to use it in this case.

My proof is this:

Let's say that if $2|a$ and $9|a$ then $18 ∤ a$ (antistatement). That is a contradiction as we can, for example, take $ a=36$. Then $2|a$, $9|a$, but also $18|a$. So the original statement must be true.

Is this proof enough or if something is missing then what?

Alma Arjuna
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MineFacer
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  • Are you allowed to use the Fundamental Theorem of Arithmetic? – Shaun Oct 11 '20 at 13:57
  • Your proof only shows that the original claim is true for the special case $a=36$. -- namely, if your method were flawless, you could use it (again with $a=36$) shoprove the "theorem" that $18\mid a$ if and only if $2\mid a$ and $3\mid a$. – Hagen von Eitzen Oct 11 '20 at 14:01
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    The contrapositive of the desired claim is not "if $2$ and $9$ both divide $a$ then $18$ can't divide $a$". As you point out, that statement is clearly false. Rather, the contrapositive is "there exists some $a\in \mathbb N$ such that $2$ and $3$ both divide $a$ but $18$ does not." Thus just checking the contrapositive for one particular $a$, like $a=36$ doesn't help. – lulu Oct 11 '20 at 14:09
  • Hint : if $a|n$ and $b|n$, then $lcm(a,b)|n$. – TheSilverDoe Oct 11 '20 at 14:21
  • @Shaun It's not forbidden but i'd not use it if possible – MineFacer Oct 11 '20 at 15:13

1 Answers1

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If $\;18\mid a\;,\;$ then there exists $\;b\in\mathbb{Z}\;$ such that $\;a=18b\;,\;$ hence $\;a=2(9b)=9(2b)\;,$

consequently there exist $\;b_1=9b\in\mathbb{Z}\;,\;b_2=2b\in\mathbb{Z}\;$ such that $\;a=2b_1\;$ and $\;a=9b_2\;,$

therefore $\;2\mid a\;$ and $\;9\mid a\;.$

Conversely, if $\;2\mid a\;$ and $\;9\mid a\;,\;$ there exist $\;c\;,\;d\in\mathbb{Z}\;$ such that $\;a=2c\;$ and $\;a=9d\;,\;$ hence

$2c=9d\;.$

$d\;$ has to be an even number, otherwise $\;9d\;$ would be odd but it is impossible because $\;2c=9d\;$ is even.

Since $\;d\;$ is an even number, there exists $\;d_1\in\mathbb{Z}\;$ such that $\;d=2d_1\;,\;$ consequently

$a=9d=9\cdot 2d_1=18d_1\;,$

$a=18d_1\;,\;$ therefore

$18\mid a\;.$

Angelo
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