0

So I'm practicing writing proofs and starting simple. I'm doing a proof by cases, but something seems off about it that I can't place my finger on, could someone verify this with me? I'm not sure if it's ok to say "follows the same reasoning" or not in this instance, I'm presuming little about my audience.

Theorem: $a|b$ if and only if $a|(-b)$ if and only if $(-a)|b$.

Proof: $a|b$ is equivalent to $b = ax_1$. $a|(-b)$ is equivalent to $-b = ax_1$, and multiplying through by $-1$, $b = (-1)ax_1$, so $b = a(-1)x_1$, or $b= a(-1x_1)$ and so $b = ax_2$, which is to say that $a|(-b)$ implies $a|b$. That $a|b$ implies $a|(-b)$ follows the same reasoning.

Now it will be shown that $a|(-b)$ implies $(-a)|b$. Again, from earlier $a|(-b)$ implies $a|b$, so $b = ax_1$, $(-a)|b$ is to say $b = (-a)x_1$, and multiplying by $-1$ we have $-b = ax_1$ which is equivalent to the statement $a|(-b)$ and also to the statement $a|b$. To show that $(-a)|b$ implies $a|(-b)$ follows the same reasoning.

So, is there a better way to do this proof?

jaypowers
  • 527
  • 1
    "a|b is equivalent to $b=ax_1$" No, it's not. $x_1$ hasn't been introduced. And if we do introduce then if $a|b$ means $b = ax_1$ that would mean $2b = 2ax_1 \ne ax_1$. Does that mean $a\not \mid 2b$? Need to be clear in your language..... – fleablood Mar 20 '20 at 20:12
  • It might be easier to say $m|n$ means there exists and integer $k$ so that $n=km$. If we assume $n\ne 0$ then this is true iff and only if $\frac nm=k$ for some integer $k$. So, assuming $a \ne 0$ then $a|b \iff \frac ba$ is an integer and $a|(-b)\iff \frac {-b}a$ is an integer and $(-a)|b\iff \frac{b}{-a}$ is an integer and $(-a)|(-b)\iff \frac {-b}{-a}$. As $\frac{-b}{-a}=\frac ab$ and $\frac {-b}{a}=\frac {b}{-a}=-\frac ba$ and as $\frac ab$ is an integer $\iff -\frac ba$ is an integer these are all equivalent. – fleablood Mar 20 '20 at 20:19
  • 1
    @DietrichBurde yes, I do need to state that they are integers. Thanks! – jaypowers Mar 20 '20 at 20:35
  • @fleablood Would introducing x1 as any integer not 0 solve the problem then? And I do agree, stating n=km for some integer k, n also not 0, would be clearer, Thanks! – jaypowers Mar 20 '20 at 20:35
  • The point I was getting at is you must introduce $x_1$ and what it means. $a|b$ if and only if there is an integer $x_1$ so that $b=x_1 a$. – fleablood Mar 20 '20 at 21:42
  • @fleablood Ok, I think I get it now, I need to introduce $x_1$ as any integer not equal to zero, and define $a|b$ as $a$ divides $b$ if $a$ and $b$ are integers and $b = ax$. If that's ok, then I think I can close this out, what do you think? – jaypowers Mar 20 '20 at 22:29
  • You still have not said that $a\mid b$ is in the ring of integers. – Dietrich Burde Mar 20 '20 at 22:32
  • Why is this down voted? – jaypowers Jul 18 '20 at 12:58

2 Answers2

0

$$a|b \iff b/a\in \Bbb Z \iff$$ $$\iff -(b/a)\in \Bbb Z $$ $$\iff (-b)/a\in \Bbb Z \iff $$ $$ \iff a|(-b)\iff$$ $$\iff (-b)/a \in \Bbb Z \iff$$ $$\iff b/(-a)\in \Bbb Z \iff$$ $$\iff (-a)|b.$$

DanielWainfleet
  • 59,529
  • This was aleady mentioned 3 hours prior in the comments on the question (by fleablood). Of course the case $a = 0$ needs to also be handled (separately). – Bill Dubuque Mar 20 '20 at 23:54
-1

So, is there a better way to do this proof?

Yes, first prove the Lemma below that scaling by units (invertibles) preserves divisibility, thus

$$ a(\color{#0a0}{-1})\mid b\iff a\mid b\iff a\mid b(\color{#c00}{-1})$$

Lemma $\, $ if $\ u,v\mid 1\ $ then $\, a\mid b\iff a\color{#0a0}u\mid b\color{#c00}v,\ $ for $\,a,b,u,v\in \Bbb Z\ $ (or any commutative ring)

Proof $\ $ Same as the proof(s) in your argument.

Bill Dubuque
  • 282,220