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The hint for this question is "Use existence of $s,t$ such that $sa + tb = 1$"

So, the relevant definitions and corollaries that I have collected for this problem are:

  1. Def of divisibility - Since $a|bx$ that means that $(bx) = qa$ with $q \in \mathbb{Z}$
  2. Def of relatively prime - GCD($a,b$) = $1$
  3. Corollary to (2.) - $\exists$ $s,t$ such that $sa + tb = 1$

Now, conceptually I believe I understand the logic for this statement: We know that $a$ divides the product $bx$ but we also know that $a$ and $b$ are relatively prime and hence $a$ cannot divide $b$ since the greatest divisor between them is $1$. This only leaves the factor of $x$ for which $a$ can divide in order to divide the product $bx$.

I'm having a bit of difficulty putting this into "math terms":

  • I started by writing out the relevant definitions above and seeing if there was any way I could combine them to show that $a$ divides $x$ or doesn't divide $b$. I started by writing out the def of relatively prime integers $sa + tb = 1$ I then multiplied by $x$ in order to introduce an $x$ into the equations then I subtracted the $tbx$ to get: $sa = x - tbx = x(1 - tb)$ The $x$'s just cancel and that brings us in a round-about way back to a single manipulation of the original equation $sa = 1 - tb$ which I don't think really says anything valuable or makes any headway to our goal.

  • Next, I tried starting from the equation $sax + tbx = x$ again but this time substituting the def of divisibility above to get $sax + t(qa) = x$. From here, I pulled out the $a$ to get $a(sx + tq) = x$ which does look promising because if we notice that $(sx + tq) = k \in \mathbb{Z}$ then we can see that $ak = x$ is in the form for the definition of divisibility for $x$. So can we say that this proves that $a$ divides $x$? Furthermore, is that sufficient, if I state (in plain english) that $a$ cannot divide $b$ because of the reasons stated above? Or do I need to show that in "math terms" as well?

Bill Dubuque
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Numerical Disintegration
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  • BTW, this is Euclid's lemma – PM 2Ring Dec 15 '21 at 04:40
  • Your argument by scaling by $,x,$ the Bezout equation for $,\gcd(a,b) = 1,$ is correct. It is the most common way to prove this from of Euclid's Lemma in $,\Bbb Z,,$ e.g. see the $\rm\color{#0a0}{Bezout}$ form here in the first dupe for a more concise presentation (and generalizations), and see the other dupes for variations on such. – Bill Dubuque Dec 15 '21 at 10:03

2 Answers2

2

Your second solution is great, and you can state that $a$ cannot divide $b$ because of the reasons above.

One tiny thing I would suggest you add, depending on how harsh the graders might be (if this were graded), would be to state why $sx$ and $tq$ are each integers, and why $k$ is therefore an integer.

I am a person
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We first observe that $\gcd(a, b) = 1$ implies $a \not \mid b$ unless in fact $a = 1$; otherwise we would have $\gcd(a, b) = a \ne 1$.

Given that

$\gcd(a, b) = 1, \tag 1$

it follows from Bexout's identity that there exist

$s, t \in \Bbb Z \tag 2$

with

$as + bt = 1; \tag 3$

we multiply this by $x$:

$asx + btx = x; \tag 4$

we further observe that

$a \mid bt \Longrightarrow \exists c \in \Bbb Z, \; bt = ac; \tag 5$

thus (4) may be transformed into

$asx + acx = x, \tag 6$

from which it immediately follows that

$a(sx + cx) = x \Longrightarrow a \mid x. \tag 7$

QED.

I agree with our colleague I am a person: your second solution is the way to go. In fact, it is more-or-less the "standard proof" of this "theorem".

Robert Lewis
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