Is this an acceptable proof for If $d^{2}\mid a^{2}$ , then $d\mid a$?

Question

Proof. We will prove the contrapositive. Suppose $d\nmid a$. By the division algorithm, $a = qd + r$ for some $q\geq 0 $ and $0<r<\lvert d \rvert$. Then, $a^{2} = (qd + r)^{2} = q^{2}d^{2} + 2qdr + r^{2} = d^{2}(q^{2} + \frac{2qr}{d}) + r^{2}$. Since $r < \lvert d \rvert$ ,$r^{2}< d^{2}$. Thus, there is a nonzero remainder, so $d^{2}\nmid a^{2}$. Hence if $d^{2}\mid a^{2}$, $d\mid a$.

$\square$

How can you be sure that the quantity $q^2+\frac{2qr}{d} \in \Bbb{Z}$? Also why is $|d|r^2 < d^2$? Based on $|r|<d$, the best you can say is that$|d|r^2 < d^3$. — Anurag A, Jun 28 '20 at 20:27
@AnuragA In the "Since" statement, there is a missing comma. To avoid confusion, the comma should be added between $|d|$ and $r^2$. — Geoffrey Trang, Jun 28 '20 at 20:33
@GeoffreyTrang That would make sense, but the gap still remains (first part of my comment). — Anurag A, Jun 28 '20 at 20:34

peter.petrov · Answer 1 · 2020-06-28T20:54:25.857

2

Why use contrapositive argument... This is much easier proved directly, no?

We know $d^2 | a^2$ (which on a side note means $d \ne 0$)

$$\Rightarrow a^2 = d^2.q$$

for some integer $q$

But then $q = (a/d)^2$ so $q$ is a perfect square of a rational number.
Now let's assume $(a/d) = (a_1/d_1)$ where $a_1$ and $d_1$ are coprime.

$q = (a_1/d_1)^2$
$q d_1^2= a_1^2$

$\Rightarrow d_1/d_1$ and $d_1 / a_1^2$ but these two $d_1$ and $a_1^2$ are also coprime.

So the only way this to happen is if $d_1 = 1$ or $d_1 = -1$
And this means the rational $a/d$ is also an integer.

But $q$ is also an integer.

So $q = m^2$ where $m=a/d$ is also an integer.

Then we get: $a^2 = d^2.m^2$

and then: $|a| = |d|.|m|$ This means $|d|$ divides $|a|$ and so $d$ divides $a$.

edited Jun 28 '20 at 20:54

answered Jun 28 '20 at 20:38

peter.petrov

12,568

+1. However, OP is using contrapositive not contradiction – Koro Jun 28 '20 at 20:39
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@Koro Thanks... I didn't study math in English :) – peter.petrov Jun 28 '20 at 20:40
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I think for my proof to be flawless one needs also this lemma: If an integer is perfect square of a rational, then this rational is an integer too. This should be proved separately. – peter.petrov Jun 28 '20 at 20:45
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@peter.petrov: mh, isn't this lemma equivalent to the initial claim ? – Jun 28 '20 at 20:47
@YvesDaoust Yeah... maybe :) Just filled in this gap. – peter.petrov Jun 28 '20 at 20:50

score 0 · Answer 2 · answered Jun 28 '20 at 21:01

0

Every prime factor of $d^2$ has an even multiplicity and is a prime factor of $a^2$ with non smaller multiplicity. Hence every prime factor of $d$ divides $a$.

answered Jun 28 '20 at 21:01

Is this an acceptable proof for If $d^{2}\mid a^{2}$ , then $d\mid a$?

2 Answers2