0
$\begingroup$

I'm having some trouble understanding the following problem, why can you write the following congruence:

$$3x ≡ -29 \pmod{5} $$ as

$$3x ≡ 1\pmod{5} $$

$\endgroup$
  • 9
    $\begingroup$ Because $-29\equiv 1 \pmod 5$. $\endgroup$ – lulu Jun 19 at 13:50
  • 1
    $\begingroup$ $3x+5 \equiv -24 \ldots \implies 3x+15 \equiv -14\,\ldots \implies 3x+30 \equiv 1\, \implies 3x \equiv 1\, \quad (\text{Mod}\, 5)$ $\endgroup$ – Kevin Jun 19 at 13:54
  • $\begingroup$ in short just add 30 on both side and know that $30=0 (mod 5)$ $\endgroup$ – JustWandering Jun 19 at 13:57
  • $\begingroup$ Why is this question downvoted? It's a perfectly legitimate one? $\endgroup$ – David Jun 19 at 14:10
  • $\begingroup$ @David Agreed, I upvoted. $\endgroup$ – Kevin Jun 21 at 13:54
4
$\begingroup$

We have that $1-(-29)=30$ so $5$ divides to $1-(-29)$. That means that $-29 \equiv 1 \ (\mbox {mod }5)$. By transitivity we have that if $ 3x \equiv -29 \ (\mbox {mod }5)$ and $-29 \equiv 1 \ (\mbox {mod }5)$ then $ 3x \equiv 1 \ (\mbox {mod }5).$

$\endgroup$
  • $\begingroup$ Thanks for the answer, but how did you get to −29≡1 (mod 5) thats the part i dont really understand $\endgroup$ – kokayy Jun 19 at 14:20
  • 1
    $\begingroup$ @kokayy $-29 \equiv 1 (mod 5)$ because $-29 = (-6)*5 +1$ and $1 = (0)*5 +1$ Similarly, 11, 16, -4, -9... are also congruent with 6 $\endgroup$ – David Jun 19 at 14:22
  • $\begingroup$ @David ohh makes perfect sense now! thanks alot $\endgroup$ – kokayy Jun 19 at 14:23
  • $\begingroup$ @kokayy If $m \in \mathbb{N}$ and $a,b \in \mathbb{Z}$ then $a \equiv b \ \mbox{(mod }m)$ if and only if $m$ divides to $b-a$. $\endgroup$ – Mainkit Jun 19 at 15:56
0
$\begingroup$

Keep in mind that, in $\mathbb{Z}/5\mathbb{Z}$, 5 and 0 are the same $5 \equiv 0 (\mod 5)$

Since 5 and 0 are the same, you can add $0$ to the left -hand side of the equation and $5$ to the right-hand side, because you are actually adding $0$ on both sides.

This reasoning works for every multiple of $5$, $30$ included. Similarly, you could add $6$ to one side of the equation and $13$ to the other, without altering the solutions.

Also, since $\mathbb{Z}/5\mathbb{Z}$ happens to be a field ($5$ is prime), you can do the same thing for multiplication except, of course, you are not allowed to multiply by $5$ (because it's $0$!), so it's the same rules as with equations in $\mathbb{R}$

If you were in $\mathbb{Z}/n\mathbb{Z}$, with n not prime, you would be able to multiply both sides of the equation by $m$ (without creating any new false solutions) as long as $\gcd(n,m)=1$

$\endgroup$
0
$\begingroup$

Alternative look.

In e.g. the theory of abelian groups

$$3x\equiv -29\mod5$$ expresses exactly that: $$\{3x+5n\mid n\in\mathbb Z\}=\{-29+5n\mid n\in\mathbb Z\}$$

(This especially when we are dealing with group $\mathbb Z/5\mathbb Z$)

Note that here the equivalence sign is replaced by the equality sign, which might make things more simple to grasp.

Now observe that it is not difficult to prove that $\{-29+5n\mid n\in\mathbb Z\}=\{1+5n\mid n\in\mathbb Z\}$ so that the equality can also be written as:$$\{3x+5n\mid n\in\mathbb Z\}=\{1+5n\mid n\in\mathbb Z\}$$or again as:$$3x\equiv 1\mod5$$

$\endgroup$
  • $\begingroup$ Be aware that congruence is not normally defined to be "actually a notation" for an equality of (co)sets. Rather, the usual definition is that $\,a\equiv b\pmod{\! n}\iff n\mid a-b.\ $ Usually the coset viewpoint is not considered at length until quotient rings are introduced (often much later - perhaps even in a different book/course). $\endgroup$ – Bill Dubuque Jun 19 at 14:45
  • $\begingroup$ @BillDubuque Yes, you are right. I have decided to delete this answer within some minutes. $\endgroup$ – drhab Jun 19 at 14:48
  • 1
    $\begingroup$ You could instead just mention it as another viewpoint instead of "actually a notation" and that would help to avoid any confusion. No doubt that it is a useful viewpoint. $\endgroup$ – Bill Dubuque Jun 19 at 14:49
  • $\begingroup$ @BillDubuque So I did then. Thank you for your comment. $\endgroup$ – drhab Jun 19 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.