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This is an example from Discrete Mathematics and its Applications enter image description here

This is example 1 that this example references enter image description here

And here's Theorem 1 that the example references enter image description here

Example 1 makes sense. We have to determine if the inverse of 3 modulo 7 exists first of all. It will exist based on theorem 1 if m > 1, which it is in this case bc m = 7 and if gcd(3, 7) is 1 - definition of relatively prime.

To determine if gcd(3, 7) is 1, you can use Euclid's Algorithm,

7 = 3(2) + 1

3 = 1(3) + 0

In this case, 1 is the gcd because it is last remainder before the remainder goes to zero. $\equiv$

Now you know the inverse exists. To find the actual inverse, you want the inverse to be in the form

a'a $\equiv$ 1 $\bmod$ m

where a' would be the inverse.

To do this, you would need to use Bezout's Theorem that gcd(a, b) = sa + tb, so from my work in Euclid's Theorem, you can see

7 = 3(2) + 1

-2(3) + 1(7) = 1

then

-2(3) - 1 = -1(7)

-2(3) $\equiv$ 1$\bmod$(7) which is in the form of a'a $\equiv$ 1 $\bmod$ m, so a' or the inverse of a modulo m is -2.

Here's where I get lost. In example 3, the author uses the result he got from the inverse to solve the linear congruence of 3x $\equiv$ 4(mod 7).

I understand that -6 $\equiv$ 1 ($\bmod$ 7) and -8 $\equiv$ 6 ($\bmod$ 7), but how does that lead to x $\equiv$ -8 $\equiv$ 6 ($\bmod$ 7)? What does that even mean?

committedandroider
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2 Answers2

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$\qquad\ \ 3\,x\,\equiv\, 4 \pmod 7\ $ scaled by $\ {-2}\equiv 3^{-1}_{\phantom{I_{I_{I_I}}}} $ yields

$\smash[t]{\ \ \overbrace{-2\cdot 3}^{\Large\ \equiv\, \color{#c00}1}\,x\,\equiv\,\overbrace{-2\cdot 4}^{\Large\! \equiv\, \color{#0a0}6}\pmod 7}$

$\ \Rightarrow\ \color{#c00}1\cdot x\equiv \color{#0a0}6\pmod 7,\, $ where we've used the Congruence Product Rule to scale it.

Indeed $\ x\equiv 6\,\Rightarrow\, 3x \equiv 18\equiv 4\pmod 7$

Alternatively $\ x\,\equiv\ \dfrac{4}{3}\,\equiv\, \dfrac{-3}3\,\equiv\, {-}1\,\equiv\, 6 $

Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.

Community
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Bill Dubuque
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  • For Congruence product rule A≡a, and B≡b ⇒ AB≡ab (mod m), how is −2⋅3 $\equiv$ 1? – committedandroider Feb 12 '15 at 05:15
  • @com Use $\ {-}2\equiv -2,$ times $,3x\equiv 4,,$ which scales the latter by $,-2,$ as in the answer. – Bill Dubuque Feb 12 '15 at 05:19
  • Can you elaborate on that? I thought the whole point of this $\equiv$ sign was to go along with a modulus expression like 9 $\equiv$ 4(mod 5) is saying the difference between 9 and 4 is a multiple 5. What does it mean when you don't have the modular expression? – committedandroider Feb 12 '15 at 05:23
  • @com Some writers omit the $!\pmod m,$ if the modulus is clear from the context. That is not advised if one is working with congruences wrt more than one modulus (possible ambiguity). – Bill Dubuque Feb 12 '15 at 05:25
  • I don't get how -2 * 3 is scaled to 1 and -2 * 4 is scaled to 6? – committedandroider Feb 12 '15 at 05:32
  • @com No scaling there, just modular arithmetic $,-2(3)\equiv -6\equiv 7-6\equiv 1,,$ and $,-2(4)\equiv -8\equiv 2(7)-8\equiv 6.\ $ Recall $, a\equiv a+7n\pmod 7,\ $ since $,7,$ divides their difference. $\ $ – Bill Dubuque Feb 12 '15 at 05:37
  • Why do you do 2 * 7 from −2(4)≡−8≡2(7)−8 but just 7 in −2(3)≡−6≡7−6? – committedandroider Feb 12 '15 at 05:48
  • @com I keep adding $,7,$ till I get a rep $,n,$ in the standard range $,0\le n < 7 = m.\ \ $ – Bill Dubuque Feb 12 '15 at 05:50
  • Oh because you want remainder to be non zero right. So essentially -6 mod 7 is -1 remainder 1 and -8 mod 7 is -2 remainder 6? – committedandroider Feb 12 '15 at 06:02
  • @com The integers $, -8+7n,$ are all $\equiv -8\pmod 7.,$ From that set we usually choose the least nonnegative element as the canonical representative, here $, -8+7(2) = 6.\ \ $ – Bill Dubuque Feb 12 '15 at 16:39
  • How did you go from x≡6 to 3x≡18? Based on what you talked about previously, wouldn't you want to use modular arithmetic to break the 18 in 3x ≡ 18(mod 7) into 4? because 18 = 7(2) + 4? – committedandroider Feb 13 '15 at 03:24
  • @com As above, we use the Congruence Product Rule to scale $,x\equiv 6,$ by $,3.,$ Yes, we reduce $,18,$ to its least positive rep $,4,$ by subtracting $,7,$ twice, which is equivalent to computing the remainder left when dividing $,18,$ by $,7,,$ i.e. $,4 = 18\ {\rm mod}\ 7.\ \ $ – Bill Dubuque Feb 13 '15 at 03:31
  • oh the two ≡ signs in 3x≡18≡4(mod7) has tripped me up for a while now. So you can just simply this down to 3x ≡4(mod7)? – committedandroider Feb 13 '15 at 03:37
  • @com Right, just like an ordinary equality, $,a\equiv b\equiv c,$ means $,a\equiv b,\ b\equiv c,\ $ so $\ a\equiv c,$ follows by transitivity (congruence is an equivalence relation). Again, you want to think of congruence equations just like ordinary equations, and learn to manipulate these generalized equations just as skillfully, by exploiting their many common properties. – Bill Dubuque Feb 13 '15 at 03:46
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You are trying to find solutions to the equivalence $3x \equiv 4 \pmod{7}$. What the author is saying is that both $-8$ and $6$ are solutions to the equivalence, which we can verify by direct substitution.

If we substitute $-8$ for $x$, we obtain \begin{align*} 3(-8) & \equiv -24 \pmod{7}\\ & \equiv -28 + 4 \pmod{7}\\ & \equiv -4 \cdot 7 + 4 \pmod{7}\\ & \equiv 4 \pmod{7} \end{align*} If we substitute $6$ for $x$, we obtain \begin{align*} 3 \cdot 6 & \equiv 18 \pmod{7}\\ & \equiv 14 + 4 \pmod{7}\\ & \equiv 2 \cdot 7 + 4 \pmod{7}\\ & \equiv 4 \pmod{7} \end{align*} Moreover, \begin{align*} -8 & \equiv -14 + 6 \pmod{7}\\ & \equiv -2 \cdot 7 + 6 \pmod{7}\\ & \equiv 6 \pmod{7} \end{align*} In fact, if $n \in \mathbb{Z}$, then $6 + 7n$ is a solution to the equivalence since \begin{align*} 3(6 + 7n) & \equiv 18 + 21n \pmod{7}\\ & \equiv 14 + 4 + 21n \pmod{7}\\ & \equiv 7(2 + 3n) + 4 \pmod{7}\\ & \equiv 4 \pmod{7} \end{align*} The author found the particular solution $-8$ (which corresponds to the choice $n = -2$). However, we usually wish to express the solution of an equivalence modulo $n$ as one of the residues $0, 1, \ldots, n - 1$. Since $n = 7$, the answer in the set of residues $\{0, 1, 2, 3, 4, 5, 6\}$ that is equivalent modulo $7$ to $-8$ is $6$.

N. F. Taussig
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  • Quick question: Why did you break −24 into -28 + 4? – committedandroider Feb 13 '15 at 03:31
  • So -8 + 7n would also be a solution then? – committedandroider Feb 13 '15 at 08:21
  • I broke $-24$ into $-28 + 4$ since $-28 = -4 \cdot 7$, so $-28 \equiv 0 \pmod{7}$, from which we can conclude that $-24 \equiv -28 + 4 \equiv -4 \cdot 7 + 4 \equiv 4 \pmod{7}$. In answer to your second question, $-8 + 7n$ is another way of writing the solution. I wrote it in the form $6 + 7n$ since we usually express the solution of an equivalence modulo $m$ as a residue from the set ${0, 1, 2, \ldots, m - 1}$. For $m = 7$, this means we usually choose the appropriate residue from the set ${0, 1, 2, 3, 4, 5, 6}$ of remainders obtained when an integer $k = 7q + r$, with $0 \leq r < 7$. – N. F. Taussig Feb 13 '15 at 11:32
  • Oh ok so both are accepted solutions then? But then isn't remainder defined as a value r that >= 0 and < m? Based off that I would think that my solution -8 + 7n isn't valid because I use a negative remainder and your solution is valid because you use a positive remainder, 6. – committedandroider Feb 13 '15 at 23:24
  • The best way to think of it is that $-8$ is a member of the residue class of $6 \pmod{7}$ since $-8 = 6 - 2 \cdot 7$. By the residue class of $6 \pmod{7}$, I mean all integers $n$ that differ from $6$ by an integer multiple of $7$. Since $-8$ and $6$ are in the same residue class, the solution $-8 + 7m, m \in \mathbb{Z}$ is valid since it produces the same set of integers as $6 + 7n, n \in \mathbb{Z}$ (take $m = n + 2$). That said, if $n$ is an integer such that $n = qk + r$, with $0 \leq r < k$, we usually think of $n$ as belonging to the residue class of the remainder $r \pmod{k}$. – N. F. Taussig Feb 13 '15 at 23:53