We can solve $8x \equiv 9 \pmod {15} $ by multiplying both sides by $8^{-1} \equiv 2 \pmod {15}$, getting $x \equiv 3 \pmod {15}$ so $x=3$. I know that the multiplicative inverse of a number is unique; but I am struggling conceptually to justify why this is a unique solution to $8x \equiv 9 \pmod {15} $. How can we show that there exists no other solution for x here?
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Bill Dubuque
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Princess Mia
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2Your argument already shows this. You assumed $x$ was a solution to $8x\equiv 9 \mod 15$, and then proved using that assumption that $x$ must satisfy $x\equiv 3 \mod 15$. – M W Aug 24 '23 at 01:10
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1If $8x \equiv 8y \pmod {15}$, then $8(x-y) \equiv 0 \pmod {15}$. But $8$ is not a zero divisor in $\Bbb Z / 15 \Bbb Z$ (that's why it has an inverse), so $8(x-y) \equiv 0 \pmod {15} \Rightarrow x-y \equiv 0 \pmod {15} \Rightarrow x \equiv y \pmod {15}$. This proof is slightly more general than yours because it doesn't require $8$ to have an inverse, just that it not be a zero divisor. Of course, in a finite commutative ring, the two conditions are equivalent. – Robert Shore Aug 24 '23 at 01:12
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2Strictly speaking, the solution is not unique. It is only unique modulo $15$. – Brauer Suzuki Aug 24 '23 at 04:17
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[Recall $ $ "solutions are unique $\!\bmod n$" means that if $x_1\,$ is a solution then every other solution $x_2$ is congruent to it $\,x_2\equiv x_1\pmod{\!n}\,$ or, equivalently, solutions of $\,8x=9\,$ are unique in $\,\Bbb Z_{15}$]
The proof uses only existence (not uniqueness) of inverses, i.e. suppose $\,a'\,$ is some inverse of $\,a.\,$ Scaling $\,ax\equiv b\,$ by $\,a'\,$ yields $\,x\color{#c00}{\equiv a'b}.\,$ Uniqueness follows immediately: if $\,x_1,x_2\,$ are roots then $\,x_1\color{#c00}{\equiv a'b\equiv} x_2\,$ (which implictly exploits the fact that congruence, like equality, is an equivalence relation so it is transitive, i.e. $\,x_1\color{#c00}{\equiv y}\equiv x_2\,\Rightarrow\, x_1\equiv x_2)$.
Existence also follows immediately since scaling an equation or congruence by a unit (invertible) is an invertible operation so it yields an equivalent equation or congruence. Thus the above scaling is reversible: scaling $\,x\equiv a'b\,$ by $\,a\,$ yields $\,ax\equiv b,\,$ so $\,a'b\,$ is indeed a root. Summarizing
$${\rm if}\ \ \ a'a \equiv 1 \equiv aa'\ \ \ {\rm then}\ \ \ ax\equiv b\iff x\color{#c00}{\equiv a'b}\qquad$$
Remark $ $ That the proof uses only the existence (not uniqueness) of some inverse $\,a'\,$ of $\,a\,$ is a point often misunderstood, perhaps because some (very) popular textbooks mislead students by wrongly claiming that $\,x\,$ is unique because inverses of $\,a\,$ are unique, e.g. see here.
Though it is completely obvious with hindsight, it is often overlooked that uniqueness is an immediate consequence of an explicit solution $\,x\equiv\, \ldots\,$ (as in the direction $(\Rightarrow)$ above). See here for further discussion of another textbook example (additive case). Another example occurred here a couple weeks ago.
The above shows that scaling a congruence by a unit (invertible) yields an equivalent congruence. This is the congruence analog of a fact that is well know for equations, e.g. scaling a polynomial equation to an equivalent equation that is monic (lead coefficient $= 1),\,$ e.g. choosing monic minimal polynomials, or unit-normalized polynomial gcds in polynomial rings $\,k[x]$.
Bill Dubuque
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That scaling a congruence yields another true congruence is the Congruence Product Rule. Congruences, like equality, are equivalence relations, so we can work with congruences in much the same way as we do equations, esp. since they are compatible with addition and multiplication (by the Congruence Sum & Product Rules). So we can think of congruences as generalized equations (they can be reified as equations when one learns about quotient rings, here $,\Bbb Z/m\Bbb Z\cong \Bbb Z_m)$ – Bill Dubuque Aug 24 '23 at 03:18
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The question was originally written with solution $x=3$. It has now been edited to read $x \equiv 3$
The following refers to the question as it was originally asked with x=3, which technically refers specifically and exactly to only 3.
It's not. For example, $x= 18$ is also a solution. In fact $x=3+ 15n$ for $n\in \mathbb Z$ are also solutions.
Unless you meant the entire equivalence class {3 + 15n} in which case our answers are the same.
nickalh
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Hmmm, am I missing something? There are two other comments suggesting I'm incorrect. – nickalh Aug 24 '23 at 01:15
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1The tacit assumption is that we're looking at elements in $\Bbb Z / 15 \Bbb Z$. In that ring, all of your proposed solutions are identical. – Robert Shore Aug 24 '23 at 01:16
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1Uniqueness of roots of $f(x)\equiv 0\pmod{m},$ means if $x_1,x_2$ are roots then $,x_1\equiv x_2\pmod{m},,$ i.e. the roots are equal in $\Bbb Z_m.,$ It's best to delete this answer to avoid further confusion. – Bill Dubuque Aug 24 '23 at 01:57
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1It’s not mod 3. It should technically be $x\equiv 3 \mod 15$ or $x=[3]_{15}$, but it’s tacitly understood what they’re trying to show. – Eric Aug 24 '23 at 02:45
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1No, $,x\pmod{!3},$ is incorrect notation (cf. comments here). When working with congruence relations the correct way to express the solution is $,x\equiv 3\pmod{!15}$ – Bill Dubuque Aug 24 '23 at 03:08
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Oops, my comment had a typo. I meant $x=3$ is incorrect notation.
I meant to suggest your correct notation $x \equiv 3$ or even $x \equiv 3 (mod 15)$My previous comment simply used the wrong tag. pmod instead of equiv.
– nickalh Aug 24 '23 at 04:57 -
About deleting my answer- students and people come to mathSE with a wide variety of different levels of mathematical understanding. Most of the participants are graduate level and above. However, many undergraduate students and a few high school students come here. They will be significantly confused by BillDubuque 's answer. Unit's, monic minimal poly's gcd's, etc. particularly. Sometimes a simpler answer is useful particularly for students at earlier levels. – nickalh Aug 24 '23 at 05:43