Congruences and Modular Arithmetic: Examples and Counterexamples

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-07-31Open item page →

The system $x\equiv2\pmod3$ , $x\equiv3\pmod5$ , $x\equiv2\pmod7$ has the unique solution $x\equiv23\pmod{105}$

Example

The system

$x\equiv2\pmod3,\qquad x\equiv3\pmod5,\qquad x\equiv2\pmod7$

has exactly the solutions $x\equiv23\pmod{105}$ .

Facts & Assumptions

Given: The three displayed congruences.

[L1]

For a finite pairwise-coprime list of positive moduli, prescribed residues determine one class modulo their product (Chinese remainder theorem for a finite pairwise-coprime list: simultaneous residues determine one class modulo the product, and the resulting bijection preserves addition and multiplication).

[L2]

The relation $u\equiv v\pmod n$ means $n\mid(u-v)$ (Congruence modulo an integer: $a\equiv b\pmod n$ when $n\mid(a-b)$ , including the moduli $0$ and $1$ ).

Verification

technique · constructive

1.1

The numbers $3,5,7$ are pairwise coprime and have product $105$ . The complementary products $35,21,15$ satisfy $2\cdot35\equiv1\pmod3$ , $21\equiv1\pmod5$ , and $15\equiv1\pmod7$ , while each is divisible by the other two moduli.

L2

2.1

Therefore $x:=2(2\cdot35)+3(21)+2(15)=233$ has the prescribed three residues, and $233=23+2\cdot105$ gives $x\equiv23\pmod{105}$ .

constructstep 1.1L2

3.1

By [L1], all simultaneous solutions form one class modulo $3\cdot5\cdot7=105$ ; since step 2.1 exhibits $23$ in that class, it is exactly $[23]_{105}$ .

step 2.1L1discharge-construct∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-07-31Open item page →

The congruence $14x\equiv30\pmod{100}$ has exactly the two solutions $x\equiv45,95\pmod{100}$

Example

The congruence $14x\equiv30\pmod{100}$ has exactly the solution classes

$[45]_{100}\quad\text{and}\quad[95]_{100}.$

Facts & Assumptions

Given: The congruence $14x\equiv30\pmod{100}$ .

[L1]

A linear congruence $ax\equiv b\pmod n$ is solvable exactly when $\gcd(a,n)\mid b$ , and then has exactly $\gcd(a,n)$ solution classes modulo $n$ (For $n\ge1$ , $ax\equiv b\pmod n$ is solvable exactly when $\gcd(a,n)\mid b$ , and then has exactly $\gcd(a,n)$ solution classes modulo $n$ ).

[L2]

The extended Euclidean algorithm produces gcds and Bézout coefficients (The extended Euclidean algorithm: the same descent produces integers $x, y$ with $ax + by = \gcd(a,b)$ , so Bézout coefficients are computed and not merely shown to exist).

[L3]

Congruence modulo $n$ means divisibility of a difference by $n$ (Congruence modulo an integer: $a\equiv b\pmod n$ when $n\mid(a-b)$ , including the moduli $0$ and $1$ ).

Verification

technique · direct

1.1

The Euclidean algorithm gives $\gcd(14,100)=2$ , and $2\mid30$ . Cancelling the common factor reduces the congruence to $7x\equiv15\pmod{50}$ .

L1L2L3

1.2

Since $7\cdot43=301\equiv1\pmod{50}$ , multiplication by $43$ gives $x\equiv43\cdot15=645\equiv45\pmod{50}$ .

L3

1.3

The two lifts of this class modulo $100$ are $45$ and $45+50=95$ , and both verify the original congruence: $14\cdot45-30=600$ and $14\cdot95-30=1300$ are divisible by $100$ .

L3

2.1

By [L1] there are exactly two solution classes modulo $100$ , so the two distinct classes in step 1.3 exhaust them.

step 1.1step 1.2step 1.3L1∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-07-31Open item page →

$(\mathbb{Z}/12)^\times=\{[1],[5],[7],[11]\}$ and $\varphi(12)=4$

Example

$(\mathbb Z/12)^\times=\{[1]_{12},[5]_{12},[7]_{12},[11]_{12}\},\qquad\varphi(12)=4.$

Every displayed unit is its own inverse.

Facts & Assumptions

Given: The quotient $\mathbb Z/12$ and its unit group.

[L1]

The classes modulo $12$ have the unique standard representatives $0,\ldots,11$ (For $n\ge 1$ , every class in $\mathbb{Z}/n$ has one representative $r$ with $0\le r<n$ , so $\lvert\mathbb{Z}/n\rvert=n$ ; while $\mathbb{Z}/0$ is in bijection with $\mathbb{Z}$ ).

[L2]

The class $[a]_{12}$ is a unit exactly when $\gcd(a,12)=1$ , and $\varphi(12)$ counts these units (For $n\ge1$ , $[a]_n$ is a unit if and only if $\gcd(a,n)=1$ , The unit group $(\mathbb{Z}/n)^\times$ and Euler's totient $\varphi(n)=\lvert(\mathbb{Z}/n)^\times\rvert$ for $n\ge1$ ).

[L3]

Euler's finite prime-divisor product gives $\varphi(n)=\prod_{p^k\parallel n}(p^k-p^{k-1})$ (Euler's product formula $\varphi(n)=n\prod_{p\mid n}(1-1/p)=\prod_{p^k\parallel n}(p^k-p^{k-1})$ for $n\ge1$ , stated through a finite injective list of its prime divisors).

[F1]

Equality of residue classes is congruence of representatives (The congruence class $[a]_n$ and the quotient set $\mathbb{Z}/n$ ), and congruence means divisibility of their difference (Congruence modulo an integer: $a\equiv b\pmod n$ when $n\mid(a-b)$ , including the moduli $0$ and $1$ ).

[F2]

Products of residue classes are computed by multiplying representatives: $[a]_{12}[b]_{12}=[ab]_{12}$ (Addition and multiplication on $\mathbb{Z}/n$ by $[a]_n+[b]_n=[a+b]_n$ and $[a]_n[b]_n=[ab]_n$ ).

Verification

technique · direct

1.1

Among $0,\ldots,11$ , exactly $1,5,7,11$ have gcd $1$ with $12$ : every other representative is divisible by $2$ or $3$ . Thus [L1] and [L2] give the displayed unit group and $\varphi(12)=4$ .

L1L2

1.2

The congruences $5^2=25\equiv1$ , $7^2=49\equiv1$ , and $11^2=121\equiv1\pmod{12}$ show that the three nonidentity units, as well as $[1]_{12}$ , are self-inverse.

L2F1F2

2.1

Since $12=2^2\cdot3$ , [L3] independently gives $\varphi(12)=(2^2-2)(3-1)=2\cdot2=4$ , agreeing with the list.

step 1.1L3∎

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

The Chinese remainder bijection $\mathbb{Z}/12\to\mathbb{Z}/3\times\mathbb{Z}/4$ computed on all standard representatives

Example

For standard representatives, the Chinese remainder map is

$a$	$([a]_3,[a]_4)$	$a$	$([a]_3,[a]_4)$
$0$	$([0]_3,[0]_4)$	$6$	$([0]_3,[2]_4)$
$1$	$([1]_3,[1]_4)$	$7$	$([1]_3,[3]_4)$
$2$	$([2]_3,[2]_4)$	$8$	$([2]_3,[0]_4)$
$3$	$([0]_3,[3]_4)$	$9$	$([0]_3,[1]_4)$
$4$	$([1]_3,[0]_4)$	$10$	$([1]_3,[2]_4)$
$5$	$([2]_3,[1]_4)$	$11$	$([2]_3,[3]_4)$

Facts & Assumptions

Given: The map $\Phi([a]_{12}):=([a]_3,[a]_4)$ .

[L1]

Every class modulo a positive integer has one standard representative (For $n\ge 1$ , every class in $\mathbb{Z}/n$ has one representative $r$ with $0\le r<n$ , so $\lvert\mathbb{Z}/n\rvert=n$ ; while $\mathbb{Z}/0$ is in bijection with $\mathbb{Z}$ ).

[L2]

Since $3,4$ are coprime, $\Phi$ is a bijection and preserves addition and multiplication componentwise (Chinese remainder theorem for a finite pairwise-coprime list: simultaneous residues determine one class modulo the product, and the resulting bijection preserves addition and multiplication).

[L3]

Modular addition and multiplication are computed by adding and multiplying representatives (Addition and multiplication on $\mathbb{Z}/n$ by $[a]_n+[b]_n=[a+b]_n$ and $[a]_n[b]_n=[ab]_n$ ).

Verification

technique · direct

1.1

Reducing each integer $0,\ldots,11$ modulo $3$ and modulo $4$ gives the displayed table. Its entries are all distinct, and [L1] shows that they exhaust the domain and codomain standard pairs, in agreement with [L2].

L1L2

1.2

For a nontrivial addition check, $[7]_{12}+[8]_{12}=[3]_{12}$ , and the table gives $([1]_3,[3]_4)+([2]_3,[0]_4)=([0]_3,[3]_4)$ .

L3

1.3

For a nontrivial multiplication check, $[5]_{12}[7]_{12}=[11]_{12}$ , and the table gives $([2]_3,[1]_4)([1]_3,[3]_4)=([2]_3,[3]_4)$ .

L3

2.1

The table computes the whole bijection, while steps 1.2 and 1.3 instantiate its componentwise arithmetic preservation.

step 1.1step 1.2step 1.3L2∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-07-31Open item page →

$(\mathbb{Z}/8)^\times=\{[1],[3],[5],[7]\}$ is not cyclic because every element squares to $[1]$

Example

$(\mathbb Z/8)^\times=\{[1]_8,[3]_8,[5]_8,[7]_8\}$

is not cyclic: every element has square $[1]_8$ .

Facts & Assumptions

Given: The unit group $(\mathbb Z/8)^\times$ .

[L1]

The standard representatives of units are exactly the integers $a$ with $0\le a<8$ and $\gcd(a,8)=1$ (For $n\ge 1$ , every class in $\mathbb{Z}/n$ has one representative $r$ with $0\le r<n$ , so $\lvert\mathbb{Z}/n\rvert=n$ ; while $\mathbb{Z}/0$ is in bijection with $\mathbb{Z}$ , For $n\ge1$ , $[a]_n$ is a unit if and only if $\gcd(a,n)=1$ , The unit group $(\mathbb{Z}/n)^\times$ and Euler's totient $\varphi(n)=\lvert(\mathbb{Z}/n)^\times\rvert$ for $n\ge1$ ).

[L2]

A finite group is cyclic when one element generates it, and an element generates a finite group exactly when its order is the group cardinality (The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , and cyclic groups, The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity, If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

[L3]

Natural powers in a group satisfy $g^0=e$ and $g^2=gg$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[F1]

Equality of residue classes is congruence of representatives (The congruence class $[a]_n$ and the quotient set $\mathbb{Z}/n$ ), and congruence means divisibility of their difference (Congruence modulo an integer: $a\equiv b\pmod n$ when $n\mid(a-b)$ , including the moduli $0$ and $1$ ).

Verification

technique · direct

1.1

The odd standard representatives $1,3,5,7$ are precisely those coprime to $8$ , giving the displayed unit group by [L1].

L1

1.2

Their squares are $1^2=1$ , $3^2=9\equiv1$ , $5^2=25\equiv1$ , and $7^2=49\equiv1\pmod8$ . Hence every element has order at most $2$ .

L3F1

2.1

The group has cardinality $4$ by step 1.1, but no element has order $4$ by step 1.2. Therefore no element generates it, and [L2] shows that it is not cyclic.

step 1.1step 1.2L2∎

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)verified 2026-08-04 (gpt-5.6-sol-codex-subscription)Open item page →

$\varphi(360)=96$ by both inclusion-exclusion and the prime-power product formula

Example

Euler's totient at $360$ is $96$ , obtained either by excluding multiples of $2,3,5$ or by multiplying the prime-power contributions.

Facts & Assumptions

Given: The positive integer $360$ .

[L1]

Its canonical prime factorisation is $360=2^3\cdot3^2\cdot5$ (For $n \ge 1$ and any injective list $p : r \to \mathbb{Z}$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i<r} p_i^{\,v_{p_i}(n)}$ ; the exponents are determined by $n$ , and $v_q(n) = 0$ for every prime $q$ outside the list).

[L2]

Euler's formula gives $\varphi(n)=\prod_{i<r}\bigl(p_i^{k_i}-p_i^{k_i-1}\bigr)$ over the distinct prime divisors of $n$ and, after carrying the natural numbers into $\mathbb{Q}$ , the equivalent form $\varphi(n)=n\prod_{i<r}\left(1-\frac1{p_i}\right)$ (Euler's product formula $\varphi(n)=n\prod_{p\mid n}(1-1/p)=\prod_{p^k\parallel n}(p^k-p^{k-1})$ for $n\ge1$ , stated through a finite injective list of its prime divisors). Multiplying out that second product over the subsets of $\{p_0,\ldots,p_{r-1}\}$ is the inclusion-exclusion display used below; the cited theorem states the two products, not that display.

Verification

technique · direct

1.1

Inclusion-exclusion over the multiples of $2,3,5$ gives $\varphi(360)=360-180-120-72+60+36+24-12=96$ .

L1L2

1.2

The product form gives $\varphi(360)=(2^3-2^2)(3^2-3)(5-1)=4\cdot6\cdot4=96$ .

L1L2

2.1

Both computations therefore give the same value, $\varphi(360)=96$ .

step 1.1step 1.2∎

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

The system $x\equiv1\pmod2$ , $x\equiv2\pmod4$ has no solution, so coprimality in the Chinese remainder theorem cannot be dropped

Statement refuted

Refuted claim: prescribed residue classes always determine a simultaneous class even when the positive moduli are not coprime.

The system $x\equiv1\pmod2$ , $x\equiv2\pmod4$ is a counterexample.

Facts & Assumptions

Given: The two displayed congruences with moduli $2$ and $4$ .

[L1]

Congruence $a\equiv b\pmod n$ means $n\mid(a-b)$ (Congruence modulo an integer: $a\equiv b\pmod n$ when $n\mid(a-b)$ , including the moduli $0$ and $1$ ).

[L2]

For positive moduli $m,n$ , simultaneous residues $a,b$ are compatible exactly when $\gcd(m,n)\mid(a-b)$ ; pairwise coprimality makes this automatic (For coprime positive $m,n$ , every pair of residue classes modulo $m$ and $n$ has a unique simultaneous class modulo $mn$ , Chinese remainder theorem for a finite pairwise-coprime list: simultaneous residues determine one class modulo the product, and the resulting bijection preserves addition and multiplication).

Counterexample

technique · direct

1.1

If $x\equiv1\pmod2$ , then $x-1$ is even and $x$ is odd. If $x\equiv2\pmod4$ , then $x-2$ is divisible by $4$ and $x$ is even. No integer is both odd and even, so the system has no solution.

L1

1.2

Equivalently, $\gcd(2,4)=2$ does not divide $1-2=-1$ , so the compatibility criterion in [L2] fails.

L2

2.1

The claim without coprimality admits the unsolvable system in step 1.1 and is therefore false.

step 1.1step 1.2∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-07-31Open item page →

A modulus $n>1$ is composite exactly when two nonzero classes in $\mathbb{Z}/n$ have product $[0]_n$

Example

For every integer $n>1$ , the following are equivalent:

$n$ is composite.
There are nonzero classes $u,v\in\mathbb Z/n$ with $uv=[0]_n$ .

Facts & Assumptions

Given: An integer $n>1$ .

[L1]

A positive integer greater than $1$ is composite exactly when it has a divisor $d$ with $1<d<n$ ; otherwise it is prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L2]

Products in $\mathbb Z/n$ are computed on representatives, and the quotient has standard representatives $0,\ldots,n-1$ (Addition and multiplication on $\mathbb{Z}/n$ by $[a]_n+[b]_n=[a+b]_n$ and $[a]_n[b]_n=[ab]_n$ , For $n\ge 1$ , every class in $\mathbb{Z}/n$ has one representative $r$ with $0\le r<n$ , so $\lvert\mathbb{Z}/n\rvert=n$ ; while $\mathbb{Z}/0$ is in bijection with $\mathbb{Z}$ ).

[L3]

A class $[a]_n$ is a unit exactly when $\gcd(a,n)=1$ , and the gcd is a nonnegative common divisor that is positive when its arguments are not both zero (For $n\ge1$ , $[a]_n$ is a unit if and only if $\gcd(a,n)=1$ , Common divisor, and the greatest common divisor $\gcd(a,b)$ , with the convention $\gcd(0,0) := 0$ ).

Verification

technique · direct

1.1

If $n$ is composite, write $n=ab$ with $1<a<n$ and $1<b<n$ . Then $[a]_n$ and $[b]_n$ are nonzero standard classes, but $[a]_n[b]_n=[ab]_n=[n]_n=[0]_n$ .

L1L2

1.2

Conversely, suppose nonzero classes have product zero and choose their standard representatives $a,b$ , so $0<a,b<n$ and $n\mid ab$ . The class $[a]_n$ cannot be a unit, because multiplying an equation $[a]_n[b]_n=[0]_n$ by its inverse would give $[b]_n=[0]_n$ . Hence $d:=\gcd(a,n)>1$ by [L3]. Since $d\mid a$ and $0<a<n$ , one also has $d<n$ , so $d$ is a nontrivial divisor of $n$ and $n$ is composite by [L1].

L1L2L3

2.1

Steps 1.1 and 1.2 prove the equivalence.

step 1.1step 1.2∎

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

False: $[a]_n[b]_n=[0]_n$ always implies $[a]_n=[0]_n$ or $[b]_n=[0]_n$

Statement

False claim: for every natural modulus $n$ ,

$[a]_n[b]_n=[0]_n\quad\Longrightarrow\quad[a]_n=[0]_n\ \text{or}\ [b]_n=[0]_n.$

Facts & Assumptions

Given: The quotient $\mathbb Z/6$ and the classes $[2]_6,[3]_6$ .

[L1]

Products in $\mathbb Z/n$ are computed by multiplying representatives (Addition and multiplication on $\mathbb{Z}/n$ by $[a]_n+[b]_n=[a+b]_n$ and $[a]_n[b]_n=[ab]_n$ ).

[L2]

Composite moduli admit nonzero classes with zero product; moreover $[a]_n=[b]_n$ exactly when $a\equiv b\pmod n$ , which means $n\mid(a-b)$ (A modulus $n>1$ is composite exactly when two nonzero classes in $\mathbb{Z}/n$ have product $[0]_n$ , The congruence class $[a]_n$ and the quotient set $\mathbb{Z}/n$ , Congruence modulo an integer: $a\equiv b\pmod n$ when $n\mid(a-b)$ , including the moduli $0$ and $1$ ).

Refutation

technique · direct

1.1

In $\mathbb Z/6$ , $[2]_6[3]_6=[6]_6=[0]_6$ by [L1].

L1

1.2

Neither $6\mid2$ nor $6\mid3$ , so $[2]_6\ne[0]_6$ and $[3]_6\ne[0]_6$ .

L2

2.1

Thus the antecedent of the claimed implication holds while both alternatives in its conclusion fail, so the claim is false.

step 1.1step 1.2∎

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-07-31Open item page →

$\mathbb{Z}/1$ has one element and satisfies $[0]_1=[1]_1$ , so it is not a field

Statement refuted

The modular operations do not make $\mathbb Z/n$ a field for every positive modulus $n$ : the boundary modulus $n=1$ is a counterexample.

Facts & Assumptions

Given: The quotient $\mathbb Z/1$ with its modular operations.

[L1]

The quotient $\mathbb Z/1$ has exactly one class (For $n\ge 1$ , every class in $\mathbb{Z}/n$ has one representative $r$ with $0\le r<n$ , so $\lvert\mathbb{Z}/n\rvert=n$ ; while $\mathbb{Z}/0$ is in bijection with $\mathbb{Z}$ ).

[L2]

Its modular operations satisfy the abelian-group, commutative-monoid and distributive identities, with additive identity $[0]_1$ and multiplicative identity $[1]_1$ (For every natural $n$ , $(\mathbb{Z}/n,+)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L3]

A field must have distinct additive and multiplicative identities (Field).

Counterexample

technique · direct

1.1

By [L1], all integers determine the same class modulo $1$ , so in particular $[0]_1=[1]_1$ .

L1

2.1

Although the other algebraic identities hold by [L2], the equality in step 1.1 violates the distinct-identities clause [L3]. Hence $\mathbb Z/1$ is not a field.

step 1.1L2L3∎

Congruences and Modular Arithmetic: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The system x≡2(mod3)x\equiv2\pmod3x≡2(mod3), x≡3(mod5)x\equiv3\pmod5x≡3(mod5), x≡2(mod7)x\equiv2\pmod7x≡2(mod7) has the unique solution x≡23(mod105)x\equiv23\pmod{105}x≡23(mod105)

Example

Facts & Assumptions

Verification

The congruence 14x≡30(mod100)14x\equiv30\pmod{100}14x≡30(mod100) has exactly the two solutions x≡45,95(mod100)x\equiv45,95\pmod{100}x≡45,95(mod100)

Example

Facts & Assumptions

Verification

(Z/12)×={[1],[5],[7],[11]}(\mathbb{Z}/12)^\times=\{[1],[5],[7],[11]\}(Z/12)×={[1],[5],[7],[11]} and φ(12)=4\varphi(12)=4φ(12)=4

Example

Facts & Assumptions

Verification

The Chinese remainder bijection Z/12→Z/3×Z/4\mathbb{Z}/12\to\mathbb{Z}/3\times\mathbb{Z}/4Z/12→Z/3×Z/4 computed on all standard representatives

Example

Facts & Assumptions

Verification

(Z/8)×={[1],[3],[5],[7]}(\mathbb{Z}/8)^\times=\{[1],[3],[5],[7]\}(Z/8)×={[1],[3],[5],[7]} is not cyclic because every element squares to [1][1][1]

Example

Facts & Assumptions

Verification

φ(360)=96\varphi(360)=96φ(360)=96 by both inclusion-exclusion and the prime-power product formula

Example

Facts & Assumptions

Verification

The system x≡1(mod2)x\equiv1\pmod2x≡1(mod2), x≡2(mod4)x\equiv2\pmod4x≡2(mod4) has no solution, so coprimality in the Chinese remainder theorem cannot be dropped

Statement refuted

Facts & Assumptions

Counterexample

A modulus n>1n>1n>1 is composite exactly when two nonzero classes in Z/n\mathbb{Z}/nZ/n have product [0]n[0]_n[0]n​

Example

Facts & Assumptions

Verification

False: [a]n[b]n=[0]n[a]_n[b]_n=[0]_n[a]n​[b]n​=[0]n​ always implies [a]n=[0]n[a]_n=[0]_n[a]n​=[0]n​ or [b]n=[0]n[b]_n=[0]_n[b]n​=[0]n​

Statement

Facts & Assumptions

Refutation

Z/1\mathbb{Z}/1Z/1 has one element and satisfies [0]1=[1]1[0]_1=[1]_1[0]1​=[1]1​, so it is not a field

Statement refuted

Facts & Assumptions

Counterexample

Sources

Standard references

The system $x\equiv2\pmod3$ , $x\equiv3\pmod5$ , $x\equiv2\pmod7$ has the unique solution $x\equiv23\pmod{105}$

The congruence $14x\equiv30\pmod{100}$ has exactly the two solutions $x\equiv45,95\pmod{100}$

$(\mathbb{Z}/12)^\times=\{[1],[5],[7],[11]\}$ and $\varphi(12)=4$

The Chinese remainder bijection $\mathbb{Z}/12\to\mathbb{Z}/3\times\mathbb{Z}/4$ computed on all standard representatives

$(\mathbb{Z}/8)^\times=\{[1],[3],[5],[7]\}$ is not cyclic because every element squares to $[1]$

$\varphi(360)=96$ by both inclusion-exclusion and the prime-power product formula

The system $x\equiv1\pmod2$ , $x\equiv2\pmod4$ has no solution, so coprimality in the Chinese remainder theorem cannot be dropped

A modulus $n>1$ is composite exactly when two nonzero classes in $\mathbb{Z}/n$ have product $[0]_n$

False: $[a]_n[b]_n=[0]_n$ always implies $[a]_n=[0]_n$ or $[b]_n=[0]_n$

$\mathbb{Z}/1$ has one element and satisfies $[0]_1=[1]_1$ , so it is not a field