These first questions were all with o4-mini. The theorem statements are all from this book. I found ChatGPT’s proofs more didactic than the originals.
For any fixed natural number m greater than 1, we say that the integer a is congruent to the integer b modulo m if a − b is divisible by m. Show that the congruence relation (modulo m) is transitive.
To prove transitivity, start with the hypotheses:
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$a \equiv b \pmod m$. This means $m$ divides $a-b$, so there exists an integer $k$ such that
$$ a - b = km. $$
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$b \equiv c \pmod m$. This means $m$ divides $b - c$, so there exists an integer $ℓ$ such that
$$ b - c = ℓm. $$
Now consider $a - c$. We can write
$$ a - c \;=\; (a - b) + (b - c) \;=\; km + ℓm \;=\; (k + ℓ)\,m. $$
Since $k + ℓ$ is an integer, this shows $m$ divides $a-c$. Hence
$$ a \equiv c \pmod m, $$
which is exactly the statement of transitivity for congruence modulo $m$.
Show that when a and b are nonnegative integers, the relationship a ≡ b (mod m) is equivalent to a and b leaving equal remainders upon division by m.
To show the two statements are equivalent, we prove each implication in turn.
1. If $a \equiv b \pmod m$, then $a$ and $b$ have the same remainder upon division by $m$.
Write the Euclidean divisions of $a$ and $b$ by $m$:
$$ \begin{aligned} a &= q_a\,m + r_a, \quad 0 \le r_a < m,\\ b &= q_b\,m + r_b, \quad 0 \le r_b < m. \end{aligned} $$
Then
$$ a - b = (q_a - q_b)\,m + (r_a - r_b). $$
If $a \equiv b \pmod m$, by definition $m$ divides $a - b$. But
$$ a - b = (q_a - q_b)\,m + (r_a - r_b) $$
and $(q_a - q_b)\,m$ is already a multiple of $m$. For the whole sum to be a multiple of $m$, the “leftover” term $r_a - r_b$ must also be a multiple of $m$. Since
$$ |r_a - r_b| < m, $$
the only multiple of $m$ in that range is $0$. Hence
$$ r_a - r_b = 0 \quad\Longrightarrow\quad r_a = r_b. $$
2. If $a$ and $b$ leave the same remainder upon division by $m$, then $a \equiv b \pmod m$.
Again write
$$ a = q_a\,m + r,\quad b = q_b\,m + r \quad\text{with }0 \le r < m. $$
Then
$$ a - b = (q_a\,m + r) - (q_b\,m + r) = (q_a - q_b)\,m, $$
which is manifestly divisible by $m$. Therefore $a \equiv b \pmod m$.
Combining (1) and (2) shows the two characterizations are logically equivalent.
Show that for a given modulus m, each integer is congruent to exactly one of the numbers in the set {0, 1, 2, …, m − 1}.
Proof. Fix an integer $m>1$. We must show that for any integer $a$ there is one—and only one—remainder $r$ with
$$ 0 \le r \le m-1 $$
such that
$$ a \equiv r \pmod m. $$
Existence
By the Division Algorithm, for any integer $a$ there exist unique integers $q$ and $r$ with
$$ a = q\,m + r \quad\text{and}\quad 0 \,\le\, r \,<\, m. $$
Rewriting gives
$$ a - r = q\,m, $$
so $m$ divides $a-r$, i.e.\ $a\equiv r\pmod m$. Thus some $r$ in $\{0,1,\dots,m-1\}$ is congruent to $a$.
Uniqueness
Suppose $r$ and $s$ are two remainders in $\{0,1,\dots,m-1\}$ with
$$ a \equiv r \pmod m \quad\text{and}\quad a \equiv s \pmod m. $$
Then $m$ divides both $a-r$ and $a-s$, so it also divides their difference:
$$ m \mid \bigl[(a-r) - (a-s)\bigr] \;\Longrightarrow\; m \mid (s - r). $$
But $r$ and $s$ lie in the range $0\le r,s< m$, so $|s-r|<m$. The only multiple of $m$ in that range is $0$. Hence
$$ s - r = 0 \quad\Longrightarrow\quad s = r. $$
Combining these two parts shows each integer $a$ is congruent to exactly one element of $\{0,1,2,\dots,m-1\}$.
Show that if a ≡ b (mod m) and c ≡ d (mod m), then
(i) (a + c) ≡ (b + d) (mod m), and
(ii) ac ≡ bd (mod m).
Proof. Let $m>1$ be fixed, and suppose
$$ a \equiv b \pmod m \quad\text{and}\quad c \equiv d \pmod m. $$
By definition, there exist integers $k$ and $\ell$ such that
$$ a - b = k\,m, \qquad c - d = \ell\,m. $$
(i) Addition
Consider
$$ (a + c) \;-\;(b + d) = (a - b) + (c - d) = k\,m + \ell\,m = (k + \ell)\,m. $$
Since $(k+\ell)$ is an integer, $m$ divides $(a+c)-(b+d)$. Hence
$$ a + c \equiv b + d \pmod m. $$
(ii) Multiplication
Consider
$$ a\,c \;-\; b\,d = a\,c \;-\; b\,c \;+\; b\,c \;-\; b\,d = c\,(a - b) \;+\; b\,(c - d). $$
Substituting the expressions above,
$$ c\,(a - b) + b\,(c - d) = c\,(k\,m) + b\,(\ell\,m) = (c\,k + b\,\ell)\,m, $$
which shows $m$ divides $ac - bd$. Therefore
$$ ac \equiv bd \pmod m. $$
Thus congruence modulo $m$ is compatible with both addition and multiplication.
Show that if a ≡ b (mod m), then a^n ≡ b^n (mod m), for every natural number n.
Proof by induction on $n$.
Fix a modulus $m>1$ and suppose $a \equiv b \pmod m$. We show $a^n \equiv b^n\pmod m$ for every natural number $n$.
Base case ($n=1$)
Trivially,
$$ a^1 = a \equiv b = b^1 \pmod m. $$
Inductive step
Assume for some $k\ge1$ that
$$ a^k \equiv b^k \pmod m. $$
We must show
$$ a^{k+1} \equiv b^{k+1} \pmod m. $$
Notice
$$ a^{k+1} \;=\; a^k \cdot a, \qquad b^{k+1} \;=\; b^k \cdot b. $$
By the inductive hypothesis $a^k \equiv b^k\pmod m$, and by our original assumption $a\equiv b\pmod m$. From the compatibility of congruence with multiplication (proved earlier), we conclude
$$ a^{k+1} = a^k\cdot a \;\equiv\; b^k\cdot b = b^{k+1} \pmod m. $$
By induction, $a^n \equiv b^n \pmod m$ for all natural numbers $n$. This completes the proof.
Show that every natural number is congruent to the sum of its digits modulo 9.
Proof. Every natural number $N$ can be written in its decimal expansion as
$$ N = d_0 + d_1\cdot 10 + d_2\cdot 10^2 + \cdots + d_k\cdot 10^k, $$
where each $d_i$ is one of the digits $0,1,\dots,9$. Call
$$ S = d_0 + d_1 + d_2 + \cdots + d_k $$
the sum of its digits.
We will show
$$ N \equiv S \pmod 9. $$
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Observe that $10 \equiv 1 \pmod 9$. Indeed, $10-1=9$ is divisible by 9.
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Use the fact that congruence is compatible with powers and multiplication. From $10\equiv1\pmod9$ it follows by induction (or by repeatedly applying the multiplicative property) that
$$ 10^i \equiv 1^i = 1 \pmod 9 \quad\text{for every }i\ge0. $$
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Apply congruence to each term in the expansion of $N$.
$$ N = \sum_{i=0}^k d_i\,10^i \equiv \sum_{i=0}^k d_i\cdot1 = \sum_{i=0}^k d_i = S \pmod 9. $$
Since every step uses only the rules of congruence (namely that if $x\equiv y$ and $u\equiv v\pmod m$, then $xu\equiv yv\pmod m$ and sums likewise agree), we conclude
$$ N \equiv S \pmod 9, $$
i.e.\ every natural number is congruent modulo 9 to the sum of its digits.
For this question I used o4-mini-high.
Summarize and contextualize the knowledge about congruences we have seen so far.
Here’s a bird’s‑eye view of what we’ve built up so far in our study of congruences modulo $m$, and why it matters:
1. Congruence as an Equivalence Relation
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Definition:
$$ a \equiv b \pmod m \quad\Longleftrightarrow\quad m \mid (a - b). $$
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Properties:
- Reflexive: $a\equiv a\pmod m$.
- Symmetric: If $a\equiv b$, then $b\equiv a$.
- Transitive: If $a\equiv b$ and $b\equiv c$, then $a\equiv c$.
Together, these show congruence partitions the integers into equivalence classes.
2. Equivalence Classes and Canonical Representatives
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Every integer $a$ is congruent to exactly one of
$$ \{0,1,2,\dots,m-1\}, $$
its remainder upon division by $m$.
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Thus the set of all integers splits into $m$ classes, often denoted $\Bbb Z/m\Bbb Z = \{[0],[1],\dots,[m-1]\}$.
3. Arithmetic in $\Bbb Z/m\Bbb Z$
Congruence “respects” the usual operations: if
$$ a\equiv b\pmod m, \quad c\equiv d\pmod m, $$
then
- Addition: $(a+c)\equiv(b+d)\pmod m.$
- Multiplication: $a\,c\equiv b\,d\pmod m.$
- Powers: $a^n\equiv b^n\pmod m$ for all $n\in\Bbb N$.
Because of this, one can do arithmetic “mod $m$” and get a well‑defined algebraic system (a ring).
4. A Classic Application: The “Divisibility by 9” Test
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Writing a decimal number $N$ as $\sum d_i\,10^i$, and using $10\equiv1\pmod9$, shows
$$ N \equiv \sum d_i \pmod9. $$
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Hence a number is divisible by 9 exactly when the sum of its digits is.
5. Why This Matters
- Number‑theoretic tools: solving linear congruences, understanding invertibility (“units” in $\Bbb Z/m\Bbb Z$), working toward Fermat’s Little Theorem and Euler’s Theorem.
- Chinese Remainder Theorem: decomposing problems mod $m$ into simpler ones mod its prime‑power factors.
- Cryptography & coding theory: RSA, Diffie–Hellman, checksums all rely on modular arithmetic.
- Practical divisibility tests: beyond 9, there are tests for 3, 11, 7, etc., all built on the same congruence ideas.
In short, congruences give us a way to “collapse” the infinite set of integers into finitely many residue classes while still doing meaningful arithmetic—and that simple idea underpins huge swaths of both pure and applied mathematics.
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