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Theorem: Let m be a positive integer. If a ≡ b (mod m) and c ≡ d (mod m), then a + c ≡ b + d (mod m) and ac ≡ bd (mod m).

Proof:We use a direct proof. Because a ≡ b (mod m) and c ≡ d (mod m), by Theorem 4 there are integers s and t with b = a + sm and d = c + tm. Hence, b + d = (a + sm) + (c + tm) = (a + c) + m(s + t) and bd = (a + sm)(c + tm) = ac + m(at + cs + stm). Hence, a + c ≡ b + d (mod m) and ac ≡ bd (mod m).

how does b + d = (a + sm) + (c + tm) = (a + c) + m(s + t) and bd = (a + sm)(c + tm) = ac + m(at + cs + stm). show that a + c ≡ b + d (mod m) and ac ≡ bd (mod m).? and shouldnt it be a = b + sm not b = a + sm since its a ≡ b (mod m) ?

Bill Dubuque
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user114300
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2 Answers2

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The proofs are clearer by divisibility, i.e. $\,m\mid a,b\,\Rightarrow\, m\mid ja+kb\,$ for any integers $\,j,k,\,$ see below.

Congruence Sum Rule $\rm\qquad\quad A\equiv a,\quad B\equiv b\ \Rightarrow\ \color{#0a0}{A+B\,\equiv\, a+b}\ \ \ (mod\ m)$

Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a) + (B\!-\!b)\ =\ \color{#0a0}{A+B - (a+b)} $

Congruence Product Rule $\rm\quad\ A\equiv a,\ \ and \ \ B\equiv b\ \Rightarrow\ \color{#c00}{AB\equiv ab}\ \ \ (mod\ m)$

Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a)\ B + a\ (B\!-\!b)\ =\ \color{#c00}{AB - ab} $

Congruence Power Rule $\rm\quad\ A\equiv a\ \Rightarrow\ A^n\equiv a^n\ \ (mod\ m)$

Proof $\ $ It is true for $\rm\,n=1\,$ and $\rm\,A\equiv a,\ A^n\equiv a^n \Rightarrow\, A^{n+1}\equiv a^{n+1},\,$ by the Product Rule, so the result follows by induction on $\,n.$

Polynomial Congruence Rule $\ $ If $\,f(x)\,$ is polynomial with integer coefficients then $\ A\equiv a\ \Rightarrow\ f(A)\equiv f(a)\,\pmod m.$

Proof $\ $ By induction on $\, n = $ degree $f.\,$ Clear if $\, n = 0.\,$ Else $\,f(x) = f(0) + x\,g(x)\,$ for $\,g(x)\,$ a polynomial with integer coefficients of degree $< n.\,$ By induction $\,g(A)\equiv g(a)\,$ so $\, A g(A)\equiv a g(A)\,$ by the Product Rule. Hence $\,f(A) = f(0)+Ag(A)\equiv f(0)+ag(a) = f(a)\,$ by the Sum Rule.

Beware $ $ that such rules need not hold true for other operations, e.g. the exponential analog of above $\rm A^B\equiv a^b$ is not generally true (unless $\rm B = b,\,$ so it reduces to the Power Rule, so follows by inductively applying $\,\rm b\,$ times the Product Rule).

Bill Dubuque
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Since $b + d = (a + sm) + (c + tm) = (a + c) + m(s + t)$,

$\;\;\;(a+c)-(b+d)=m(-s-t)$ so $a+c\equiv b+d\pmod m$ since $m\vert((a+c)-(b+d))$.

Since $bd = (a + sm)(c + tm) = ac + m(at + cs + stm)$,

$\;\;\;ac-bd=m(-at-cs-stm)$ so $ac\equiv bd\pmod m$ since $m\vert(ac-bd)$.

user84413
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