In the proof of the theorem of inverse modulo. I encountered this problem and unable to comprehend it.
sa + tm ≅ 1 (mod m)
Then why it is equal to
sa ≡ 1 (mod m)
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1Because $m \equiv 0 \pmod m$. – Alex Provost Jan 07 '21 at 08:16
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$,\color{#c00}{m\equiv 0}\Rightarrow t\cdot\color{#c00}m\equiv t\cdot \color{#c00}0\equiv 0,$ by the linked Congruence Product Rule. so $, sa + tm \equiv sa + 0\equiv sa,$ by the linked Congruence Sum Rule. – Bill Dubuque Jan 07 '21 at 08:42
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$\equiv$ is a congruence relation and $m \equiv 0 \pmod m$, so $$sa + tm \equiv sa + t\cdot 0 \equiv sa \pmod m.$$
Alex Provost
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I do know that
m ≡ 0 (mod m)but I want to know that what is the rule through which we can replace m by 0 in the congruencesa + tm ≡ 1 (mod m)? – Ali Husnain Jan 07 '21 at 08:24 -
1@AliHusnain $m \equiv 0 !\pmod{m}$ implies $tm \equiv 0 !\pmod{m}$, and then add $sa$ on both sides to get $1 \equiv sa+tm \equiv sa+0 = sa !\pmod{m}$. More generally, if $a \equiv b \pmod{n}$, then $f(a) \equiv f(b) \pmod{n}$ for any polynomial $f(x)$ with integer coefficients. – azif00 Jan 07 '21 at 08:39
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@AliHusnain These are immediate consequences of the basic rules of congruence arithmetic (sum & product rules) = see the linked dupe for their proofs. – Bill Dubuque Jan 07 '21 at 08:45