1

Just what the title says, I have searched but haven't found anything, either for yes or no. Does somebody know?

My reasoning is:

$a \equiv b \pmod m$

$\implies a = b + mk$, for some $k$

$\implies$ multiply both sides by $m$ and I get $am = (b + mk)m$

$\implies am = bm + (m^2)k$

$\implies am \equiv bm \pmod {m^2}$

Is that right? Am I missing something?

Bill Dubuque
  • 272,048
G25
  • 19
  • 5
    This follows directly from the definition: $a \equiv b \bmod m$ if and only if $a-b = mx$ for some integer $x$. This is equivalent to your argument, which is fine. – Randall Dec 23 '23 at 00:25
  • 1
    Indeed, $a\equiv b\bmod m\iff an\equiv bn\bmod mn$. – coiso Dec 23 '23 at 00:25
  • 2
    For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. It's best for site health to delete this dupe question (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Dec 23 '23 at 00:55

0 Answers0