lean_workbook_plus_47303

Let $a_1,a_2$ be integers such that $a_1\equiv a_2\pmod{6}$ and let $b$ be an integer. Show that\n\n

\n\nWhat if $b=6$ ? Then the expression would be \n\n

If we use the proof we derived in Chapter 14, then $a_1b\equiv a_2b\pmod{1}$ , and there must be a problem here, since there would be infinite possibilities to this congruence.\n\nBy definition, all integers are congruent to each other in modulo $1$ . \n\nSince $a_1$ , $a_2$ and $b$ are all integers, this means that $a_1b\equiv a_2b\pmod1$ , then this means $6a_1b\equiv6a_wb\pmod6$ .\n\nSuppose that we have: \n\n

If we multiply both sides by $6$ , we discover that $6a_1$ is divisible by $6$ , which means that $6a_1\equiv0\pmod6$ . Similiarly, $6a_2\pmod6$ . \n\nThis shows us that: \n\n

which is what we wanted to prove. \n\nThis is like multiplying both sides by $0$ . Suppose that: \n\n

If we multiply both sides by $0$ , then $0\cdot x=0$ and $0\cdot y=0$ , which means that: \n\n

or \n\n

which is still true, even though there may be infinitely many possibilities to it.