Question

Let m be a positive integer. Show that a mod m - b mod m t a - b (mod m) Drag the necessary statements and drop them into the appropriate blank to build your proof (mod m Dag the mecesary eemnes a ohem int the approprite Proof method: Proof assumptions), at-qm + Proof by contradiction aaandh mam it Implication(s) and deduction(s) resulting from the assumption(s): a mk + bmk Hqm tr a-(k + q)m+ r Conclusion(s) from implications and deductions: a mod m b mod m Counterexample m+ (a - b) a b (mod m) or a & (mod m) a mod m + b mod m r a mod m m(mod m) nd Direct proof 4 mod 2 Proof tby contradiction ak E Z: a mk"m + qm t 0

Answer 1

Direct Proof :-

Proof Assumption :- $\exists k \in \mathbb{Z}:a-b = mk$

Implications and deduction resulting from assumption :-

$\Rightarrow m | (a-b) \text{ and }b=qm+r$

$\Rightarrow a = mk+qm+r$

$\Rightarrow a = (k+q)m+r$

$\Rightarrow r = a \mod m$

$\Rightarrow a = b ( \mod m)\text{ and }r= b \mod m$

Conclusion from implication and deductions:

$a \mod m = b \mod m$

Please comment for any clarification.