Divisibility is an existential integer relation
Objective: Without looking at the worked lines, prove or verify this exact claim: “If a∣b and a∣c, then a∣(mb+nc) for all integers m,n.” Then solve “Show 7 divides 3·35−2·14 and give the quotient witness.” and check the result in the original equation, congruence, factorization, or algorithm.
Writing a∣b means a scales by some integer to equal b; the quotient witness matters, and zero cases must be handled explicitly.
Read the notation before using it: For integers a,b, a∣b means there exists k∈Z with b=ak. The statement to establish is: If a∣b and a∣c, then a∣(mb+nc) for all integers m,n. Every quotient, remainder, coefficient, exponent, residue, or factor that makes a claim true is kept as a checkable witness.
The worked question is: Show 7 divides 3·35−2·14 and give the quotient witness. Follow the numbered exact-arithmetic lines, then compare the result with the original statement. The failure boundary is shown separately so that a familiar formula is not used after one of its hypotheses has disappeared.
If a∣b and a∣c, then a∣(mb+nc) for all integers m,n.
Choose witnesses b=ar and c=as.
Substitute into mb+nc=a(mr+ns).
Since mr+ns is an integer, the definition gives divisibility.
Show 7 divides 3·35−2·14 and give the quotient witness.
- Compute 35=7·5 and 14=7·2.
- Then 3·35−2·14=7(15−4).
- The expression equals 7·11=77.
Result: Yes; the quotient witness is 11.