i claim a necessary and sufficient condition for the set

A = {m in N: m = 3j+8k, with j,k in N}

to be inductive, is it contain 3 consecutive integers: and that the desired n of the original problem is, in this case (that is, when A is inductive): min(A) = 14.

sufficiency should be clear: if A is inductive, it must contain 3 consecutive integers.

the necessity is the tough part: if A contains 3 consecutive integers, r, r+1, and r+2, so that:

r = 3j+8k

r+1 = 3j'+8k'

r+2 = 3j"+8k",

then A contains r+q for any non-negative integer q. why? because any integer (not just non-negative ones) can be written either as q = 3s, 3s+1, or 3s+2.

case 1) r+q = r+3s --> r+q = 3(s+j) + 8k

case 2) r+q = r+(3s+1) --> r+q = (r+1)+3s = 3(s+j') + 8k'

case 3) r+q = r+(3s+2) --> r+q = (r+2)+3s = 3(s+j") + 8k".

so it remains to be shown that min(A) = 14.

14 = 3(2) + 8(1)

15 = 3(5) + 8(0)

16 = 3(0) + 8(2) so these 3 consecutive integers are in A.

13 = 3j+8k --> k < 2 (since for k ≥ 2, j ≥ 0 3j + 8k ≥ 16 > 13).

13 = 3j + 8 --> 5 = 3j, no integer solution, since 3 does not divide 5.

13 = 3j, no integer solution since 3 does not divide 13.

thus 13 is not in A, but 14 is. so min(A) = 14, when A is inductive.

one can, and some may prefer to, use strong induction on cases 1-3 above to show that r in A reduces to either 14,15 or 16 in A.

*****

in response to MoeBlee:

in all fairness, it's hard to know what level of detail is required here. to me, at an informal level, a proof is as follows: we can't make 13, we can make 14,15 and 16. just add 3-cent stamps to one of those numbers, and you can get any bigger integer you like. any formalism beyond that, is just "mathese" for the...an appropriate term might get me infracted, so i will just say..."purists". the original problem is stated in "lay terms", honestly, there's a lot of over-thinking going on.

the orginal poster made a perfectly good argument in his first go, i merely pointed out he left a "gap", in that he had two cases: and he did not show that after applying one of them, he would always still have case 1 or case 2 (in case 2, he could always subsequently apply case 1, but in case 1, he did not adequately show that either case 1 or case 2 MUST apply. he remedied that adequately in a later post).

i think you are confusing elegance with clarity. while from a purely aesthetic standpoint, elegance might be preferrable, it's not necessary. your argument isn't very clear. it makes sense, once it's unravelled, but it places more of a demand on the reader than Sneaky's does. i stand by my original assessment, i don't find it straight-forward. it's not so dense as to be unintelligible, it's just not straight-forward.

by your own admission, you make an existential and universal quantification on the same variable, q. understanding WHY this is permissible (and not logically indefensible) makes QUITE a demand on a reader, especially for an induction proof, which are often quite elementary in nature. in reading your proof for the first time, i had to go back and tell myself: "oh, he doesn't mean the same q in the same way HERE, as he does THERE".

you also seem to be reacting to this quite personally. the fact is, i don't know you, and i certainly don't have any feelings about you one way or the other. if i had to guess, i'd say you probably have a better command of logic and proof than i do (i am rusty, and out of practice, and have forgotten most of what i once knew).