See the first lemma in this http://marauder.millersville.edu/~bi...visibility.pdf
If a|b and a|c, show that a|(bm+cn).
I have already proved that if x|y, and y|z, then x|z.
So, if a|b, then b|bm, so a|bm.
Also, if a|c, then c|cn, so a|cn.
So I think I need to show that a|aj+ai for some integers j=bm, i=cn.
ax = aj+ai
ax = a(j+i)
Let x = (j+i), thus ax = ax.
Could someone let me know if this proof is correct?
Also, as was noted, you did not end up with what you wanted to prove.
If you know modular arithmetic notation, this proof is a lot shorter. We have
Of course this relies on the properties of the congruence relation, which would be proven separately.
Also, you misuse the word "then" marked in blue. Of course "p implies q" holds if "q" is always true, but it is misleading in your proof.
implying there exists j such that aj = bm.
implying aj = bm for some j.
But take a look at HallsofIvy's post for a short proof that doesn't use the congruence notation I mentioned.
I understand your intent, but I still think it's an abuse of notation. Consider:
1: Let . Suppose is a perfect square. Then .
2: Let be a perfect square. Now let . Then .
Note that (1) and (2) essentially are the same, but with the order of declaration changed.
You see how this wording suggests that any integer satisfies , rather than some integer ?