Lemma:
pf. Suppose j|a and j|b. Then
\(a=rj\) and \(b=sj\) \(\exists \ r,s \in \mathbb{Z}\)
\(b-a = rj-sj \ \implies \ (b-a)/j\)
\(mn=a\)
\(n|mn\)
By way of contradiction, suppose \(n|(mn+k)\) => \((mn+k)=b\)
By Lemma, \(n|(b-a)\) => \(n|((mn+k)-(mn))\) => \(n|k\)
\(n|k\) = \(np\) = k,
\(\exists \ p \in \mathbb{Z}\)
\(p\) = \(k/n\) => since \(n \ne 0\)
0 < \(k/n\) < 1
Since by way of contradiction \(n|(mn+k)\) is possible, then \((mn+k)|n\) is also possible.
This is looking pretty good! I corrected some LaTeX issues. Here are some suggestions to perfect your proof:
ReplyDelete(1) State the lemma as well as giving its proof. The statement of the lemma should be: "If \(j \in \mathbb{N} \) divides \(a,b \in \mathbb{N}\), then \(j\) divides \(b-a\).
(2) In the third line, \( (b-a)/j \) should be \( j \vert (b-a) \).
(3) Remove the lines
\( mn = a \)
\( n \vert nm \)
since they are somewhat confusing. Replace with "Let \(n,m \in \mathbb{N} \), and let \(k<n\).
(4) In the line beginning "By way of contradiction,..." remove "\(implies (mn+k) =b \)" because this is somewhat confusing. I see that you want to invoke the lemma here, which is the right thing to do. But it would be better to do so a bit more explicitly, like "Then by the Lemma, taking \( b = (mn+k) \) and \( a = mn \), we have \( n \vert k \), so \( np = k \ \exists \ k \in \mathbb{Z} \).
(5) Start a new sentence with "Since \( n=0\), \( p = \frac{k}{n} \)..." and after this, remove the last line, which is confusing, and replace with a clarified version: point out the contradiction and then conclude that hence \( n \nmid (mn+k) \).
Overall good work!
PS Also please tag your proof with labels relevant to the method (contradiction) and topics (divisibility)
ReplyDelete