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Prove that for all integers "a" and "b" there is an integer "c" such that a|c and b|c
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Does 1 counts?
what is that symbol denoting?
division!!! :-)
\[2 \nmid 1\], so c does not work. Try \(c=a\cdot b\)
That should read \(c=1\) does not work. But if we let \(c=a\cdot b\), we basically have the proof by definition of divisibility right here. \(a|c\) since there exists some integer \(k\) (in this case \(k=b\)) such that \(c=a\cdot k\). It follows from a symmetrical argument that \(b|c\).
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No idea how to prove it. Using strong inductions?
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