write converse given statement if number divisible by 6 than its divisible by 2 and 3
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The point was to prove separately inverse, converse and contrapositive statements of the given statement: "for all integers n, if n is divisible by 6, then n is divisible by 3 and n is divisible by 2". I have the proof for converse and inverse similar to that given in comments. I have trouble only with the proof that integer not divisible by 2 or 3 is not divisible by 6.
As I review my proof for inverse statement, I'm not sure of it as well. "For all integers n, if n is not divisible by 6, n is not divisible by 3 or n is not divisible by 2."
n = 6*x where x in not an integer
n = 2*3*x
n/2 = 3*x and n/3 = 2*x where 2x or 3x is not an integer,
so n is not divisible by 2 or 3
3
One line with contrapositive. What have you tried, and where are you having difficulty? Please edit your question to include something more than a problem statement; this is not a do-my-homework site. – user296602 Oct 7 '18 at 22:11
By proving that if it is divisible by 6 then it is divisible by both. – fleablood Oct 7 '18 at 22:35
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If 6∣n, then n=6⋅k for some k. So n=2⋅3⋅k. Thus n is divisible by 2 and 3.