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So in general, the contrapositive of the statement (\(\,p\implies q\)) is \((\neg \,q\implies \neg \,p\)). In this case, both your p and q are other sentences. Can you tell me what the "p" is in your conditional statement?
n is a prime number
Right. And the "q"?
n is an odd number
Almost. The "q" in your statement is "n=2 or n is an odd number." Notice that there's an "or" in this statement, so even this can be broken down into something like "q OR r," where q is "n=2", and r is "n is an odd number."
So the statement "n is a prime number implies that n=2 or n is an odd number" can be rewritten as \[p\implies (q \vee r)\]
if n is not a odd number then n is not a prime number or n = 2
You're very close, and certainly have the right idea. The contrapositive of \(p\implies (q\vee r)\) is \(\neg(q\vee r)\implies \neg p\). Can you tell me what \(\neg(q\vee r)\) can be rewritten as?
This is a thing called "De Morgan's Law." It says that \[\neg(q\vee r) \Longleftrightarrow (\neg \,q) \wedge(\neg\,r) \]Can you translate this back into our statements?
So if q is "n=2," what is ~q?
Right. And if r is "n is an odd number," what is ~r?
n is not an odd number
Bingo. So if we then combine those two statements, ~q AND ~r can be written as "n is not 2, and n is not an odd number." Make sense?
Great. Then the contrapositive of your original statement should be "If n is not 2, and n is not an odd number, then n is not prime."
that the answer it was that simple?
That was it. Did it all make sense?
You almost had it the first time, you just got a little mixed up with the second half of the statement.
yes but what does it mean by Your answer must contain a conjunction in its premise.
That just means that it has to have "and" somewhere in the statement. So the way it's written, it's fine. If we had written it as "If n is an even integer greater than two, then n is not prime" instead, which says the same thing, it would not be the correct solution because it does not contain "and."
oh i see thanks so much!!