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Question 1184245: TRUE or FALSE
Let p: The set of odd numbers is closed under the operation of addition.
q: (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
1.p ^ q
2.p V q
3.q -> p
4.p <-> q
5. ~p
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!
p is false because you can add your way out of the odd numbers.
3+5=8. 3 and 5 are odd, but when we add them we get 8, an even number.
So I was in the odd numbers with 3 and 5. But then I added my way out,
so the operation is OPEN (not closed) because I added my way out.
[You can't add your way out of the even numbers.]
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q is true because that's the binomial expansion for (x + y)^3.
1.p ^ q ( F) ^ (T) = (F) [ ^ is (F) except when both sides are (T)]
2.p V q (F) V (T) = (T) [ V is (T) exeept when both sides are (F)]
3.q -> p (T) -> (F) = (F) [ -> is (T) except when (T)->(F), then (F)]
4.p <-> q (F) <-> (T) = (F) [ <-> is (T) only when both sides are alike
5. ~p ~(F) = (T) [ ~ is (F) if followed by (T) and vice
versa
Edwin
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