"converse" and "contrapositive" refer to pretty standard modifications, they're not really something you have to wonder about.

for the statement:

the converse is:

the contrapositive is:

so for:

converse:

contrapositive:

for every integer, n, either n is odd or n is evenI need to write this statement using the quantifiers "for all (or for every)" and "there exists".

An integer cannot be both even and odd.

What I could muster up was "For all integers, there exists only an odd or only an even". This really doesn't make much sense, but I really couldn't get much out of it.

the negation of "every", "for all", "for every" etc is "there exists." see hereNegations. I never know what I need to negate...

Every integer is divisible by a prime. Would this be "Not every integer is divisible by a prime"?

thus, here we have: there exists an integer that is not divisible by a prime

see the link above. (there are better resources out there also, try a google search).If n is an integer, n/(n+1) is not an integer. Would this be "If n is not an integer, n/(n+1) is not an integer"? I could use an explanation for this one!

the negation of an implication is as follows: to negate we write . (you can use "but" instead of "and" if it makes the phrase sound better)

thus we get: n is an integer and n/(n + 1) is an integer.

False.And finally....

I need to answer true or false and supply a direct proof or a counter-example to each of the following:

There exists an integer n ≠ 0 such that nq is an integer for every rational number q.

Proof:Fix , . And choose for , , such that q is in lowest terms and . Then, for and . (and remember, is fixed at this value).

Now choose for .

Then

QED

True.For every rational number q, there exists an integer n ≠ 0 such that nq is an integer.

Proof:Let be a rational number. then we can write for and

. Choose . then and . thus is an integer

QED