for the statement:
the converse is:
the contrapositive is:
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.
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 .
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