so that and
(where Q is set of all rationals, and Z is set of all integers)
So my approach was to write out the contrapositive, which is the same statement logically. Which is (I think):
so that either or
I'm not sure about the and statement. Is this contrapositive correct?
(if it is correct, then)
So now I just need to let q = a rational number and find a counter example r so that one of the statements in the or fails. So for example, let q = 0 which is rational, and r = 1, q + r is an integer, but qr is also an integer, however since q+r is an integer, the or statement is true, which should mean the original statement is false?
Is my logic correct? I'm not asking for how to prove this, I'm just asking if I can form a proof from what I've established so far.
Suppose for any integer and any rational number , .
There exist some rational number such that . Then
, which is rational for all and in . Hence , and
, which is an integer in .
Consequently, the proposition, , is true.
I tried to prove it true, but I found at least one counter example, namely the rational numbers q =0. So the statement is not always true, and a counter example is sufficient, see the proof shown by the poster immediately above me.
Note: a counter example and a contrapositive statement are not synonymous.
For this proof to work, don't you have to prove that for all r, instead of just one example of r. Logically, since it says there exists an r, showing that (2k+1)/2 doesn't work just means it doesn't work for this particular example.
So in this proof, q = 1/2, so r is an arbitrary rational number a/b where a and b are integers and b is not 0.
1/2 + a/b is not an integer if a/b is not an integer, if a/b is not an integer then a/b*1/2 = a/2b, now i have to prove a/2b is an integer...
i dunno, i think i'm confusing myself further...
Or am I mistaken?
To prove a counter example is just proving the negation is true, right?
This means that for each element q (in S) there is at least 1 element r in the set S with some property P(n). The property might be is not a whole number, but it could be some other property such as "r divides q" etc.
It means that to prove the proposition is false you only have to find one example of q, i.e. one element of S, for which there is no (or can be no) r satisfying the property.