Results 1 to 12 of 12

Math Help - Is my logic for this proof okay?

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    38

    Is my logic for this proof okay?

    <br />
\forall q \in \textbf{Q}, \exists r \in\textbf{Q} so that  q + r \notin \textbf{Z} and  qr \in \textbf{Z} <br />
    (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):

    <br />
\exists q \in \textbf{Q}, \forall r \in\textbf{Q} so that either  q + r \in \textbf{Z} or  qr \notin \textbf{Z} <br />

    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.

    Thanks!
    Last edited by swtdelicaterose; February 2nd 2010 at 08:48 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    Quote Originally Posted by swtdelicaterose View Post
    <br />
\forall q \in \textbf{Q}, \exists r \in\textbf{Q} so that  q + r \notin \textbf{Z} and  qr \in \textbf{Z} <br />
    (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):

    <br />
\exists q \in \textbf{Q}, \forall r \in\textbf{Q} so that either  q + r \in \textbf{Z} or  qr \notin \textbf{Z} <br />

    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.

    Thanks!


    You seem to have been negating the given statement, whereas the contrapositive of a statement " A => B" is "~B => ~A".
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2009
    Posts
    38
    Whoops! Yeah, I meant to do the contrapositive. But, if I prove the negation is true, doesn't that prove the original to be false?

    I'll rewrite the contrapositive tomorrow and see if I can solve it. Gotta sleep now, thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2009
    Posts
    90
    <br />
(\forall q)(\exists r)(\exists z)[z | z = q + r \wedge z = q \cdot r], q \in \mathbb{Q}, z \in \mathbb{Z}<br />
    Let  q = \frac{1}{2}
    following the above there exists some r so that (q + r) is an integer.
    <br />
\frac{1}{2} + \frac{2m + 1}{2} = z, m \in \mathbb{Z}<br />
    then we can test the second condition (q.r) is integer:
    <br />
\frac{1}{2} \cdot \frac{2m + 1}{2} = \frac{2m + 1}{4}<br />
    Contradiction: \frac{2m + 1}{4} \notin \mathbb{Z}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by bmp05 View Post
    <br />
(\forall q)(\exists r)(\exists z)[z | z = q + r \wedge z = q \cdot r], q \in \mathbb{Q}, z \in \mathbb{Z}<br />
    Let  q = \frac{1}{2}
    following the above there exists some r so that (q + r) is an integer.
    <br />
\frac{1}{2} + \frac{2m + 1}{2} = z, m \in \mathbb{Z}<br />
    then we can test the second condition (q.r) is integer:
    <br />
\frac{1}{2} \cdot \frac{2m + 1}{2} = \frac{2m + 1}{4}<br />
    Contradiction: \frac{2m + 1}{4} \notin \mathbb{Z}
    You are to find any rational number q and some rational number r so that the proposition is true, not the contrary.

    Suppose for any integer m,n \in \mathbb{Z}-\{0\} and any rational number q \in \mathbb{Q}, q = \frac{m}{n} .

    There exist some rational number r \in \mathbb{Q} such that r = \frac{m}{1} . Then

    q+r = \frac{m}{n}+\frac{n}{1}=\frac{m+n^2}{n} , which is rational for all m and n in \mathbb{Z}-\{0\} . Hence q+r \notin \mathbb{Z} , and

    qr=\frac{m}{n} \cdot \frac{n}{1}=m , which is an integer in \mathbb{Z} .

    Consequently, the proposition, \forall q \exists r (q,r \in \mathbb{Q}, q+r \notin \mathbb{Z} \wedge qr \in \mathbb{Z}) , is true.
    Last edited by novice; February 3rd 2010 at 11:15 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Sep 2009
    Posts
    502
    Sorry Archie Meade,

    My proof cannot stand because m=0 is a counter example.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by swtdelicaterose View Post
    <br />
\forall q \in \textbf{Q}, \exists r \in\textbf{Q} so that  q + r \notin \textbf{Z} and  qr \in \textbf{Z} <br />
    (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):

    <br />
\exists q \in \textbf{Q}, \forall r \in\textbf{Q} so that either  q + r \in \textbf{Z} or  qr \notin \textbf{Z} <br />

    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.

    Thanks!
    I stared at this and found that the proposition is not a logical implication. If it's not a logical implication, it cannot have a contrapositive statement.

    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    You should be ok, novice!

    It works for any non-zero rational number.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Banned
    Joined
    Sep 2009
    Posts
    502
    Archie Meade,

    The q = 0 really bothers me because \forall q \in \mathbb{Q} does not exclude q=0.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by Archie Meade View Post
    You should be ok, novice!

    It works for any non-zero rational number.
    Thank you for the link to Wolfram math in regard to the rational numbers. It's an interesting reading. I have learned something new.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Nov 2009
    Posts
    38
    Quote Originally Posted by bmp05 View Post
    <br />
(\forall q)(\exists r)(\exists z)[z | z = q + r \wedge z = q \cdot r], q \in \mathbb{Q}, z \in \mathbb{Z}<br />
    Let  q = \frac{1}{2}
    following the above there exists some r so that (q + r) is an integer.
    <br />
\frac{1}{2} + \frac{2m + 1}{2} = z, m \in \mathbb{Z}<br />
    then we can test the second condition (q.r) is integer:
    <br />
\frac{1}{2} \cdot \frac{2m + 1}{2} = \frac{2m + 1}{4}<br />
    Contradiction: \frac{2m + 1}{4} \notin \mathbb{Z}

    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?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Mar 2009
    Posts
    90
    <br />
(\forall q)(\exists r)[P(n)], q, r \in \mathbb{S} <br />

    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 (q + r) 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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Logic and Proof help!
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: January 26th 2011, 08:57 AM
  2. p and q logic proof
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: December 24th 2010, 03:59 AM
  3. logic proof
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: February 12th 2010, 08:23 AM
  4. Logic Proof
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 26th 2008, 02:24 PM
  5. logic and proof
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: June 23rd 2007, 12:30 PM

Search Tags


/mathhelpforum @mathhelpforum