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Contrapositive statement

  1. Sep 1, 2009 #1
    Fact: If a and b are non-negative numbers, then ab is non-negative.

    What is the equivalent contrapositive statement of the above?
    I think it is:
    If ab<0, then at least one of a and b <0.
    Am I right?

    But this implication doesn't seem quite right to me...shouldn't the correct statement be...
    If ab<0, then exactly one of a and b <0 ?

    Could someone please explain?

    Thank you!
     
  2. jcsd
  3. Sep 1, 2009 #2
    "If ab<0, then at least one of a and b <0" is the contrapositive, and it is perfectly true. "If ab<0, then exactly one of a and b < 0" is also true, but it is not the contrapositive (it says more than the contrapositive does).
     
  4. Sep 1, 2009 #3
    Some things to consider:
    1) The statement S
    "If x then y"
    has the following http://www.acm.org/crossroads/xrds10-3/gfx/img1.gif":
    For x false, y false the statement "If x then y" is true (vacuous truth)
    For x false, y true the statement "If x then y" is true (vacuous truth)
    For x true, y false the statement "If x then y" is false
    For x true, y true the statement "If x then y" is true

    (Instead of "If x then y" we also write "x => y")


    2) The negation of "x and y"
    is "(not x) or (not y)".
    ---------------------------

    Now, to your statement:
    [tex](a,b \geq 0) \Rightarrow (ab \geq 0)[/tex]
    or equivalently
    [tex](a \geq 0 \text{ and } b \geq 0) \Rightarrow (ab \geq 0)[/tex]

    The contrapositive is:
    [tex]\text{not }(ab \geq 0) \Rightarrow \text{not }( a \geq 0 \text{ and } b \geq 0)[/tex]
    or equivalently
    (ab < 0) => (a<0 or b<0)


    Cases:
    We will check for every case of a,b whether the statement
    (ab <0) => (a<0 or b<0) is true.
    (Let's call this statement S as in the beginning.
    The x corresponds to "ab<0" and y corresponds to "a<0 or b<0".)


    Examine statement S: "If (ab<0) then (a<0 or b<0)"

    Case 1) a<0, b<0:
    (ab<0) is false, (a<0 or b<0) is true
    From our truth table we can conclude that S is true.

    Case 2) a<0, b=0:
    (ab<0) is false, (a<0 or b<0) is true
    From our truth table we can conclude that S is true.

    Case 3) a<0, b>0:
    (ab<0) is true, (a<0 or b<0) is true
    From our truth table we can conclude that S is true.

    Case 4) a=0, b<0:
    (ab<0) is false, (a<0 or b<0) is true
    From our truth table we can conclude that S is true.

    Case 5) a=0, b=0:
    (ab<0) is false, (a<0 or b<0) is false
    From our truth table we can conclude that S is true.

    Case 6) a=0, b>0:
    (ab<0) is false, (a<0 or b<0) is false
    From our truth table we can conclude that S is true.

    Case 7) a>0, b<0:
    (ab<0) is true, (a<0 or b<0) is true
    From our truth table we can conclude that S is true.

    Case 8) a>0, b=0:
    (ab<0) is false, (a<0 or b<0) is false
    From our truth table we can conclude that S is true.

    Case 9) a>0, b>0:
    (ab<0) is false, (a<0 or b<0) is false
    From our truth table we can conclude that S is true.

    As you can see the statement S is true for all cases of a,b.
     
    Last edited by a moderator: Apr 24, 2017
  5. Sep 1, 2009 #4
    What you are stumbling on is the fact that the converse of the original statement happens to be true also and its contrapositive along with the one you're being asked to find together make the "exaclty one" part true. But the contrapositive of the implication you're given is "at least one," and no further.

    --Elucidus
     
  6. Sep 1, 2009 #5
    Thanks for the helpful comments!
     
  7. Sep 8, 2009 #6
    You are partially correct. It seems obvious to me that ab<0 (P) iff exactly one of (a,b) < 0 and exactly one of (a,b)>0 (Q). The contrapositive of this is ~P-->~Q, ie: not(ab<0) iff not(exactly one of (a, b) < 0 and exactly one of (a,b)>0).

    I don't see where ab<0 if at least one of (a,b)<0 is true since it doesn't exclude the possibility of (a,b) having the same sign. It's obvious that (a,b) cannot have the same sign and neither (a,b) can be zero if ab<0.
     
    Last edited: Sep 8, 2009
  8. Sep 8, 2009 #7
    Yes, but the (false) statement "ab < 0 if at least one of a and b is less than 0" was never mentioned. Remember that "P if Q" means "if Q, then P."
     
  9. Sep 8, 2009 #8

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Since "NOT non-negative" is "negative", the direct contrapositive to "If a and b are non-negative numbers then ab is non-negative" is "If ab is negative then it is not the case that a and b are non-negative numbers".

    But "not P and Q" is "not P or not Q" so "NOT (a and b are non-negative)" is "either a is negative or b is negative".

    So the contrapositive of "if a and b are non-negative numbers then ab is non-negative" is "if ab is negative then either a is negative or b is negative".

    While it is true that a and b can't both be negative, that fact does NOT follow from the original statement. That requires more information.

    For example, suppose statement were "if a and b are non-even numbers then ab is non-even". What is the contrapositive of that?
     
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