Results 1 to 9 of 9

Math Help - Proposition conclusion

  1. #1
    Member
    Joined
    Oct 2010
    Posts
    95

    Proposition conclusion

    In my book it says: On the nth statement of a list of 100 statements, it says "There are exactly n number of false statements in this list."

    It is said that this concludes that the 99th statement is true while the others are false.
    But how does it tell that? n could be just any arbitary numbers, isn't it? If n=20, the 99th statement may not be true. Statements #5, #8, #34, and other random statements could be true but not the 99th one. Moreover, n could also be not 20 but 5, 50, or even 100. What should my thought process be when coming to this conclusion?

    Then, it also says that if the nth statement becomes "At least n number of statements in this list are false."
    It is said that statements 1 to 50 are true and 51 to 100 are false.
    But again, how is this possible? n could again be any number, not neccessarily be right in the middle of the list of 100 statements. Even if it is n is 50, it is just 50 false statements in possibly different order. It does not need to be the first or last 50 of the list of statements.

    Am I missing something that I don't understand how these conclusions were drawn?

    thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    For the first problem: I think your list of statements all look like this:

    Statement # 1: There is exactly one false statement in this list.
    Statement # 2: There are exactly two false statements in this list.
    Statement # 3: There are exactly three false statements in this list.
    .
    .
    .
    Statement # 98: There are exactly 98 false statements in this list.
    Statement # 99: There are exactly 99 false statements in this list.
    Statement # 100: There are exactly 100 false statements in this list.

    Obviously, at most one of these statements can be true, because they are all contradictory. You have two cases:

    Case 1: None of the statements are true. Where does this lead?

    Case 2: One of the statements is true. Where does this lead?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2010
    Posts
    95
    Case 1: None of the statements are true. Where does this lead?
    This will include the last Statement #100 to be false, which this statement #100 has to be true for all 100 statements to be false.

    Case 2: One of the statements is true. Where does this lead?
    If one of the statements is true, only the statement #99 can be true as from statement #1 to #98 + statement #100 = 99 statements are false, statement #99 will be true.

    wow thanks... i didn't know i should read it this way. And I didn't thought of the case that one of the statements is true.

    then for the second part,
    At least one number of statements in this list are false.
    At least two number of statements in this list are false.
    At least three number of statements in this list are false.
    At least four number of statements in this list are false.
    .
    .
    .
    At least 99 number of statements in this list are false.
    At least 100 number of statements in this list are false.

    Isn't it also 99 statements are false and statement #99 is true?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Right. So case 1 leads to a contradiction, because it makes Statement 100 to be both true and false at the same time. Case 2 is true, which, as you've seen, leads to making Statement 99 true. The generalization of the first problem would be that in such a constructed list of m entries, the (m-1)th statement is the only true one.

    For the second problem: I would put in a few more particular statements in your list:

    At least one statement in this list are false.
    At least two statements in this list are false.
    At least three statements in this list are false.
    At least four statements in this list are false.
    .
    .
    .
    At least 49 statements in this list are false.
    At least 50 statements in this list are false.
    At least 51 statements in this list are false.
    .
    .
    .
    At least 99 statements in this list are false.
    At least 100 statements in this list are false.

    I would just trot through each of the list statements, and ask yourself this question: what if this statement were true? What if it were false? And if a particular statement were true, which statements would be the false ones? Where does that lead you?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Another hint to the second problem: work through the list backwards.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2010
    Posts
    95
    hmm..
    from the last statement: At least 100 statements in this list are false.
    For the #100 statement to be true, it needs to have at least 100 statements, including itself, to be false, which is a contradiction.
    On statement #99: At least 99 statements in this list are false.
    For #99 to be true, statement #1 to 98 and #100 must be false. But for at least 99 statements to be false, statement #98, which says "At least 98 statements in this list are false", is automatically true, causing statement #99 to be false.
    This repeats all the way until statement #51.
    From statement #50, "At least 50 statements in this list are false" is true because all the 50 statements below itself are false.
    So on statement #49 is automatically true since statement #50 is already true.
    This repeats all the way up to statement #1.
    Therefore, statements #1 to #50 are true while #51 to #100 are false.

    woah...this is pretty crazy... i feel like i am literally cracking my mind. i don't know why i am not getting much intuition in such thinking.
    thank you so much!!!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    You're welcome! Have a good one!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Nov 2010
    From
    Staten Island, NY
    Posts
    451
    Thanks
    2
    Great problem. I haven't seen this one before.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Quote Originally Posted by DrSteve View Post
    Great problem. I haven't seen this one before.
    Wholeheartedly agree. That's why I thanked the OP.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: July 24th 2010, 03:21 AM
  2. Logical conclusion
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: March 24th 2010, 11:59 AM
  3. Premise and Conclusion
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: January 12th 2010, 04:52 PM
  4. Replies: 1
    Last Post: September 14th 2008, 07:05 PM
  5. triangle conclusion
    Posted in the Geometry Forum
    Replies: 2
    Last Post: June 21st 2007, 07:51 PM

Search Tags


/mathhelpforum @mathhelpforum