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Math Help - Proof by contradiction.

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    Proof by contradiction.

    Someone PLEASE help me!
    I'm trying to write a contradiction to the statement:
    If three positive real numbers x,y,z are chose between 0 and 1 with 0<x<y<z<1, then the distance from x to y is less than .5 or the distance from y to z is less than .5.

    I belive it is:

    If three positive real numbers x,y,z are chose between 0 and 1 with0<x<y<z<1, then the distance from x to y is not less than .5 and thedistance from y to z is not less than .5.


    But my teacher tells me this is wrong.
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    Re: Proof by contradiction.

    Quote Originally Posted by Aquameatwad View Post
    Someone PLEASE help me!
    I'm trying to write a contradiction to the statement:
    If three positive real numbers x,y,z are chose between 0 and 1 with 0<x<y<z<1, then the distance from x to y is less than .5 or the distance from y to z is less than .5.

    I belive it is:

    If three positive real numbers x,y,z are chose between 0 and 1 with0<x<y<z<1, then the distance from x to y is not less than .5 and thedistance from y to z is not less than .5.


    But my teacher tells me this is wrong.
    I suggest you review your terminology. What you have given is not (even close to) a proof. All you've done is write the negation of the original statement.

    To prove a statement by contradiction, you need to write the negation of your original statement, then go through a series of logical steps to arrive at a contradiction. This shows that the negation is false, and therefore that the original statement is true.
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    Re: Proof by contradiction.

    That is what i am trying to do, is just find the correct statement. i have not proved it yet.
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    Re: Proof by contradiction.

    the way i want to prove it, is if the distance from x to y is not less than .5 and the distance from y to z is not less than .5.
    then we assume z - y >= 0.5 and y - x > =0.5. Adding the two we get z - x >= 1, which is impossible for two numbers between 0 and 1. I guess im confused as to what kind of proof im actually using
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    Re: Proof by contradiction.

    Quote Originally Posted by Aquameatwad View Post
    the way i want to prove it, is if the distance from x to y is not less than .5 and the distance from y to z is not less than .5.
    then we assume z - y >= 0.5 and y - x > =0.5. Adding the two we get z - x >= 1, which is impossible for two numbers between 0 and 1. I guess im confused as to what kind of proof im actually using
    I think you're on the right track.

    I would say something like...

    \displaystyle  \begin{align*}y - x &\geq 0.5 \\ y - x &= 0.5 + m\textrm{ for some }m \geq 0\end{align*}

    and

    \displaystyle \begin{align*}z - y &\geq 0.5 \\ z - y &= 0.5 + n\textrm{ for some } n \geq 0 \end{align*}

    So

    \displaystyle \begin{align*} (z - y ) + (y - x) &= (0.5 + m) + (0.5 + n) \\ z - x &= 1 + m + n \\ z - x &\geq 1 \end{align*}

    which is where our contradiction lies, since \displaystyle 0 < x < y < z < 1 \implies z - x \leq 1.
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    Re: Proof by contradiction.

    all right, so what would be my statement? Since my teacher keeps telling my contradiction statement is wrong.
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    Re: Proof by contradiction.

    Am i even using proof by contradiction?
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    Re: Proof by contradiction.

    Quote Originally Posted by Aquameatwad View Post
    Am i even using proof by contradiction?
    I don't see any problem with the statement of the negation.
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    Re: Proof by contradiction.

    So its proof by negation?... Is that valid?
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    Re: Proof by contradiction.

    Quote Originally Posted by Aquameatwad View Post
    So its proof by negation?... Is that valid?
    I suggest you reread Post 2.
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    Re: Proof by contradiction.

    Then why does my teacher tell me its wrong?!?!
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    Re: Proof by contradiction.

    Quote Originally Posted by Aquameatwad View Post
    I'm trying to write a contradiction to the statement:
    If three positive real numbers x,y,z are chose between 0 and 1 with 0<x<y<z<1, then the distance from x to y is less than .5 or the distance from y to z is less than .5.

    I belive it is:

    If three positive real numbers x,y,z are chose between 0 and 1 with0<x<y<z<1, then the distance from x to y is not less than .5 and thedistance from y to z is not less than .5.
    You forgot to change the universal quantification to existential quantification.
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    Re: Proof by contradiction.

    So my statement

    If positive real numbers x,y,z are chosen between 0 and 1 with 0<x<y<z<1, then the distance from x to y is less than .5 or the distance from y to z is less than .5
    .


    Then i have the statement P then Q where Q=(q1 v q2)
    P= Positive real numbers x,y,z between 0 and 1 with 0<x<y<z<1
    Q = (q1=the distance from x to y is less than .5) v (q2=the distance from y to z is less than .5)

    Okay so now i use contradiction to prove it
    The statement then must turn into P and ~Q ( P and not Q)
    So
    Positive real numbers x,y,z are between 0 and 1 with 0<x<y<z<1, AND the distance from x to y is not less than .5 and the distance from y to z is not less than .5

    Then the statement can be written as :
    Positive real numbers x,y,z are chosen between 0 and 1 with 0<x<y<z<1, AND the distance from x to y is greater than or equal to .5 and the distance from y to z is greater than or equal to .5.

    This doesn't sound right to me
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    Re: Proof by contradiction.

    Should my statement be written as:
    If Real numbers x,y,z are 0<x<y<z1, then the distance from x to y is less than .5 or the distance from y to z is less than .5.

    The contradiction is then
    Real numbers x,y,z are 0<x<y<z1, AND the distance from x to y is greater than or equal to .5 and the distance from y to z is greater than or equal to .5.

    What do you think?
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    Re: Proof by contradiction.

    Quote Originally Posted by Aquameatwad View Post
    Should my statement be written as:
    If Real numbers x,y,z are 0<x<y<z1, then the distance from x to y is less than .5 or the distance from y to z is less than .5.
    The negation of that statement is:
    That exist three numbers \{a,b,c\}\subset (0,1) such 0<a<b<c<1\text{ and }c-b\ge 0.5\text{ and }b-a\ge 0.5.
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