1. ## Proof by contradiction.

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.

2. ## Re: Proof by contradiction.

Originally Posted by Aquameatwad
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.

3. ## Re: Proof by contradiction.

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

4. ## 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

5. ## Re: Proof by contradiction.

Originally Posted by Aquameatwad
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 $\displaystyle 0 < x < y < z < 1 \implies z - x \leq 1$.

6. ## Re: Proof by contradiction.

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

7. ## Re: Proof by contradiction.

Am i even using proof by contradiction?

8. ## Re: Proof by contradiction.

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

9. ## Re: Proof by contradiction.

So its proof by negation?... Is that valid?

10. ## Re: Proof by contradiction.

Originally Posted by Aquameatwad
So its proof by negation?... Is that valid?
I suggest you reread Post 2.

11. ## Re: Proof by contradiction.

Then why does my teacher tell me its wrong?!?!

12. ## Re: Proof by contradiction.

Originally Posted by Aquameatwad
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.

13. ## 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

14. ## 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?

15. ## Re: Proof by contradiction.

Originally Posted by Aquameatwad
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.

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