# Prove by method of contradiction

1. Jul 10, 2015

### agent1594

1. The problem statement, all variables and given/known data
Using the method of contradiction prove the following.

There's no real number x such that |x-2|+|x-3|=1/2

2. Relevant equations

3. The attempt at a solution
I can see that the least possible value that LHS can take is 1. That is when x=2.5 which is the middle value of 2 and 3. Then the LHS≥1. So LHS=1/2 can not happen.
But I have no idea how to write a proof by contradiction for this. If it was given as |x-2|+|x-3|=0, I can how to prove it as follows,
Suppose x is a real number. ------------------------------(1)
Since |x-2|>o and |x-3|>0, |x-2|+|x-3|>o -----------(2)
But |x-2|+|x-3|=0
This is a contradiction by (2)
So there is no real number x such that |x-2|+|x-3|=0
QED​

Thanks.

Last edited: Jul 10, 2015
2. Jul 10, 2015

### fzero

There are a couple of ways to do this. First, you can note that, if we did not have the absolute value signs, then we could do algebraic manipulation of the equation. So we could consider the 3 cases: $x<2$; $2\leq x < 3$; and $x\geq 3$. For instance, if $x\geq 3$ then $|x-2|=x-2$ and $|x-3| = x-3$. We can then solve the equation that was assumed to be true for $x$.

Second, we can use the positivity argument that you use above, but now for $|x-3| = -|x-2| + 1/2$. Here we need to consider the bounds on $|x-2|$ and what they imply for the bounds on $|x-3|$. You then have to check compatibility with the equation that is assumed.

3. Jul 10, 2015

### SammyS

Staff Emeritus
First of all: The instructions say to use the method of contradiction . You have not used that. -- not in your attempt. -- not in your example for a slightly easier problem..

By the way: Your proof for that simpler case, namely: for showing that there is no solution to |x-2|+|x-3|=0 , is flawed.
|x-2|≥0 and |x-3|≥0
Weak rather than strong inequalities should be used . This makes your proof fall apart. ​

4. Jul 12, 2015

### andrewkirk

I think they want you to use the triangle inequality: |a| + |b| >= |a+b|

Your LHS is the sum of two absolute values, just like in the triangle inequality. If you play around with the signs of a and b you can find a combination that makes |a+b|=1, so the triangle inequality will then contradict the assertion that |a|+|b|=1/2.

5. Jul 13, 2015

### RUber

Remember that |x-3| = |x -2 -1|.
If you assume that |x-2|+|x-3| = 1/2, then clearly, since both are positive,
|x-2| <= 1/2
What does that tell you about |x-3|?
Can the sum ever be 1/2?
That's where you will find the contradiction.