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agent1594

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## Homework Statement

Using the method of contradiction prove the following.

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

## Homework Equations

## 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

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.*

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