# Thread: Analysis Absolute Value Proofs

1. ## Analysis Absolute Value Proofs

Ok, so I'm having some issues working on these proofs:

a) |xy| = |x| * |y|
b) If |x-y| < c, then |x| < |y| + c
and
c) If |x-y| < E for all E > 0, then x = y.

Any help would be greatly appreciated!

2. Originally Posted by auwesty
Ok, so I'm having some issues working on these proofs:

a) |xy| = |x| * |y|
b) If |x-y| < c, then |x| < |y| + c
and
c) If |x-y| < E for all E > 0, then x = y.

Any help would be greatly appreciated!
a.) $\displaystyle |xy|=\sqrt{(xy)^2}=\sqrt{x^2y^2}=\sqrt{x^2}\sqrt{y ^2}=|x|\,|y|$

b.) $\displaystyle |x|=|y+x-y|\leq|y|+|x-y|<|y|+c$

c.) Assume $\displaystyle x\neq y$. Then by the properties of a metric, $\displaystyle |x-y|=d>0$. Thus, taking $\displaystyle E=\frac{d}{2}$ gives us an $\displaystyle E>0$ such that $\displaystyle |x-y|>E$, which is a contradiction. So $\displaystyle x=y$. $\displaystyle \square$

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500th Post!

3. Thanks!

That's all we have to do? All the examples we did in class had 2 - 4 cases we needed to prove (based on the definition of absolute value and the problem in question) in order to make this work.

4. Originally Posted by auwesty
Thanks!

That's all we have to do? All the examples we did in class had 2 - 4 cases we needed to prove (based on the definition of absolute value and the problem in question) in order to make this work.
Yes. You could prove (a) by cases (assume $\displaystyle x$ is positive and $\displaystyle y$ is negative, etc.) but it's unnecessary if you know that $\displaystyle |x|=\sqrt{x^2}$.

For the other two proofs, all you need to know is that $\displaystyle |x-y|$ is a metric, and then you can use properties of metrics to complete the proof. No cases required!

5. ahhh ok, we haven't learned metrics yet that's why i was a little confused

6. Originally Posted by auwesty
b) If |x-y| < c, then |x| < |y| + c

c) If |x-y| < E for all E > 0, then x = y.
b) $\displaystyle \left| x \right| = \left| {x - y + y} \right| \leqslant \left| {x - y} \right| + \left| y \right| < c + \left| y \right|$

c) Suppose that $\displaystyle x\ne y$ then that means $\displaystyle |x-y|>0$.
So let $\displaystyle E=|x-y|>0$ then you have $\displaystyle |x-y|<|x-y|.$
That is a contradiction to $\displaystyle x\ne y.$