Analysis Absolute Value Proofs

**auwesty** · October 14th 2009, 05:03 PM

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!

**redsoxfan325** · October 14th 2009, 07:05 PM

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.) $|xy|=\sqrt{(xy)^2}=\sqrt{x^2y^2}=\sqrt{x^2}\sqrt{y ^2}=|x|\,|y|$

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

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

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500th Post!

**auwesty** · October 17th 2009, 06:55 PM

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.

**redsoxfan325** · October 17th 2009, 07:25 PM

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 $x$ is positive and $y$ is negative, etc.) but it's unnecessary if you know that $|x|=\sqrt{x^2}$ .

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

**auwesty** · October 18th 2009, 06:27 AM

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

**Plato** · October 18th 2009, 08:05 AM

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) $\left| x \right| = \left| {x - y + y} \right| \leqslant \left| {x - y} \right| + \left| y \right| < c + \left| y \right|$

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

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