Difference between revisions of "Triangle inequality"

Revision as of 16:11, 7 March 2015 (view source) Alec (Talk | contribs) m (Alec moved page Triangle Inequality to Triangle inequality without leaving a redirect: To keep with naming conventions) ← Older edit		Revision as of 12:06, 14 April 2015 (view source) Alec (Talk | contribs) m Newer edit →
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−	The triangle inequality takes a few common forms {{M||d(x,z)\le d(x,y)+d(y,z)}} of which <math>|x-z|\le|x-y|+|y-z|</math> is a special case.	+	The triangle inequality takes a few common forms, for example: <math>d(x,z)\le d(x,y)+d(y,z)</math> (see [[Metric space|metric space]]) of which <math>|x-z|\le|x-y|+|y-z|</math> is a special case.
−	Another common way of writing it is <math>|a+b|\le |a|+|b|</math>, notice if we set {{M|a=x-y}} and {{M|b=y-z}} then we get <math>|x-y+y-z|\le|x-y|+|y-z|</math> which is just <math>|x-z|\le|x-y|+|y-z|</math>	+	Another common way of writing it is <math>|a+b|\le |a|+|b|</math>, notice if we set {{M|1=a=x-y}} and {{M|1=b=y-z}} then we get <math>|x-y+y-z|\le|x-y|+|y-z|</math> which is just <math>|x-z|\le|x-y|+|y-z|</math>
		+
		+	==Definition==
		+	The triangle inequality is as follows:
		+	* <math>|a+b|\le |a|+|b|</math>
		+
		+	===Proof===
		+	'''Style: ''' case analysis
		+
		+	{{Todo|Take time to write it out}}
	==Reverse Triangle Inequality==		==Reverse Triangle Inequality==

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Revision as of 12:06, 14 April 2015

The triangle inequality takes a few common forms, for example:

$$d(x,z)\le d(x,y)+d(y,z)$$ (see metric space) of which $$|x-z|\le|x-y|+|y-z|$$ is a special case.

Another common way of writing it is

$$|a+b|\le |a|+|b|$$ , notice if we set $a=x-y$ and $b=y-z$ then we get $$|x-y+y-z|\le|x-y|+|y-z|$$ which is just $$|x-z|\le|x-y|+|y-z|$$

Definition

The triangle inequality is as follows:

• $$|a+b|\le |a|+|b|$$

Proof

Style: case analysis

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TODO: Take time to write it out

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Reverse Triangle Inequality

This is

$$|a|-|b|\le|a-b|$$

Proof

Take

$$|a|=|(a-b)+b|$$ then by the triangle inequality above:
$$|(a-b)+b|\le|a-b|+|b|$$ then $$|a|\le|a-b|+|b|$$ clearly $$|a|-|b|\le|a-b|$$ as promised

Note

However we see

$$|b|-|a|\le|b-a|$$ but $$|b-a|=|(-1)(a-b)|=|-1||a-b|=|a-b|$$ thus $$|b|-|a|\le|a-b|$$ also.

That is both:

• $$|a|-|b|\le|a-b|$$
• $$|b|-|a|\le|a-b|$$

Full form

There is a "full form" of the reverse triangle inequality, it combines the above and looks like:

$$|a-b|\ge|\ |a|-|b|\ |$$

It follows from the properties of absolute value, I don't like this form, I prefer just "swapping" the order of things in the abs value and applying the same result