Difference between revisions of "Triangle inequality"
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===Proof=== | ===Proof=== | ||
| − | + | We have 4 cases: | |
| − | + | # Suppose that {{M|a>0}} and {{M|b>0}} | |
| − | {{Todo| | + | #: We see immediately that {{M|1=a>0\implies a+b>0+b=b>0}} so {{M|a+b>0}} |
| + | #:* thus {{M|1=\vert a+b\vert=a+b}} | ||
| + | #: We also see that {{M|1=\vert a\vert=a}} as {{M|a>0}} | ||
| + | #: and that {{M|1=\vert b\vert=b}} for the same reason. | ||
| + | #:* thus {{M|1=\vert a\vert+\vert b\vert=a+b}} | ||
| + | #* We see that {{M|1=\vert a+b\vert=\vert a\vert+\vert b\vert}} | ||
| + | #** Notice {{M|1=\vert a+b\vert=\vert a\vert+\vert b\vert\implies\vert a+b\vert\le\vert a\vert+\vert b\vert}} in the literal sense of "if left then right" | ||
| + | #**: ({{M|\implies}} denotes [[Implies|logical implication]]) | ||
| + | # Suppose that {{M|a>0}} and {{M|b\le 0}} | ||
| + | # Suppose that {{M|a\le 0}} and {{M|b> 0}} | ||
| + | #* [[Mathematicians are lazy]], as {{M|\mathbb{R} }} is a [[field]] (an instance of a [[ring]]) we know that {{M|1=a+b=b+a}} (addition is [[commutative]]) | ||
| + | #* As {{M|\vert\cdot\vert:\mathbb{R}\rightarrow\mathbb{R} }} is a [[function]] we know that if {{M|1=x=y}} then {{M|1=\vert x\vert=\vert y\vert}} | ||
| + | #* As, again, {{M|\mathbb{R} }} is a ring, we know that {{M|1=\vert a\vert+\vert b\vert=\vert b\vert+\vert a\vert}} | ||
| + | #* So: | ||
| + | #** {{M|1=\vert a+b\vert}} | ||
| + | #**: {{M|1= =\vert b+a\vert}} by the [[Commutative|commutativity]] of addition on {{M|\mathbb{R} }} | ||
| + | #**: {{M|1=\le \vert b\vert+\vert a\vert}} by the {{M|2^\text{nd} }} case (above) | ||
| + | #**: {{M|1= =\vert a\vert+\vert b\vert}} again by commutativity of the real numbers | ||
| + | #* Thus {{M|\vert a+b\vert\le \vert a\vert+\vert b\vert}} | ||
| + | # Both {{M|a\le 0}} and {{M|b\le 0}} | ||
| + | {{Todo|Finish proof}} | ||
==Reverse Triangle Inequality== | ==Reverse Triangle Inequality== | ||
Revision as of 16:57, 28 August 2015
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) 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 [ilmath]a=x-y[/ilmath] and [ilmath]b=y-z[/ilmath] 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
We have 4 cases:
- Suppose that [ilmath]a>0[/ilmath] and [ilmath]b>0[/ilmath]
- We see immediately that [ilmath]a>0\implies a+b>0+b=b>0[/ilmath] so [ilmath]a+b>0[/ilmath]
- thus [ilmath]\vert a+b\vert=a+b[/ilmath]
- We also see that [ilmath]\vert a\vert=a[/ilmath] as [ilmath]a>0[/ilmath]
- and that [ilmath]\vert b\vert=b[/ilmath] for the same reason.
- thus [ilmath]\vert a\vert+\vert b\vert=a+b[/ilmath]
- We see that [ilmath]\vert a+b\vert=\vert a\vert+\vert b\vert[/ilmath]
- Notice [ilmath]\vert a+b\vert=\vert a\vert+\vert b\vert\implies\vert a+b\vert\le\vert a\vert+\vert b\vert[/ilmath] in the literal sense of "if left then right"
- ([ilmath]\implies[/ilmath] denotes logical implication)
- Notice [ilmath]\vert a+b\vert=\vert a\vert+\vert b\vert\implies\vert a+b\vert\le\vert a\vert+\vert b\vert[/ilmath] in the literal sense of "if left then right"
- We see immediately that [ilmath]a>0\implies a+b>0+b=b>0[/ilmath] so [ilmath]a+b>0[/ilmath]
- Suppose that [ilmath]a>0[/ilmath] and [ilmath]b\le 0[/ilmath]
- Suppose that [ilmath]a\le 0[/ilmath] and [ilmath]b> 0[/ilmath]
- Mathematicians are lazy, as [ilmath]\mathbb{R} [/ilmath] is a field (an instance of a ring) we know that [ilmath]a+b=b+a[/ilmath] (addition is commutative)
- As [ilmath]\vert\cdot\vert:\mathbb{R}\rightarrow\mathbb{R} [/ilmath] is a function we know that if [ilmath]x=y[/ilmath] then [ilmath]\vert x\vert=\vert y\vert[/ilmath]
- As, again, [ilmath]\mathbb{R} [/ilmath] is a ring, we know that [ilmath]\vert a\vert+\vert b\vert=\vert b\vert+\vert a\vert[/ilmath]
- So:
- [ilmath]\vert a+b\vert[/ilmath]
- [ilmath]=\vert b+a\vert[/ilmath] by the commutativity of addition on [ilmath]\mathbb{R} [/ilmath]
- [ilmath]\le \vert b\vert+\vert a\vert[/ilmath] by the [ilmath]2^\text{nd} [/ilmath] case (above)
- [ilmath]=\vert a\vert+\vert b\vert[/ilmath] again by commutativity of the real numbers
- [ilmath]\vert a+b\vert[/ilmath]
- Thus [ilmath]\vert a+b\vert\le \vert a\vert+\vert b\vert[/ilmath]
- Both [ilmath]a\le 0[/ilmath] and [ilmath]b\le 0[/ilmath]
TODO: Finish proof
Reverse Triangle Inequality
This is [math]|a|-|b|\le|a-b|[/math]
Proof
Take [math]|a|=|(a-b)+b|[/math] then by the triangle inequality above:
[math]|(a-b)+b|\le|a-b|+|b|[/math] then [math]|a|\le|a-b|+|b|[/math] clearly [math]|a|-|b|\le|a-b|[/math] as promised
Note
However we see [math]|b|-|a|\le|b-a|[/math] but [math]|b-a|=|(-1)(a-b)|=|-1||a-b|=|a-b|[/math] thus [math]|b|-|a|\le|a-b|[/math] also.
That is both:
- [math]|a|-|b|\le|a-b|[/math]
- [math]|b|-|a|\le|a-b|[/math]
Full form
There is a "full form" of the reverse triangle inequality, it combines the above and looks like: [math]|a-b|\ge|\ |a|-|b|\ |[/math]
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
