Difference between revisions of "Triangle inequality"

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m (Definition)
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===Proof===
 
===Proof===
'''Style: ''' case analysis
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We have 4 cases:
 
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# Suppose that {{M|a>0}} and {{M|b>0}}
{{Todo|Take time to write it out}}
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#: We see immediately that {{M|1=a>0\implies a+b>0+b=b>0}} so {{M|a+b>0}}
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#:* thus {{M|1=\vert a+b\vert=a+b}}
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#: We also see that {{M|1=\vert a\vert=a}} as {{M|a>0}}
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#: and that {{M|1=\vert b\vert=b}} for the same reason.
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#:* thus {{M|1=\vert a\vert+\vert b\vert=a+b}}
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#* We see that {{M|1=\vert a+b\vert=\vert a\vert+\vert b\vert}}
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#** 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"
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#**: ({{M|\implies}} denotes [[Implies|logical implication]])
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# Suppose that {{M|a>0}} and {{M|b\le 0}}
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# Suppose that {{M|a\le 0}} and {{M|b> 0}}
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#* [[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]])
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#* 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}}
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#* 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}}
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#* So:
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#** {{M|1=\vert a+b\vert}}
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#**: {{M|1= =\vert b+a\vert}} by the [[Commutative|commutativity]] of addition on {{M|\mathbb{R} }}
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#**: {{M|1=\le \vert b\vert+\vert a\vert}} by the {{M|2^\text{nd} }} case (above)
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#**: {{M|1= =\vert a\vert+\vert b\vert}} again by commutativity of the real numbers
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#* Thus {{M|\vert a+b\vert\le \vert a\vert+\vert b\vert}}
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# Both {{M|a\le 0}} and {{M|b\le 0}}
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{{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: d(x,z)d(x,y)+d(y,z)

(see metric space) of which |xz||xy|+|yz|
is a special case.

Another common way of writing it is |a+b||a|+|b|

, notice if we set a=xy and b=yz then we get |xy+yz||xy|+|yz|
which is just |xz||xy|+|yz|

Definition

The triangle inequality is as follows:

  • |a+b||a|+|b|

Proof

We have 4 cases:

  1. Suppose that a>0 and b>0
    We see immediately that a>0a+b>0+b=b>0 so a+b>0
    • thus |a+b|=a+b
    We also see that |a|=a as a>0
    and that |b|=b for the same reason.
    • thus |a|+|b|=a+b
    • We see that |a+b|=|a|+|b|
      • Notice |a+b|=|a|+|b||a+b||a|+|b| in the literal sense of "if left then right"
        ( denotes logical implication)
  2. Suppose that a>0 and b0
  3. Suppose that a0 and b>0
    • Mathematicians are lazy, as R is a field (an instance of a ring) we know that a+b=b+a (addition is commutative)
    • As ||:RR is a function we know that if x=y then |x|=|y|
    • As, again, R is a ring, we know that |a|+|b|=|b|+|a|
    • So:
      • |a+b|
        =|b+a| by the commutativity of addition on R
        |b|+|a| by the 2nd case (above)
        =|a|+|b| again by commutativity of the real numbers
    • Thus |a+b||a|+|b|
  4. Both a0 and b0

TODO: Finish proof



Reverse Triangle Inequality

This is |a||b||ab|

Proof

Take |a|=|(ab)+b|

then by the triangle inequality above:
|(ab)+b||ab|+|b|
then |a||ab|+|b|
clearly |a||b||ab|
as promised

Note

However we see |b||a||ba|

but |ba|=|(1)(ab)|=|1||ab|=|ab|
thus |b||a||ab|
also.

That is both:

  • |a||b||ab|
  • |b||a||ab|

Full form

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