Answer to Question #88937 in Algebra for RAKESH DEY

Question #88937
|x1-x2|= |x1|-|x2| for all x1,x2€R.
Is the statement true or false? Justify your answer.
1
Expert's answer
2019-05-01T10:11:57-0400

For all a,bRa,b\in R we have the triangle inequality

a+ba+b|a|+|b|\ge |a+b|

Setting a=x1x2a={{x}_{1}}-{{x}_{2}} and b=x2b={{x}_{2}}, we obtain

x1x2+x2x1x2+x2|{{x}_{1}}-{{x}_{2}}|+|{{x}_{2}}|\ge|{{x}_{1}}-{{x}_{2}}+{{x}_{2}}|

That is

x1x2x1x2|{{x}_{1}}-{{x}_{2}}|\ge |{{x}_{1}}|-|{{x}_{2}}|

It follows that there exist x1 and x2 for which inequality

x1x2>x1x2|{{x}_{1}}-{{x}_{2}}|>|{{x}_{1}}|-|{{x}_{2}}|

holds. For example, if x1=2 and x2=-3 we have

2(3)>23|2-\left( -3 \right)|>|2|-|-3|

or

5>23|5|>|2|-|3|

that is

5>15>-1

Hence, the statement

x1x2=x1x2|{{x}_{1}}-{{x}_{2}}|\,=\,|{{x}_{1}}|\,-|{{x}_{2}}|

is false.



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