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Answer to Question #4650 in Real Analysis for aiza

Question #4650
show that if a,b element R.then
a. max{a,b}=1/2(a+b+|a-b|) and min{a,b}=1/2(a+b-|a-b|)
b. min{a,b,c}=min{min{a,b},c}
Expert's answer
to prove a) we can consider two cases
first one a>=b then& a=max{a,b} but
1/2(a+b+|a-b|)=1/2(a+b+a-b)=1/2(2a)=a& (in this case |a-b|=(a-b) because
a>=b) so we have 1/2(a+b+|a-b|)=a=max{a,b} and for min{a,b} we have
min{a,b}=b and 1/2(a+b-|a-b|)=1/2(a+b-(a-b))=1/2(2b)=b
second one a<b
then& b=max{a,b} but 1/2(a+b+|a-b|)=1/2(a+b-(a-b))=1/2(2b)=b& (in this case
|a-b|=-(a-b) because a<b) so we have 1/2(a+b+|a-b|)=b=max{a,b} and for
min{a,b} we have min{a,b}=a and
1/2(a+b-|a-b|)=1/2(a+b-(-(a-b)))=1/2(a+b+a-b)=1/2(2b)=b

b.min{a,b,c}=min{min{a,b},c}
we will consider three cases:
1) c<=a,c<=b so min{a,b,c}=c by
definition. then min{a,b} is equal to a or b so min{a,b}>=c so
min{min{a,b},c}=c=min{a,b,c}
2) min{a,b,c}=a so a<=b, a<=c so we have
min{a,b}=a& min{min{a,b},c} =min{a,c}=a because a<=c
3) min{a,b,c}=b so
b<=a, b<=c so we have min{a,b}=b& min{min{a,b},c} =min{b,c}=b because
b<=c
So in all cases min{a,b,c}=min{min{a,b},c}

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