53 066
Assignments Done
97,9%
Successfully Done
In October 2017
Your physics homework can be a real challenge, and the due date can be really close — feel free to use our assistance and get the desired result.
Be sure that math assignments completed by our experts will be error-free and done according to your instructions specified in the submitted order form.
Our experts will gladly share their knowledge and help you with programming homework. Keep up with the world’s newest programming trends.

Answer on Real Analysis Question 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}

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question