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- mathmath333

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- katieb

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- mathmath333

\(\large \color{black}{\begin{align} & \{a,b\}\in \mathbb{Z} ,\ \ \ a,b >2,\ a>b\ \hspace{.33em}\\~\\
& \normalsize \text{which of the following will be always true ?} \hspace{.33em}\\~\\
& a.)\ \dfrac{a+b}{2}>\dfrac{a-b}{2} \hspace{.33em}\\~\\
& b.)\ \dfrac{a+b}{2}\geq \dfrac{a-b}{2} \hspace{.33em}\\~\\
& c.)\ a^2-b^2

- anonymous

B is confusing, but I thinks it's true
The answer is E since we are dealing with positive numbers and bigger than one

- mathmath333

but in the book it is given as option d.)

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- anonymous

Not too good at calc but here it goes...
What I did was, was I basically picked random numbers that are greater than 2. (I did multiple trials).
Trail 1: a = 8 while b = 6
A. True
B. True
C. True
Trial 2: a = 12 while b = 4
A. True
B. True
C. True
So I would have to agree with @Ahmad-nedal that the answer is E. However, I am not in calc therefore I don't serve much of a purpose. Perhaps it is d because of the {a,b} E Z.

- anonymous

for a and b note that \(a+b\) is strictly greater than \(a-b\)

- anonymous

I understand the reason, some books deal with ineaqalities from different aspects.
Although the equality stated in part B will not be achieved, but the argument still holds for any a,b in z that satsifies the constraints given. So I insist on my opinion

- mathmath333

is the answer option e.) ?

- anonymous

Yes

- anonymous

it's d

- mathmath333

ok thnx

- anonymous

It is given that a and b are integers greater than 2 and that a is always greater than b.
You can choose a number.
Let as say a=4 and b=3
For a)= \[\frac{ 4+3 }{ 2 }>\frac{ 4-3 }{ 2 }=\frac{ 7 }{ 2 }>\frac{ 1 }{ 2 }\]
Similarly at c) \[4^{2}-3^{2}<4^{3}-3^{3}=7<37\]
Therefore both a and c are true.

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