Question 1199346
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If a < b, show that a < (a + b)/2 < b.
Note: The number (a + b)/2 is called the arithmetic mean of a and b.
Can someone get me started by providing the first-two steps?
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<pre>
Write this equality/(identity), which is true for any/each value of "a"

   a = {{{a/2}}} + {{{a/2}}}.     (1)


In this equality, replace the second addend {{{a/2}}} by the greater value {{{b/2}}}.


Doing this way, from equality (1) you will get then INEQUALITY

    a < {{{a/2}}} + {{{b/2}}}.    (2)



It is the same as  

    a < {{{(a+b)/2}}}.     (3)


Thus, first part of the statement is proved.


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   |  From this point, I continue for the second part. |
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Write this equality/(identity), which is true for any/each value of "b"


   {{{b/2}}} + {{{b/2}}} = b.     (4)


In this equality, replace first addend {{{b/2}}} by the lesser value {{{a/2}}}.


You will get then INEQUALITY

    {{{a/2}}} + {{{b/2}}} < b.    (5)



It is the same as  

    {{{(a+b)/2}}} < b.     (6)



Combining (3) and (6) together in one compound inequality, we get

    a < {{{(a+b)/2}}} < b,


QED.
</pre>

Solved and explained.