|
Question 1207706: If a < b, show that a < (a + b)/2 < b. The number (a + b)/2 is called the arithmetic mean of a and b.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
We want to show that a < (a + b)/2 < b
This breaks down into these two key parts
a < (a + b)/2 and (a + b)/2 < b
What I'll do is start with the first piece and follow these steps
a < (a + b)/2
2a < a + b
2a-a < a + b - a
a < b
Next, I'll reverse those steps by following the chain upward.
a < b
a+a < b+a
2a < a+b
a < (a+b)/2
We have gone from the given a < b to prove that a < (a+b)/2 is the case.
Follow similar steps for the other needed key component.
a < b
a+b < b+b
a+b < 2b
(a+b)/2 < b
We have shown that if a < b, then
a < (a + b)/2 and (a + b)/2 < b
in which those two pieces glue together to get
a < (a + b)/2 < b
--------------------------------------------------------------------------
Another approach
Let c = a+b
Add 'a' to both sides of the original inequality.
a < b
a+a < b+a
2a < a+b
2a < c
Return back to the original inequality.
This time add b to both sides.
a < b
a+b < b+b
c < 2b
Since 2a < c and c < 2b, we know that 2a < c < 2b.
Then,
2a < c < 2b
2a < a+b < 2b
2a/2 < (a+b)/2 < 2b/2
a < (a+b)/2 < b
|
|
|
| |
