help with this problem?

Question

Prove the following statements are equivalent for any two distinct real numbers a and b.
(i). b is larger than a.
(ii). Their average, (a+b)/2, is larger than a.

Answer 1

Actually statement ii is a logical corrolary of statement i.

If b is > a, it follows that the average of a and b (a + b)/2, will be > a. This is true if a and b are real numbers.

b - a will be more than 0 if b > a

b - a / 2 will also be > 0 in that case.

So (a + b)/2 (the average) will be larger than a by b - a /2 which is +ve value greater than 0. That proves that b > a.

Answer 2

b - a = x where x > 0
therefore b = a + x

now (a + b) / 2 = (a + (a + x) ) / 2
= (a + a + x) / 2
= (2a + x ) / 2
= a + x/2
> a since x > 0 and 2x > 0

Answer 3

If b is larger than a then we could say that b = x + a, where x is the part of b that makes it bigger than a.

Using substitution, we could write (a+b)/2 as (a + (x + a))/2, which is (2a+x)/2.

Breaking apart (2a+x)/2 into two fractions, we get 2a/2 + x/2, where 2a/2 = a.

So, we now have the equation a + x/2, which is bigger than a.

I hope that helps...

Answer 4

above ones r right