Proving the Inequality of Two Real Numbers

[bonfire09]

Homework Statement

Prove that for every two distinct real numbers a and b, either (a+b)/2>a or (a+b)/2>b

Homework Equations

The Attempt at a Solution

Proof:
if two distinct numbers a and b then (a+b)/2>a
Since a≠b and a,bεR, (a+b)/2>a=a+b>2a=b>a. Therefore (a+b)/2>a if b>a.
and
if two distinct numbers a and b then (a+b)/2>b
Since a≠b and a,bεR, (a+b)/2>b=a+b>2b=a>b.
Therefore (a+b)/2>b if a>b.

would this suffice as a proof or no?

 

[cryora]

Something is missing. I think you ought to state that a≠b implies either a>b or b>a
and then show that
a>b is an equivalent statement to (a+b)/2>b
and
b>a => (a+b)/2>a

Would you have to prove that a≠b => a>b or b>a?

 

[Mark44]

Homework Statement

Prove that for every two distinct real numbers a and b, either (a+b)/2>a or (a+b)/2>b

Homework Equations

The Attempt at a Solution

Proof:
if two distinct numbers a and b then (a+b)/2>a
Since a≠b and a,bεR, (a+b)/2>a=a+b>2a=b>a. Therefore (a+b)/2>a if b>a.
and
if two distinct numbers a and b then (a+b)/2>b
Since a≠b and a,bεR, (a+b)/2>b=a+b>2b=a>b.
Therefore (a+b)/2>b if a>b.

would this suffice as a proof or no?

No. You're assuming part of what you need to prove.

Since a≠b and a,bεR, (a+b)/2>a ...

Also, this makes no sense: (a+b)/2>a=a+b>2a=b>a
You're saying that (a + b)/2 > a, which equals a + b, which is greater than 2a, which equals b, which is greater than a. The problem is that you are apparently connecting inequalities (such as (a+b)/2>a and a + b > 2a), with =. That's not the right symbol. Equations and inequalities aren't equal to anything; they might be equivalent, or one might imply another, but they're not equal.

 

[bonfire09]

yeah i knew this was wrong. I think the problem is that I am not approaching the conclusion of the result correctly. let me try this again

 

[Mark44]

Break it up into two cases, along the lines of what cryora suggests.

Case 1: Suppose a < b.
Show that (a + b)/2 > a.

Case 2: Suppose that b < a.
Show that (a + b)/2 > b.

 

[bonfire09]

proof: Since a and b are distinct numbers this implies that either a>b or b>a

Case 1: Let a>b. Then a/2>b/2. a/2+b/2>2b/2. (a+b)/2>b. Since a>b then (a+b)/2>b.

Something like this. Damn proving is tough. Has anyone used a transition to advanced math by chartrand? I am trying to learn how to do proofs from this book. I can do about 70%-80% of the problems but some of them are tricky like this one. Anyone have other books that are good at showing different kinds of proofs that I can use to supplement this book?

 

[Mark44]

proof: Since a and b are distinct numbers this implies that either a>b or b>a

Case 1: Let a>b. Then a/2>b/2. a/2+b/2>2b/2. (a+b)/2>b. Since a>b then (a+b)/2>b.

Something like this. Damn proving is tough. Has anyone used a transition to advanced math by chartrand? I am trying to learn how to do proofs from this book. I can do about 70%-80% of the problems but some of them are tricky like this one. Anyone have other books that are good at showing different kinds of proofs that I can use to supplement this book?

Here are a couple that I think would be helpful.
How to Read and Do Proofs (http://books.google.com/books/about/How_to_read_and_do_proofs.html?id=K3itQwAACAAJ)
The Nuts and Bolts of Proofs (https://www.amazon.com/dp/0123822173/?tag=pfamazon01-20)

 

[bonfire09]

ok thanks. I'll look into these books.