# Proof help

1. Mar 10, 2005

### Kamataat

This isn't homework or anything. I'm just interested in learning to prove things in mathematics, so I took a piece of paper and did the following:

For every a < b it is true that a < (a+b)/2 < b (1). First I separated the inequality: a < (a+b)/2 and (a+b)/2 < b. Then I did this:

a < (a+b)/2
2a < a + b
2a - a < b
a < b

and

(a+b)/2 < b
a + b < 2b
a < 2b - b
a < b

So for the inequality (1) to be true, it must be true that a < b. So we have proven the statement in the first sentence.

Right? Wrong? Not rigorous enough?

- Kamataat

2. Mar 10, 2005

### Icebreaker

I would have done it the other way around:

Let $$\{a,b\} \in R$$ such that $$a<b$$.

$$a<b$$

$$a+a-a<b$$

$$2a-a<b$$

$$2a<a+b$$

$$a<\frac{a+b}{2}$$

and

$$a<b$$

$$a<b+b-b$$

$$a<2b-b$$

$$a+b<2b$$

$$\frac{a+b}{2}<b$$

And therefore $$a<\frac{a+b}{2}<b$$.

Last edited by a moderator: Mar 10, 2005
3. Mar 10, 2005

### HallsofIvy

There is a small difficulty: you started with what you want to prove and wound up with the hypothesis. Strictly speaking, that's wrong- you can't ASSUME what you want to prove.

What you CAN do is write everything in reverse. That is, a< b so, adding a to both sides, 2a< a+ b. Divide both sides by 2: a< (a+b)/2.

The reason I said "strictly speaking" is that, in fact, that's sometimes called "synthetic proof" and is commonly used, for example, in proving trig identities.

As long as it is clear that everything you do is "reversible" you're okay. But the "rigorous" proof is actually going the other way.

4. Mar 11, 2005

### Kamataat

ic, tnx both of you

- Kamtaat