# Is this sufficiently proven? ( A set being a convex set)

1. Jun 28, 2013

### Dazed&Confused

Hi, just a few details prior: I'm trying to study techniques for maths proofs in general after having completed A level maths as I feel it will be of benefit later when actually doing more advanced maths/physics. With this question what is important is the proof is correct which means I don't actually know what a convex set is or what it is useful for...

P.S sorry if this is the wrong sub forum.

1. The problem statement, all variables and given/known data
Prove that if a and b are real numbers, then the set C= { real numbers x: ax ≤ b} is a convex set.

2. Relevant equations

tx + (1-t) y ( for convex sets)

ax ≤ b

3. The attempt at a solution

To show that C = x ≤ $\frac{b}{a}$ is a convex set let x and x' be members of C.

The inequality must hold:

tx + (1-t)x' ≤ $\frac{b}{a}$

Case 1)

Assume x ≤ x' ≤ $\frac{b}{a}$

Then

t(x-x') ≤ 0

and hence t(x-x') +x' ≤ $\frac{b}{a}$

Case 2)

assume x' ≤ x ≤ $\frac{b}{a}$

Then

(1-t)x' ≤ $\frac{b}{a}$ -tx ≤ (1-t)$\frac{b}{a}$

(1-t)x' ≤ (1-t)$\frac{b}{a}$

And hence the inequality holds for both cases Q.E.D

Last edited by a moderator: Jun 28, 2013
2. Jun 28, 2013

### LCKurtz

How do you get that middle term from (1-t)x?

Why not just calculate $a(tx + (1-t)x')$ directly and show it is $\le b$? No separate cases needed.

3. Jun 28, 2013

### Dazed&Confused

I thought about that but not sure how to go about it. I mean wouldn't you essentially be doing the same thing?

and sorry I'm don't know what term you are referring to. Do you mean the x' term instead of x?

4. Jun 28, 2013

### Ray Vickson

Note: you need to be careful! The inequalities ax ≤ b and x ≤ b/a are equivalent only if a > 0. If a < 0 you need to reverse one of the inequality signs, and if a = 0, one of them does not make sense.

5. Jun 28, 2013

### Dazed&Confused

whoops :/ .. well say I kept the inequality in the question and changed all the other inequalities accordingly e.g.

at(x-x') + ax' ≤ b

etc... does that then make the proof valid?

6. Jun 28, 2013

### LCKurtz

I'm referring to the expression in red.

No. It's really simple. Expand it and use what you know about ax and ax'.

7. Jun 28, 2013

### Dazed&Confused

rearraged tx + (1-t)x' ≤ b/a

although now thinking about it.. looks like circular reasoning to me

8. Jun 28, 2013

### LCKurtz

So try my suggestion...

9. Jun 28, 2013

### Dazed&Confused

so tax + (1-t)ax' ≤ tb + (1-t)b = b ?

10. Jun 28, 2013

### LCKurtz

So simple it leaves you Dazed&Confused, eh? You should start with $a(tx + (1-t)x')=$ on the left to be complete.

11. Jun 28, 2013

### Dazed&Confused

yh pretty simple... makes me feel like a simpleton.. thank you anyway