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

[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.

Homework Statement

Prove that if a and b are real numbers, then the set C= { real numbers x: ax ≤ b} is a convex set.

Homework Equations

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

ax ≤ b

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

 

[LCKurtz]

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.

Homework Statement

Prove that if a and b are real numbers, then the set C= { real numbers x: ax ≤ b} is a convex set.

Homework Equations

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

ax ≤ b

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}

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.

 

[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?

Thanks for reply btw.

 

[Ray Vickson]

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.

Homework Statement

Prove that if a and b are real numbers, then the set C= { real numbers x: ax ≤ b} is a convex set.

Homework Equations

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

ax ≤ b

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

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.

 

[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?

 

[LCKurtz]

assume x' ≤ x ≤ \frac{b}{a}

Then
(1-t)x' ≤ \color{red}{\frac{b}{a} -tx} ≤ (1-t)\frac{b}{a}

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

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

I'm referring to the expression in red.

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

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

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

 

[Dazed&Confused]

rearraged tx + (1-t)x' ≤ b/a

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

 

[LCKurtz]

rearraged tx + (1-t)x' ≤ b/a

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

So try my suggestion...

 

[Dazed&Confused]

so tax + (1-t)ax' ≤ tb + (1-t)b = b ?

 

[LCKurtz]

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

 

[Dazed&Confused]

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