How to prove α - 2β < 0 from β > α

[Bashyboy]

Hello,

I am given that β > α, which can be written as β - α > 0. What justification would I have to use in order to conclude that α - 2β < 0, given that the preceding propositions are true? Could someone possibly help me?

 

[Ray Vickson]

Hello,

I am given that β > α, which can be written as β - α > 0. What justification would I have to use in order to conclude that α - 2β < 0, given that the preceding propositions are true? Could someone possibly help me?

This is not necessarily true: if α = -3 and β = -2 we have α < β but 2β < α.

Anyway, you are supposed to show your work.

 

[Bashyboy]

This isn't actually a homework problem. It was something I came across and found curious, when I was drawing the slopes fields of a differential equation.

I also forgot to mention that alpha and beta are both positive.

 

[Bashyboy]

Note: I can not provide my attempt at solving this problem, for I never did attempt at solving it; the reason being, that I do not know enough about inequalities to solve this problem. As I said, this is not a homework problem, it is a side-tracking.

 

[PeroK]

So, to re-phrase the question. You have two positive numbers. You start with the smaller one and subtract twice the larger one and you want to prove that the answer is less than zero?

 

[Bashyboy]

PeroK, if I read your sentences correctly, yes that is what I would like to prove, rather, I would like to know how to prove, as I have never done any proves involving inequalities.

α > 0 and β > 0, and β > α, and I would like to know if we can conclude that α - 2β < 0.

 

[PeroK]

PeroK, if I read your sentences correctly, yes that is what I would like to prove, rather, I would like to know how to prove, as I have never done any proves involving inequalities.

α > 0 and β > 0, and β > α, and I would like to know if we can conclude that α - 2β < 0.

Well, if you had $α in the bank. And you withdrew $β. Then you withdrew $β again. Do you think you might be overdrawn? If β > α.

 

[Bashyboy]

I think there might be a misunderstanding. I understand that statement α - 2β < 0 is true; I am looking for a little more mathematically rigorous proof of the fact.

 

[PeroK]

I think there might be a misunderstanding. I understand that statement α - 2β < 0 is true; I am looking for a little more mathematically rigorous proof of the fact.

What about:

α - 2β = α - β - β < α - β (as β > 0) < 0 (as α < β)

 

[gopher_p]

I think if you take a step back and examine PeroK's last post (edit: the one talking about bank accounts), you might find that it is VERY suggestive of a proof. Very suggestive.

 

[PeroK]

Actually, I like this proof better:

α < β => α < 2β => α - 2β < 0

 

[Ray Vickson]

Actually, I like this proof better:

α < β => α < 2β => α - 2β < 0

You have assumed what he is trying to prove: he want to show that for a,b >0, a < b => a < 2b.

 

[PeroK]

You have assumed what he is trying to prove: he want to show that for a,b >0, a < b => a < 2b.

What about:

α < β => α < β + β => α < 2β => α - 2β < 0

I guess it's not that obvious that 2β = β + β. Hopefully that repairs the proof.