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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?

- Thread starter Bashyboy
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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?

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Ray Vickson

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

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?

Anyway, you are supposed to show your work.

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

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

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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.

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

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What about:

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

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Actually, I like this proof better:

α < β => α < 2β => α - 2β < 0

α < β => α < 2β => α - 2β < 0

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Ray Vickson

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You have assumed what he is trying to prove: he want to show that for a,b >0, a < b => a < 2b.Actually, I like this proof better:

α < β => α < 2β => α - 2β < 0

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

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.

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