## Verification of simple inequality proof

The question given is:
If a < b, prove that a < (a+b)/2 < b

The book had a different proof than the one I came up with. I understand the book's proof, I just want to know if my proof is also ok.

I did the following:

a < (a+b)/2 < b
2a < a+b < 2b
a < b < 2b-a
a-b < 0 < b-a

Since it was given that a < b, a-b must be less than 0, and b-a must be greater than zero, so the inequality a < (a+b)/2 < b is true if a < b.

Is this ok?

Thanks,
-GeoMike-
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 It looks a bit like you did the proof in reverse. It seems that you started with a<(a+b)/2
 really you should show the direction of the implications you're using, so what you mean is actually [tex]a<(a+b)/2

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## Verification of simple inequality proof

Proof by reverse is great, but you have to show that the steps can logically be reversed at the end
 Recognitions: Gold Member Science Advisor Staff Emeritus That's sometimes called "synthetic proof". It's often used to prove trig identitities. Start with what you want to prove and work back to an obviously true statement. It's valid as long as every stepe is reversible,.
 Thank you for the replies! -GeoMike-