
#1
Jul3106, 09:43 PM

P: 70

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 < 2ba ab < 0 < ba Since it was given that a < b, ab must be less than 0, and ba must be greater than zero, so the inequality a < (a+b)/2 < b is true if a < b. Is this ok? Thanks, GeoMike 



#2
Jul3106, 09:49 PM

P: 1,076

It looks a bit like you did the proof in reverse. It seems that you started with a<(a+b)/2<b and then got to a<b rather than starting with a<b and proving a<(a+b)/2<b like the problem seems to want.




#3
Jul3106, 10:21 PM

P: 998

really you should show the direction of the implications you're using, so what you mean is actually
[tex]a<(a+b)/2<b \Longleftarrow 2a<(a+b)<2b \Longleftarrow a<b<2ba \Longleftarrow ab<0<ba \Longleftarrow a<b,[/tex] and not the other way. It doesn't matter much here since all the inequalities there are equivalent (so the implications work both ways, ie. in fact [itex]a<b \Longleftrightarrow a<(a+b)/2<b[/itex]). 



#4
Jul3106, 11:16 PM

Mentor
P: 4,499

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




#5
Aug106, 05:29 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,900

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,.




#6
Aug106, 12:22 PM

P: 70

Thank you for the replies!
GeoMike 


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