Prove that if a and b are real numbers such that a<b, then a<(a+b)/2<b

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• i'm still not following. srz. Proofs are my weakness. Can you word that differenly maybe? I mean really dumb it down. haha

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  by chelsea on September 5th, 2009

• okay, chels, learning to think in proofs isn't easy, but once you do it a few times, you'll get the hang of it.
  i see i made a typing error. I wrote 2a/a=a. that's not true. 2a/2=2.
  think of it this way. you have a small candy bar, a, and a big one, b. cut each in half. the 2 small pieces together, a/2+a/2, will still be smaller than the 2 big pieces b/2+b/2, since they're still the same candy bars, right?
  now put one small piece, a/2, with one big piece, b/2. this equals a/2+b/2. with the same denominator (bottom number), you add numerators (top numbers), and get a+b/2.
  a small piece and a big piece are greater than 2 small pieces, right again? a small piece and a big piece are less than 2 big pieces, true?
  so, a/2 (small piece)+b/2 (big piece) is greater than a (2 small pieces) but less than b(2 big pieces).
  When i became a tutor i swore i wouldn't give a student an answer, but would guide her to it. Therefore, now that you can understand the process, you must figure out how to write this in proof form. the answer is hiding in here. your job is to find it.
  
  best wishes. you can do it!
  
  btw, what grade are you in and what math class are you taking?

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  by bruceytom on September 5th, 2009

• just so you remember it, this doesn't change your proof, but both a and b can be negative numbers, even though you can't have negative candy bars. look at model proofs in your math book and you will see how to proceed. bet they don't include candy bars, which is a shame. my adult math class used m & m's and got to eat their proofs if they were right. it was fattening but effective. still, you'd better stick to numbers and symbols.

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  by bruceytom on September 5th, 2009

• Thanks for the help :) I'm in college Algebra btw. Didn't see you replied to me and ended up going to class and got some help on it. He said if we show the contrary is false then that's proving that the above statement has to be true. Following my drift? haha If you'd like I can post my answer but it's quite long. Anywho, I'm really struggling with College Algebra which is strange because I got A's in Algebra 2 in highschool and remember my teacher saying it's quite the same to College Algebra. I find them not alike at at all though. So I'm struggling mostly with application problems and like I said earlier, proofs, but otherwise I'm doing ok. Again thanks for the help. I have another question though which I would love your help on again if you don't mind. I'll post it in a bit :P It's about the distance formula. Thank you

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  by chelsea on September 9th, 2009

• You're very welcome, chelsea. Hope it all goes well. Bear in mind that every college class will be harder (sometimes a lot) than its highschool predecessor. The key is to stay calm and think it through. One thing that may help is thinking in algebra there are no complex problems. There are many simple problems clustered together. Remember your operations and their sequence and take them apart. They're not so scary lying on the floor in pieces.
  
  Your taking math makes me think you're on a serious path. Please take it seriously and do your work BEFORE you party. Best wishes.

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  by bruceytom on September 9th, 2009

• Awww thank you bruceytom for your words of encouragement. I'm currently pursuing a undergraduate degree in Biology Education. :D Am not the partying type mind you so won't have to worry about that :)

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  by chelsea on September 9th, 2009

• I know you're a grown woman, but permit me to say, 'good girl'. Keep up your studies and live carefully so that you have no regrets.

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  by bruceytom on September 10th, 2009