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Prove a^2≥0 always

  1. Apr 16, 2012 #1
    1. The problem statement, all variables and given/known data
    prove a^2 is positive

    I have tried this, but i dont know if it's conclusive:

    i) Suppose a≥0
    then a+1≥1
    then a(a+1)≥a
    then a^2+a≥a
    then a^2≥0
    ii) Suppose a≤0
    then 1+a≤1
    then a(a+1)≥a since a is negative, by a lemma
    then a^2+1≥a
    then a^2≥0
     
  2. jcsd
  3. Apr 16, 2012 #2

    Dick

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    Seems pretty ok. You aren't stating all of your assumptions and lemmas here, though. And there is a typo in your second argument.
     
  4. Apr 17, 2012 #3
    ok great, yeah i'll be more specific and i do see my typo: a+1 instead of 1+a
     
  5. Apr 17, 2012 #4

    Dick

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    I actually meant turning a(a+1) into a^2+1, but I'm sure that's not news.
     
  6. Apr 17, 2012 #5
    oh gosh i didn't even realize i had typed that, haha well thanks anyways.

    btw you are always incredibly helpful, and for that i am really grateful!
     
  7. Apr 17, 2012 #6

    Curious3141

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    Why don't you just multiply both sides immediately by a?

    i.e.

    Case 1: a≥0
    a2≥0

    Case 2: a≤0
    a2≥0

    Seems a lot more direct and it's the same idea.
     
  8. Apr 17, 2012 #7
    because you dont know what a*a means yet
     
  9. Apr 17, 2012 #8

    Curious3141

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    What do you mean, exactly?
     
  10. Apr 17, 2012 #9
    well im taking a beginning real analysis course and basically we're building the basic structure of algebra from scratch from a couple of real number axioms.
     
  11. Apr 17, 2012 #10

    Curious3141

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    OK. But saying you don't know the implication of a*a means you don't know the implication of a*(a+1) either.

    You're already making an implicit assumption when you reverse the sign of the inequality upon multiplying by negative a. Multiplying a≤0 by a assumes no more than multiplying a+1≤1 by a.
     
  12. Apr 17, 2012 #11
    the multiplying done by the negative constant was proved in a lemma but you're right, it wasn't known before. and when i did a(a+1) i guess i knew that it was a*a+a, but i didn't know what a*a was still so i had to describe it using the inequalities
     
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