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Thread: prove inequality

  1. #1
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    prove inequality

    im in a proof writing class and am having particular trouble proving that when a and b are both either positive or negative, a<b iff (1/a) > (1/b)
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  2. #2
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    Quote Originally Posted by alr621 View Post
    im in a proof writing class and am having particular trouble proving that when a and b are both either positive or negative, a<b iff (1/a) > (1/b)
    It is almost impossible to help unless we know the set of axioms you have to work with. How basic are your axioms?
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  3. #3
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    very basic. i have the field axioms and the order axioms
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  4. #4
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    Say that a & b are both positive or both negative.
    By order axoms: $\displaystyle x > y \Leftrightarrow \left( {x - y} \right) > 0$.
    So one way is:
    $\displaystyle \frac{1}{a} > \frac{1}{b} \Rightarrow \frac{1}{a} - \frac{1}{b} > 0 \Rightarrow \frac{{b - a}}{{ab}} > 0 \Rightarrow b > a$.( note that it must be true that $\displaystyle ab > 0$).

    Now you can do the other way.
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  5. #5
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    thanks! i have another too. if (a^2)<(b^2) and a, b>0, then a<b
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  6. #6
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    Proof by contradiction
    $\displaystyle a \ge b\quad \Rightarrow \quad a^2 \ge ab$.
    Then what?
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  7. #7
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    can you clarify a little more? i have to prove that a<b if (a^2)<(b^2)
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  8. #8
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    he is saying assume a is greater than or equal to b and see where that takes you.. you will reach an inherent contradiction, finding no way to prove that a>= b and thus a must be < b
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  9. #9
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    okay. last one. if a<b and c<d, then a+c < b+d
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  10. #10
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    Quote Originally Posted by alr621 View Post
    can you clarify a little more? i have to prove that a<b if (a^2)<(b^2)
    There are responders here who will supply posters with ready to turn in solutions. I must tell you that I am not one of those. I must also tell you that I doubt that any of those have enough ability to help you with this question. What I am telling you is this: I want you to learn to do these problems on your own. I am willing to give you substantial hints and direction, but you must show effort on your own. So donít ask me to do more unless you have done more yourself.
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  11. #11
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    Quote Originally Posted by Plato View Post
    There are responders here who will supply posters with ready to turn in solutions. I must tell you that I am not one of those. I must also tell you that I doubt that any of those have enough ability to help you with this question. What I am telling you is this: I want you to learn to do these problems on your own. I am willing to give you substantial hints and direction, but you must show effort on your own. So donít ask me to do more unless you have done more yourself.
    I'm not asking for direct answers to all of my problems... I was simply asking for clarification on that specific situation because I didn't understand it fully.
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