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Math Help - Help with True/False algebra inequality

  1. #1
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    Help with True/False algebra inequality

    Hi. I had the problem below for homework. I think it is true, but I don't know why. Can anyone give me a reason why this is true, because I need to provide an explination. Thanks!

    True or False:
    If a<b then a3 <b3

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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by sleigh View Post
    Hi. I had the problem below for homework. I think it is true, but I don't know why. Can anyone give me a reason why this is true, because I need to provide an explination. Thanks!

    True or False:
    If a<b then a3 <b3
    0 < b3 - a3

    true if a and b are positive
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  3. #3
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    Quote Originally Posted by sleigh View Post
    Hi. I had the problem below for homework. I think it is true, but I don't know why. Can anyone give me a reason why this is true, because I need to provide an explination. Thanks!
    True or False:
    If a<b then a3 <b3
    Quote Originally Posted by 11rdc11 View Post
    0 < b3 - a3 true if a and b are positive
    It is true period. The function f(x)=x^3 is increasing for all x.
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  4. #4
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by sleigh View Post
    Hi. I had the problem below for homework. I think it is true, but I don't know why. Can anyone give me a reason why this is true, because I need to provide an explination. Thanks!

    True or False:
    If a<b then a3 <b3
    I assume you mean: if a < b then a^3 < b^3.

    It is actually true, but needs a fair amount of work to demonstrate.

    First, you can show it's true when both a and b are positive, as follows.

    You know that if a, b, c < 0 then if a < b then a c < b c.

    Okay, so a < b implies a^2 < a b (putting c = a in the above).

    Then a < b implies ab < b^2 (putting c = b in the above).

    So you've shown a < b implies a^2 < b^2.

    Do the same (similar) thing to prove for a^3 < b^3.

    You go through a similar process when a and b are both negative, and again when a is negative and b positive.


    An alternative approach makes use of the fact that the cube function is monotonically increasing - but that's really on a calculus course.
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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Plato View Post
    It is true period. The function f(x)=x^3 is increasing for all x.
    Thanks I didn't realise he meant cubed
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  6. #6
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    a < b = (a<b)^3 = \boxed{a^3 < b^3}
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  7. #7
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by Mozart View Post
    a < b = (a<b)^3 = \boxed{a^3 < b^3}
    You can't just put an inequality into a cube like that, surely?
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  8. #8
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    I trust this works?
    Attached Thumbnails Attached Thumbnails Help with True/False algebra inequality-lim.gif  
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  9. #9
    Super Member Matt Westwood's Avatar
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    How sure are we that a^2 + ab + b^2 > 0?
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  10. #10
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    a^{2}+ab+b^{2}=\frac{4a^{2}+4ab+4b^{2}}{4}=\frac{(  2a+b)^{2}+3b^{2}}{4}>0.

    Though you can't manipulate what you want to prove.
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