Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Square an inequality if both sides are positive

  1. Sep 16, 2010 #1
    I know that we can square an inequality if both sides are positive.
    But can we cube an inequality provided both the sides are positive?
    If no then why?
     
  2. jcsd
  3. Sep 16, 2010 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Inequality

    The question of whether you can perform an operation on both sides of an inequality is one of whether the function is increasing or decreasing. the function f(x)=x2 is not increasing (when x is negative), so will not preserve order.

    On the other hand, for just the positive set of numbers, f(x)=x2 is increasing, so f(x)<f(y) if and only if x<y.

    The function f(x)=x3 is increasing everywhere, so x3<y3 if and only if x<y This means that you can cube both sides of an inequality whenever you want
     
  4. Sep 16, 2010 #3
    Re: Inequality

    Suppose a<c and b<c (all are positive)
    and a,b,c are related by the equation a^2 + b^2 =c^2
    I proceeded in 2 ways-
    1) a^3 < c^3
    and b^3 <c^3
    i.e. (a^3 + b^3 )/2 < c^3

    2) a^3=a.a^2<c.a^2
    similarly b.b^2<c.b^2
    i.e. a^3 + b^3 < c^3 (a^2 + b^2=c^2)

    why are 1 and 2 different?
     
  5. Sep 16, 2010 #4

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Inequality

    1 and 2 are different because you did different things. When you manipulate an inequality you aren't guaranteed to get the best possible new inequality; in this case the first one just happens to be worse than the second one.

    Notice you didn't use the fact about the sum of a and b squared when doing (1), which means that it's true for any choices of a, b and c, whereas in (2) a, b and c need to satisfy a2+b2=c2
     
  6. Sep 17, 2010 #5

    HallsofIvy

    User Avatar
    Science Advisor

    Re: Inequality

    If 0< x< y, then, multiplying both sides by the positive number x, [itex]0< x^2< xy[/itex]. Multiplying 0< x< y by the positive number y, [itex]0< xy< y^2[/itex]. Since "<" is "transitive", [itex]0< x^2< xy< y^2[/itex] so [itex]x^2< y^2[/itex].

    Do that again to get cubes.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook