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Problem with simplification in Mathematica!

  1. Jul 19, 2010 #1
    Assuming[a>0 && b>0 && c>0, FullSimplify[a+Sqrt[b+c]-Sqrt[a^2+b+c+2 a Sqrt[b+c]]]] does not simplify it to zero. Although output of FullSimplify[a+Sqrt[b+c]-Sqrt[a^2+b+c+2 a Sqrt[b+c]]==0] is true, even without any assumptions. Can any one please resolve this puzzle.....Thanks
  2. jcsd
  3. Jul 19, 2010 #2


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    The reason it doesn't simplify to zero is that this would throw away one possibility. Basically you are asking it to simplify (x-Sqrt(x^2)). Since Sqrt(x^2) can be either x or -x, (x-Sqrt(x^2)) can be either 0 or 2x. By simplifying this to zero, you are throwing away the second possibility. Mathematica will keep all possibilities unless you explicitly tell it otherwise.
  4. Jul 19, 2010 #3
    Your statement is not true as i have specified a>0,b>0,c>0
    If you ask mathematica to simplify x-Sqrt[x^2] with assumption x>0, it indeed gives zero!
  5. Jul 19, 2010 #4


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    I used the simplified example with x to illustrate the problem. In your problem, just because a,b,c>0 doesn't mean that Sqrt[b+c]>0, or that Sqrt[a^2+b+c+2 a Sqrt[b+c]]>0. Even if you tell it that both of these are >0, that still isn't enough. You have to either explicitly tell it to use the positive square root, like this:

    FullSimplify[ a + Sqrt[b + c] - Sqrt[a^2 + b + c + 2 a Sqrt[b + c]] /.
    Sqrt[a^2 + b + c + 2 a Sqrt[b + c]] -> a + Sqrt[b + c]]

    or else square both terms to eliminate the sign ambiguity, like this:

    FullSimplify[(a + Sqrt[b + c])^2 - (Sqrt[
    a^2 + b + c + 2 a Sqrt[b + c]])^2, {a > 0, b > 0, c > 0}]

    It's only a computer program, after all, so it does have its limitations.
  6. Jul 19, 2010 #5
    If you're only interested a, b and c being real, then you could use something like:

    In[1]:= Reduce[a+Sqrt[b+c]-Sqrt[a^2+b+c+2 a Sqrt[b+c]]==0,{a,b,c},Reals]
    Out[1]= (a<=0&&c>=a^2-b)||(a>0&&c>=-b)

    In[2]:= Simplify[%,a>0&&b>0&&c>0]
    Out[2]= True
  7. Jul 21, 2010 #6
    Then why Assuming[x>0,Sqrt[x^2] //Simplify] is x? I think mathematica only considers positive sqrt.
  8. Jul 21, 2010 #7
    The step comes in a intermediate step of a long program. Unless the answer in simplification is zero, the final answer is too complicated to read. Any suggestion?
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