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Homework Help: Easy algebra problem

  1. Feb 16, 2008 #1
    [SOLVED] easy algebra problem

    1. The problem statement, all variables and given/known data
    Why is the following equation impossible:


    where a and b are rational numbers and b is not 0. It seems so obvious... Feel free to use group, ring, or field theory in your answer.

    2. Relevant equations

    3. The attempt at a solution
    EDIT: I can prove that [tex]\sqrt{3}[/tex] is irrational. Square both sides and rearrange to get
    [tex]\frac{3-a^2-2b^2}{2ab}=\sqrt{2} [/tex]
    which is impossible because the rationals form a field (which is closed under addition, multiplication, and division).
    Last edited: Feb 16, 2008
  2. jcsd
  3. Feb 16, 2008 #2


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    If 3= a+ b[itex]\sqrt{2}[/itex] then 3- a= b[/itex]\sqrt{2}[/itex] so [itex]\sqrt{2}[/itex]= (3- a)/b. a and b are rational numbers so ....
  4. Feb 16, 2008 #3
    well, i can prove that any rational(-non-zero) number b times sqrt(2) is irrational. Then again the sum of a rational number and an irrational one is irrational. So here we have the left side of the eq a rational number while the left side is an irrational one, which is not possible.
    Am i even close?
  5. Feb 16, 2008 #4
    Is he asking for an answer, or he is just testing some of us? i got the feeling that the op already knows the answer, but rather wants to see who knows the reason, or am i wrong?
  6. Feb 16, 2008 #5

    Well, you edited it now, right, because before ther was a plain 3, while now ther is the square root of 3.
  7. Feb 16, 2008 #6
    Yes, I edited it, sorry. Thanks.
  8. Feb 16, 2008 #7

    Well, yeah this loos fine to me, since a,b are rationals then they will also be rationals when squared, also -,+,* in the field of rationals are closed operations, so on the left side you have a rational nr. while on the right an irrational. this seems to me like a good contradiction, let's see what halls have to say.!
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