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Proving irrationality

  1. Mar 15, 2005 #1
    how do you prove that sqrt(2) + sqrt(5) is irrational...
     
  2. jcsd
  3. Mar 15, 2005 #2

    matt grime

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    by contradiction, suppose r=sqrt(2)+sqrt(5) and r is rational. play around with that and see what you can show this implies.
     
  4. Mar 15, 2005 #3
    i did get it down to 7+2root10 = m^2/n^2 where i assumed root 2 + root 5 = m/n...and before that i proved root 10 to be irrational...so i am kinda stuck on the 7+2root10 = m^2/n^2 bit...how does that mean anything...
     
  5. Mar 15, 2005 #4

    matt grime

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    why not do rearrange the equation before squaring it so that after squaring you only have, say sqrt(2) to deal with, which i presume you know is irrational.

    i don't see why you're showing 7+ something is not of the from m^2/n^2, to be honest, either, and you certainly don't need to write r=m/n either.
     
  6. Mar 15, 2005 #5
    look at this
     
  7. Mar 15, 2005 #6

    matt grime

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    I think that's aiming way too high for this question, and besides, if you're going to go and use theorems like that then why not simply invoke the fact that sqrt(2)+sqrt(3) is an algebraic integer, and hence if it were rational it would be integral, which it clearly isn't.

    r-sqrt(2) = sqrt(5)
    square and rearrange to get sqrt(2) rational, contradiction.
     
  8. Mar 15, 2005 #7
    i had an argumetn like this

    this is where i got up to...
    now i am thinking that if i put everything in terms of root 10 then:
    so we have irrational = "(m2/n2 - 7)/2"
    so that implies that (m2/n2 - 7) is irrational and also m2/n2 is irrational and they cannot be irrationals and integers at the same time and therefore its a contradiction...makes sence??? may b not...im just insane!!!
     
  9. Mar 15, 2005 #8

    HallsofIvy

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    Assuming you already know √(2) is irrational, matt grime's suggestion, "rearrange the equation before squaring it " gives this:

    Suppose r= √(2)+ √(5) is rational. Then r- √(2)= √(5).
    Squaring both sides, r2- 2√(2)r+ 2= 5 so 2√(2)= r2- 3. Then √(2)= (r2- 3)/2. Since the rational numbers are closed under multiplication, division, addition, and subtraction, the right hand side is rational, giving the contradiction that √(2) is rational.
     
  10. Mar 15, 2005 #9

    Curious3141

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    Another way of doing it, which I think is slightly "nicer" than squaring is as follows :

    Let [tex]\sqrt{5} + \sqrt{2} = x_1[/tex] and [tex]\sqrt{5} - \sqrt{2} = x_2[/tex]

    Assume [tex]x_1[/tex] is rational and can be expressed as [tex]\frac{p}{q}[/tex], where p and q are coprime integers.

    [tex]\frac{1}{2}(x_1 - x_2) = \sqrt{2}[/tex], which is irrational, hence [tex]x_2[/tex] has to be irrational (since [tex]x_1[/tex] is rational by assumption).

    But [tex]x_1x_2 = 3[/tex], giving [tex]x_2 = \frac{3}{x_1} = \frac{3q}{p}[/tex]

    This implies [tex]x_2[/tex] is rational, contradicting the above.

    Hence [tex]x_1[/tex] is irrational (QED).
     
  11. Mar 16, 2005 #10

    HallsofIvy

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    Yes, that is nice.
     
  12. Mar 16, 2005 #11

    Curious3141

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    Thank you. :smile:

    BTW, there is a small error in your explanation of the squaring proof. It should go like so :

    [tex]r^2 - 2\sqrt{2}r + 2 = 5[/tex]
    [tex]2\sqrt{2}r = r^2 - 3[/tex]
    [tex]\sqrt{2} = \frac{1}{2}(r - \frac{3}{r})[/tex]

    And LHS is irrational, RHS is rational, contradiction. You forgot the 'r' when you typed it out. Just a minor point, doesn't change the substance of your proof at all. :biggrin:
     
  13. Mar 18, 2005 #12
    thanks to all! it kinda makes sence!
     
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