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Determine if integral domain or field

  1. Aug 5, 2011 #1
    I found the unity to be 1. I'm not sure how to determine 3) for each.

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110805_224110.jpg?t=1312602377 [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110805_224012.jpg?t=1312602392 [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 5, 2011 #2

    Dick

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    The real numbers have no zero divisors, so it's a pretty safe bet that both are at least integral domains. To determine if there is an inverse, don't start from scratch. Multiply (a+sqrt(2)*b)*(a-sqrt(2)*b) and try to figure out where to go from there.
     
  4. Aug 5, 2011 #3
    Ah! Duh. I was focusing too much on the rationals a,b and forgetting that the term represents a real number.

    So, there's no way to determine if there's an inverse if you start from scratch?

    Why did you choose this? Just a good guess?

    (a+sqrt(2)*b)*(a-sqrt(2)*b)
     
  5. Aug 5, 2011 #4

    Dick

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    Yes, good guess. It gets rid of the sqrt(2). Which you have to do if you are finally going to get just 1. Now go from there.
     
  6. Aug 5, 2011 #5
    (a+sqrt(2)*b)*(a-sqrt(2)*b) = 1

    a2 - 2b2 = 1

    a2 = 1 + 2b2

    If b = 1, then a2 = 3

    Then, a = sqrt(3), which is not a rational number. This is not a Field.
     
  7. Aug 5, 2011 #6

    Dick

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    Ok, my guess was really just a hint. a^2-2*b^2 is unlikely to be 1. Now think about (a+b*sqrt(2))*k*(a-b*sqrt(2))=1 where k is some constant in your system. Can you solve for k if you are working in the rationals?
     
    Last edited: Aug 5, 2011
  8. Aug 5, 2011 #7
    Is my counterexample correct? Did I show it's not a Field?

    k = [a2 - 2b2]-1
     
  9. Aug 6, 2011 #8

    Dick

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    Well, no. Using that value of k, I'd say the inverse of a+sqrt(2) is (a-sqrt(2))/(a^2-2), isn't it? Your main worry over the rationals is whether a^2-2*b^2 could be zero.
     
  10. Aug 6, 2011 #9
    Okay, you're right. There is a valid inverse if b = 1.

    a2 - 2b2 = 0

    a = sqrt(2)b
     
    Last edited: Aug 6, 2011
  11. Aug 6, 2011 #10

    HallsofIvy

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    [itex]\frac{1}{a+ b\sqrt{2}}[/itex]

    Rationalise the denominator.
     
  12. Aug 6, 2011 #11
    You can't?
     
  13. Aug 6, 2011 #12

    SammyS

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    I'm sure he can !!!

    But the point is that you do it.
     
  14. Aug 6, 2011 #13
    No. I'm saying you can't rationalize the denominator because there will always be root two down there. Is that right?
     
  15. Aug 6, 2011 #14

    Dick

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    I thought we were getting towards being able to agree that the inverse of (a+b*sqrt(2)) is (a-b*sqrt(2))/(a^2-2*b^2)? In so far as that is defined. Is it defined over the rationals? Let's not go backwards here.
     
  16. Aug 7, 2011 #15
    Oh, sorry. I guess that's next logical conclusion.

    Is the inverse defined over the rationals? Not if sqrt(2) is in there.
     
  17. Aug 7, 2011 #16

    HallsofIvy

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    You are still missing the whole point! [itex](a+ b)(a- b)= a^2- b^2[/itex]. I'm sure you learned that long ago. So what do you get if you multiply both numerator and denominator of
    [tex]\frac{1}{a+ b\sqrt{2}}[/tex]
    by
    [tex]a- b\sqrt{2}[/tex]
     
  18. Aug 7, 2011 #17

    Hurkyl

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    I don't see why people in this thread have an aversion to "starting from scratch". Linear algebra is a very useful technique in abstract algebra, and both of the questions
    • What, when multiplied by (a + b[itex]\sqrt{2}[/itex]), gives 1?
    • What, when multiplied by (a + b[itex]\sqrt{2}[/itex]), gives 0?
    are linear systems of equations in two unknowns. (treating a and b as "knowns")

    Rationalizing denominators is a good trick, but far from required for this problem.
     
  19. Aug 7, 2011 #18

    Dick

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    What do you mean 'if sqrt(2) is in there'? Do you mean sqrt(2) is in the rationals? Can you elaborate a bit?
     
  20. Aug 7, 2011 #19
    Rationalizing is when you get rid of the square root in the denominator? I was wrong. You can rationalize the denominator with the difference of two squares trick.

    [1/sqrt(2)] = sqrt(2)/2
     
  21. Aug 7, 2011 #20
    As established earlier, nothing non-zero multiplied by a real number yields zero. Thus, that system is not possible. The inverse is [a - b*sqrt(2)] / [a2 - 2b2].
     
    Last edited: Aug 7, 2011
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