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Q is closed under division or not?

  1. Sep 10, 2011 #1
    1. The problem statement, all variables and given/known data


    Q = rational numbers

    My professor proved that it is closed under addition yesterday. I kinda understood a bit...

    How the heck does it work for when n_1 or n_2 = 0?
     
  2. jcsd
  3. Sep 10, 2011 #2

    micromass

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    Take a look at the first line. There we take [itex]r_1,r_2\in \mathbb{Q}[/itex]. Then we say that

    [tex]r_i=\frac{m_i}{n_i}[/tex]

    but obviously this doesn't work for [itex]n_i=0[/itex] since division by zero is not defined in [itex]\mathbb{Q}[/itex].

    That is [itex]\frac{m}{n}\in \mathbb{Q}[/itex] if and only if [itex]n\neq 0[/itex]...
     
  4. Sep 10, 2011 #3
    I thought it is the same logic for integers, I take 1 and 0 and take 0 to be the bottom, can't do that.

    If I take 0 and 1 and put 0 on top, that works, but it is an exception when it is the bottom?
     
  5. Sep 10, 2011 #4

    micromass

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    Yes, you can never have 0 in the denominator. By definition of a rational number.
     
  6. Sep 10, 2011 #5
    But the whole close division thing i can take two integers (including 0)?
     
  7. Sep 10, 2011 #6

    micromass

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    No, an element of [itex]\mathbb{Q}[/itex] is defined as a fraction [itex]\frac{m}{n}[/itex], where n is nonzero. So you can't take 0 is the denominator, by definition.

    You can't divide by 0.
     
  8. Sep 10, 2011 #7

    SammyS

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    No. The integers are not closed under division for reasons other than the fact that 1/0 is undefined.

    Is 4/3 an integer?

    Perhaps you should review the definition of a 'ring'.
     
  9. Sep 10, 2011 #8
    Thank you, you just added more work for me...
     
  10. Sep 10, 2011 #9

    SammyS

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    Anytime! You're welcome.

    To help out a bit more:

    Whether we're discussing the Integers, the Rationals, the Reals, or Complex Numbers (all with the usual arithmetic operations):
    It's always true that 0 times any element is 0, 0 being the identity element for the addition operation. Because of this, there is no multiplicative inverse for 0, and thus division (the operation that is the inverse of multiplication) by 0 is undefined. Therefore, when discussing whether division is closed, we exclude the case of division by zero.​
     
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