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Math Help - Irrational Numbers

  1. #1
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    Irrational Numbers

    1.) Given that I is the set of irration numbers,

    a.) Show that if a, b are in the rational numbers, then ab and a + b are elemenets of the rationals as well.

    b.) Show that if a are in the rationals, and t is in the irrationals, then a + t is in the irrationals and at is in the irrationals as long as a does not equal 0.

    c.) Part a says that the rationals is closed under addition and multilication. Are the irrationals closed under addition/multiplication? Given 2 irrationals (s and t), what are we able to say about s + t and st?

    2.) Using the above, and using that the rationals are dense in the reals a - sqrt(2) and b - sqrt(2), prove that given any 2 real numbers where a < b, there exists an irrational number t that satisfies a < t < b.
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  2. #2
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    I will help with #2.
    Given a < b \Rightarrow \quad a\sqrt 2  < b\sqrt 2 .
    There is an rational number r such that, a\sqrt 2  < r < b\sqrt 2 .
    But a < \frac{r}{{\sqrt 2 }} < b and  \frac{r}{{\sqrt 2 }} is irrational.
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  3. #3
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    Quote Originally Posted by Plato View Post
    I will help with #2.
    Given a < b \Rightarrow \quad a\sqrt 2 < b\sqrt 2 .
    There is an rational number r such that, a\sqrt 2 < r < b\sqrt 2 .
    But a < \frac{r}{{\sqrt 2 }} < b and  \frac{r}{{\sqrt 2 }} is irrational.
    Thanks Plato. And I figured out #1 too.
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  4. #4
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    Quote Originally Posted by seerTneerGevoLI View Post
    1.) Given that I is the set of irration numbers,

    a.) Show that if a, b are in the rational numbers, then ab and a + b are elemenets of the rationals as well.
    a=p/q and b=n/m so their sum is (pm+nq)/qm which is a rational number.
    Similarly ab.
    b.) Show that if a are in the rationals, and t is in the irrationals, then a + t is in the irrationals and at is in the irrationals as long as a does not equal 0.
    If a+t is rational then (a+t)-a is rational so t is rational a contradiction.
    If at is rational and a!=0 then at/a=t is rational a contradiction.
    c.) Part a says that the rationals is closed under addition and multilication. Are the irrationals closed under addition/multiplication? Given 2 irrationals (s and t), what are we able to say about s + t and st?
    No the irrationals are not. Try sqrt(2) and sqrt(8).

    2.) Using the above, and using that the rationals are dense in the reals a - sqrt(2) and b - sqrt(2), prove that given any 2 real numbers where a < b, there exists an irrational number t that satisfies a < t < b.
    This should be easy now.
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