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Math Help - Showing that a number is rational/irrational

  1. #1
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    Showing that a number is rational/irrational

    I have a bunch of questions that I don't know how to approach about whether a number is rational or irrational

    Could someone give me an idea of how to approach questions like the following:

    1 + srt(2) + srt(3/2)

    Is this rational or irrational (and why)?

    thanks
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    Re: Showing that a number is rational/irrational

    Quote Originally Posted by kinhew93 View Post
    I have a bunch of questions that I don't know how to approach about whether a number is rational or irrational

    Could someone give me an idea of how to approach questions like the following:

    1 + srt(2) + srt(3/2)

    Is this rational or irrational (and why)?
    Can you classify each of 1 ~,~ \sqrt 2 ~,\&~ \sqrt {\frac{3}{2}} as rational or irrational ?
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    Re: Showing that a number is rational/irrational

    Quote Originally Posted by Plato View Post
    Can you classify each of 1 ~,~ \sqrt 2 ~,\&~ \sqrt {\frac{3}{2}} as rational or irrational ?
    Well obviously 1 is rational, srt(2) and srt(3) are irrational, but I don't know about srt(3)/srt(2), or what happens when you sum rational and irrational numbers?
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    Re: Showing that a number is rational/irrational

    Quote Originally Posted by kinhew93 View Post
    Well obviously 1 is rational, srt(2) and srt(3) are irrational, but I don't know about srt(3)/srt(2), or what happens when you sum rational and irrational numbers?
    If \sqrt {\frac{3}{2}} were rational then what would mean ?

    Would that mean \sqrt 3  = r\sqrt 2 where r is rational ?
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    Re: Showing that a number is rational/irrational

    Quote Originally Posted by Plato View Post
    If \sqrt {\frac{3}{2}} were rational then what would mean ?

    Would that mean \sqrt 3  = r\sqrt 2 where r is rational ?
    Yes it would but I don't see the issue with that? I know that the product of a rational and irrational number is always an irrational number...
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    Re: Showing that a number is rational/irrational

    Quote Originally Posted by kinhew93 View Post
    Yes it would but I don't see the issue with that? I know that the product of a rational and irrational number is always an irrational number...
    Well, if \sqrt 3  = r\sqrt 2 where r is rational does that not mean that 3  = r\sqrt 6 where r is rational ?
    What is wrong with that?
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    Re: Showing that a number is rational/irrational

    Quote Originally Posted by Plato View Post
    Well, if \sqrt 3  = r\sqrt 2 where r is rational does that not mean that 3  = r\sqrt 6 where r is rational ?
    What is wrong with that?
    I really don't know. Is it that 3/srt(6) cannot be rational? If it is is still cannot see why
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    Re: Showing that a number is rational/irrational

    Quote Originally Posted by Plato View Post
    Well, if \sqrt 3  = r\sqrt 2 where r is rational does that not mean that 3  = r\sqrt 6 where r is rational ? What is wrong with that?
    Quote Originally Posted by kinhew93 View Post
    I really don't know. Is it that 3/srt(6) cannot be rational? If it is is still cannot see why
    You have a huge gape in understanding about rational numbers,
    If K is a non-square positive integer then \sqrt K is irrational. Can you prove that?

    So if  3  = r\sqrt 6 where r is rational what in wrong with that?
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    Re: Showing that a number is rational/irrational

    Quote Originally Posted by Plato View Post
    You have a huge gape in understanding about rational numbers,
    If K is a non-square positive integer then \sqrt K is irrational. Can you prove that?

    So if  3  = r\sqrt 6 where r is rational what in wrong with that?
    Ok gotcha. Ive actually done this question now but I still don't see what I can say about the sum of irrational numbers?

    What can be said about
    Srt(2) + srt(3) + srt(6)
    ?
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    Re: Showing that a number is rational/irrational

    Let's go back to your original question.

    If r = 1+ \sqrt{2} + \sqrt{\frac{3}{2}} is rational, surely r - 1 is rational.

    So let's focus on s = r - 1 instead. Again, if s is rational, surely s2 is rational as well.

    Now: s^2 = \left(\sqrt{2} + \sqrt{\frac{3}{2}}\right)^2 = 2 + 2\(\sqrt{2})\left(\sqrt{\frac{3}{2}}\right) + \frac{3}{2} = \frac{7}{2} + 2\sqrt{3}.

    If s2 is rational, then t = s2 - 7/2 must also be rational.

    Now t = 2\sqrt{3}, and if t is rational then t/2 = (1/2)t is also surely rational, which means that the square root of 3 is rational.

    But this is absurd.

    Working backwards, we see that t cannot be rational, which means that s2 cannot be rational, which means that s itself cannot be rational.

    But this means (finally!) that r cannot be rational, for assuming r rational leads to \sqrt{3} \in \mathbb{Q}.
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    Re: Showing that a number is rational/irrational

    The way I would do this would be to note that (x- 1- \sqrt{2}-\sqrt{3/2})(x-1+ \sqrt{2}+ \sqrt{3/2})= (x- 1)^2- (\sqrt{2}+ \sqrt{3/2})^2= (x- 1)^2- (2+ \sqrt{3}+ 3/2)= (x^2- 2x+ 1- 2- 3/2)- \sqrt{3}= (x^2- 2x- \frac{5}{2})- \sqrt{3}.

    Now multiply that by (x^2- 2x- \frac{5}{2})+ \sqrt{3} to get
    (x^2- 2x-\frac{5}{2})^3- 3= x^4- 4x^3- x^2+ 10x+ \frac{13}{4}
    Finally multiply by 4 to get 4x^5- 16x^3- 4x^2+ 40x+ 13
    a fourth degree polynomial with integer coefficients.

    Since that was created by multiplying x- \sqrt{2}- \sqrt{3/2} by other terms, x- \sqrt{2}- \sqrt{3/2} is a factor so \sqrt{2}- \sqrt{3/2} is a zero of the polynomial.

    Now, the "rational root theorem" says that if a polynomial with integer has rational zeros then they must be of the form \frac{m}{n} where the denominator, n, evenly divides the leading coefficient, here "4", and the numerator, m, evenly divides the constant term, here 13.

    The only integers that evenly divide 4 are 1, -1, 2, -2, 4, and -4. The only integers that evenly divide 13 are 1, -1, 13, and -13. Thus, the only rational number that could possibly satisfy the equation are 1, -1, 13, -13, \frac{1}{2}, -\frac{1}{2}, \frac{13}{2}, \frac{1}{4}, -\frac{1}{4}, \frac{13}{4}, and -\frac{13}{4}. But putting those into the polynomial show that, in fact, none of them are actually zeros. Therefore no zeros of that polynomial are rational. Since \sqrt{2}+ \sqrt{3}{2} is a root, it cannot be rational.
    Last edited by HallsofIvy; February 2nd 2014 at 06:08 PM.
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