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Schwarz Inequality and Irrational Numbers

  1. Aug 17, 2004 #1

    Hello everyone. I have 2 questions.

    1. Prove that the cube root (3) + sqrt (2) is irrational.

    My Solution

    Assume l is an irrational number of the form p/q where p and q are integers not equal to 0. Then

    p^6 / q^6 = [(cube root(3) + sqrt (2))]^6

    I concluded that it must be in the form x^6 +a1x^5 + ... + an(sub n) = 0, where a1..an are integers. I do not know how to prove whether x is irrational or integral. I tried using a simpler case like sqrt (2), but it did not work.

    2. State the geometrical interpretation of Schwarz Inequality for n = 2, 3.

    I know the interpretation is for any angle between two straight lines the cosine is less than the absolute value of 1. But how do you get that?

    I know Schwarz Inequality is:

    (a1b1 + a2b2 +... +anbn)^2 <= (a1^2+a2^2+...+an^2) (b1^2+b2^2+...+bn^2).

    So for n = 2, we have (a1b1 +a2b2)^2 <= (a1^2+a2^2)(b1^2+b2^2). However how do we get the interpretation mentioned above?

    3. Show that the equality sign in Schwarz Inequality holds if and only if the a's and b's are proportional; cav +dbv = 0.

    My Solution

    We know that ax^2 + 2bx + c = a(x+ b/a)^2 + ac-b^2/ a
    If b^2 - ac = 0 we can set above equation equal to 0. Then what?

    I would appreciate any responses.

  2. jcsd
  3. Aug 18, 2004 #2
    Can the sum of two irrational numbers be rational? :shy:

    If a and b are two irrational numbers, they can be represented in general by

    [tex]a = p_{1} + \sqrt{q_{1}}[/tex]
    [tex]b = p_{2} + \sqrt{q_{2}}[/tex]

    their sum is given by

    a + b = (p_{1} + p_{2}) + (\sqrt{q_{1}} + \sqrt{q_{2}})

    Think about this. I am not sure, but you can probably use this fact directly or indirectly somewhere in your proof.

    For 2, the Cauchy Schwarz inequality comes from a similar relationship for n-tuples. The geometric interpretation is easy to explain only for two or at most three dimensional spaces, for it is intuitive. For n > 3, you have to resort to an abstract sounding analogy involving n-tuples.

    I think I did 3 a few months ago. The method I read was to consider a = kb for some real constant k and proceed further.
    Last edited: Aug 18, 2004
  4. Aug 18, 2004 #3


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    That's not true. For instance, take pi or the cube root of 2.
  5. Aug 19, 2004 #4
    Okay Hurkyl, I'm sorry I got it wrong. But pi really is [tex]\frac{22}{7}[/tex] (and not exactly 3.1415926543, because this way the number of digits after the decimal is infinite). I believe pi is irrational due to a different reason though. (I think there are more fractional representations of pi...I just forget what they are right now, as I always used 22/7 in class). I actually didn't read about the cube root of 2 in your posts...forgot about it :cry:
  6. Aug 19, 2004 #5
    Hey I just looked up my math text. It says: "to prove the Cauchy Schwarz inequality for n > 3 we must define the cosine of the angle in n dimensions." Mathematically (and not logically), you could say that if the n-tuples

    [tex](a_{1}, a_{2},....,a_{n})[/tex] and
    [tex](b_{1}, b_{2},....,b_{n})[/tex]

    represent distinct points in n-dimensional space. The position vectors are, respectively

    [tex]\vec a = (a_{1}, a_{2},....,a_{n})[/tex] and
    [tex]\vec b = (b_{1}, b_{2},....,b_{n})[/tex]

    If we could define the cosine by

    [tex]\cos \theta = \frac{\vec a \bullet \vec b}{\|\vec a\| \|\vec b\|}[/tex]

    then the constraint [tex]0\leq \cos \theta \leq 1[/tex] gives the Cauchy Schwarz Inequality. The definition of the cosine for n > 3 is rather hard to visualize though since we are accustomed to at most 3 dimensions.
  7. Aug 19, 2004 #6


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    Um, pi is not 22/7; it is approximately 22/7.
  8. Aug 19, 2004 #7
    Consider pi and 4-pi . :smile:
  9. Aug 20, 2004 #8
    Yeah I got that part just after I wrote the first post (scroll below).

    There is however, another interesting proof of the Cauchy Schwarz Inequality, which I read very recently. Here goes the crux of it and perhaps you can think it through and fill in the details:

    The Cauchy Schwarz inequality states that

    [tex]a_{1}b_{1} + a_{2}b_{2} + ..... + a_{n}b_{n} \leq \sqrt{a_{1}^2 + a_{2}^2 + .... + a_{n}^2}\sqrt{b_{1}^2 + b_{2}^2 + .... + b_{n}^2}[/tex]

    Proving it is equivalent to proving the inequality [tex]C \leq \sqrt{A}\sqrt{B}[/tex] where

    [tex]A = a_{1}^2 + a_{2}^2 + .... + a_{n}^2 = \sum_{i = 1}^{n}a_{i}^2[/tex]
    [tex]B = b_{1}^2 + b_{2}^2 + .... + b_{n}^2 = \sum_{i = 1}^{n}b_{i}^2[/tex]
    [tex]C = a_{1}b_{1} + a_{2}b_{2} + ..... + a_{n}b_{n} = \sum_{i = 1}^{n}a_{i}b_{i}[/tex]

    Consider the function

    [tex]f(t) = \sum_{i = 1}^{n}(a_{i}t + b_{i})^{2}[/tex]

    Expanding it gives you

    [tex]f(t) = At^{2} + Ct + B[/tex]

    This is a quadratic expression. And I think this hint is sufficient to arouse an interest to push the proof further :-D.

  10. Aug 23, 2004 #9
    I'm sorry about a typographical error in my last post. The correct expression is

    [tex]f(t) = At^{2} + 2Ct + B[/tex]

    and not [tex]f(t) = At^{2} + Ct + B[/tex] as mentioned.

    Next assume that A>0 (for A = 0, the equality holds trivially), which is equivalent to saying at at least one of the ai's is nonzero. Now, f(t) is always positive so its discriminant must be less than or equal to zero. This proves the Cauchy Schwarz Inequality. To prove the equality strongly, consider the case when f(t) = 0. That will give you two possibilities, one of which is a contradiction since A > 0. End of proof. QED (not exactly quite easily done).

    The second thing I read was that since f(t) > 0 for all real t, its minimum is also positive. That gives the inequality too.

    Hope that helps.

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