Schwarz Inequality and Irrational Numbers

In summary, the conversation discusses the proof of the irrationality of the sum of two irrational numbers, the geometrical interpretation of Schwarz Inequality for n = 2, 3, and the conditions for the equality sign in Schwarz Inequality. The conversation also mentions the use of the Cauchy Schwarz Inequality in the proof, as well as different approaches to proving the inequality.
  • #1
courtrigrad
1,236
2
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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.

Thanks
 
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  • #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

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

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:
  • #3
maverick280857 said:
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]

That's not true. For instance, take pi or the cube root of 2.
 
  • #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:
 
  • #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.
 
  • #6
Um, pi is not 22/7; it is approximately 22/7.
 
  • #7
maverick280857 said:
Can the sum of two irrational numbers be rational? :shy:
...

Yes!
Consider pi and 4-pi . :smile:
 
  • #8
Rogerio said:
Yes!
Consider pi and 4-pi . :smile:

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.

Cheers
Vivek
 
  • #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.

Cheers
Vivek
 

1. What is the Schwarz Inequality?

The Schwarz Inequality, also known as the Cauchy-Schwarz Inequality, is a mathematical inequality that relates the dot product of two vectors to their magnitudes. It states that the absolute value of the dot product of two vectors is less than or equal to the product of their magnitudes. This inequality is commonly used in geometry, linear algebra, and functional analysis.

2. How is the Schwarz Inequality related to irrational numbers?

The Schwarz Inequality can be extended to include irrational numbers by considering them as infinite sequences of rational numbers. This allows for the inequality to hold for any two real numbers, whether they are rational or irrational.

3. What is the significance of the Schwarz Inequality in mathematics?

The Schwarz Inequality is a fundamental concept in mathematics and has many applications in various fields, including physics, engineering, and statistics. It is also used in the proof of other important theorems, such as the Cauchy-Schwarz Master Theorem and the Triangle Inequality.

4. Can the Schwarz Inequality be generalized to higher dimensions?

Yes, the Schwarz Inequality can be extended to higher dimensions. In fact, it can be extended to any number of vectors in an inner product space. This generalization is known as the generalized Schwarz Inequality or the Hölder's Inequality.

5. How is the Schwarz Inequality related to the concept of orthogonality?

The Schwarz Inequality is closely related to the concept of orthogonality, which refers to two vectors being perpendicular to each other. In fact, the Schwarz Inequality can be used to prove that two vectors are orthogonal if and only if their dot product is equal to zero. This makes the inequality a useful tool for determining orthogonality in various mathematical applications.

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