# Schwarz Inequality and Irrational Numbers

1. Aug 17, 2004

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 ehat?

I would appreciate any responses.

Thanks

2. Aug 17, 2004

### Hurkyl

Staff Emeritus
1. There's a nifty theorem about the rational solutions of polynomials:

Given the polynomal $a_n x^n + \ldots + a_1 x + a_0$, with all of the a_i integers, all rational solutions are given by p/q where p divides a_0 and q divides a_n.

2. Do you recognize a1b1 + a2b2 as the dot product of <a1, a2> with <b1, b2>?

3. Aug 17, 2004

any ideas about the third problem?

4. Aug 18, 2004