Chapter 1: Real Variables: examples II

Examples II.

1) Show that no rational number can have its cube equal to 2.

Proof 1.

Proof by contradiction. Let x=p/q. q \neq 0, p, q \in Z. (p, q have no common factors).

Let x^{3}=2. Hence, \frac{p^{3}}{q^{3}}=2. Hence, p^{3}=2q^{3}. Hence, p^{3} is even because we know that even times even is even and even times odd i also even and odd times odd is odd. Hence, p ought to be even. Let p=2m. Then, again q^{3}=4m^{3}. Hence, q^{3} is even. Hence, q is even. But, this means that p and q have a common factor 2 which contradicits our hypothesis. Hence, the proof.. QED.

Proof 2)

Let given rational fraction be \frac{p}{q}, q \neq 0, p, q \in Z.

Let \frac{p}{q}=\frac{m^{3}}{n^{3}}, n \neq 0, m,n \in Z.

Since p and q do not have any common factors, m and n also do not have any common factors.

Case I: p is even, q is odd so clearly, they do not have any common factors.

Case IIL p is odd, q is odd but with no common factors.

Case I: since m and n are without any common factors, and m^{3}, n^{3} are also in its lowest terms, we have p=m^{3}, q=n^{3}.

Case II: similar to case I above.

Proof 3.

A more general proposition, due to Gauss, includes those two above problems as special cases. Consider the following algebraic equation;

x^{n}+p_{1}x^{n-1}+p_{2}x^{n-2}+\ldots +p_{n}=0.

with integral coefficients,, cannot have a rational root but non integral root.

Proof 3:

For suppose that the equation has a root a/b, where a and b are integers without a common factor, and b is positive. Writing a/b for x, and multiply both the sides of the equation b^{n-1}, e obtain

-\frac{a^{n}}{b}=p_{1}a^{n-1}+p_{2}a^{n-2}b+\ldots +p_{n}b^{n-1},

a fraction in the lowest terms equal to an integer, which is absurd. thus, b=1, and the root is a. It is clear that a must be a divisor of p_{n}

Proof 4.

Show that if p_{n}=1 and neither of

1+p_{1}+p_{2}+p_{3}+\ldots,, 1-p_{1}+p_{2}-p_{3}+\ldots is zero, then the equation cannot have a rational root.

I will put the proof later.

Problem 5.

Find the rational toots, if any of x^{4}-4x^{3}-8x^{2}+13x+10=0.’

Solution.

The roots can only be integral and so \pm 1, \pm 2, \pm 3, \pm 5 pm 10 are the only possibilities: whether these are roots can be determined by tiral. it is clear that can in this way determine the rational roots of any equation.

More later,

Nalin Pithwa

 

 

 

 

 

 

 

 

 

 

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s