Examples II.

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

Proof 1.

Proof by contradiction. Let . (p, q have no common factors).

Let . Hence, . Hence, . 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 . Then, again . Hence, 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 , .

Let , .

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 are also in its lowest terms, we have .

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;

.

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 , e obtain

,

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

Proof 4.

Show that if and neither of

,, 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 .’

Solution.

The roots can only be integral and so 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