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