Examples II.

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

Proof #1.

Proof by contradiction.

Let, if possible, , and p and q do not have any common factor and are integers. Then, if , we have

. So, p contains a factor of 2. So, let . So, q contains a factor of 2. Hence, both p and q have a common factor have a common factor 2, contradictory to out assumption. Hence, the proof.

2) Prove generally that a rational function in its lowest terms cannot be the cube of a rational number unless p and q are both perfect cubes.

Proof #2.

Let, if possible, where m,n,p,q are integers, with n and q non-zero and p and q are in lowest terms. This implies that m and n have no common factor. Hence, .

3) A more general proposition, which is due to Gauss and includes those which precede as particular cases, is the following: an algebraical equation

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

Proof #3.

For suppose that, the equation has a root , where a and b are integers without a common factor, and b is positive. Writing

for z, and multiplying by , we obtain

,

a function in its lowest terms equal to an integer, which is absurd. Thus, , and the root is a. It is evident that a must be a divisor of .

4) Show that if and neither of

and

is zero, then the equation cannot have a rational root.

Proof #4. *Please try this and send me a solution.. I do not have a solution yet 🙂*

5) Find the rational roots, if any, of .

Solution #5.

Use problem #3.

The roots can only be integral, and so find the roots by trial and error. It is clear that we can in this way determine the rational roots of any such equation.

More later,

Nalin Pithwa

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