Question

If a, b, c are odd integers, then the equation ax2 + bx + c = 0 has no solution in the set of rational numbers.

How to prove this?

Answer 1

To prove that the equation ax² + bx + c = 0 has no solution in the set of rational numbers when a, b, and c are odd integers, we can use proof by contradiction. By assuming the equation has a solution in the set of rational numbers and then showing that it leads to a contradiction, we can determine that the equation has no rational solutions. Therefore, the equation lacks a solution in the set of rational numbers when a, b, and c are odd integers.

Step-by-step explanation:

In order to prove that the equation ax² + bx + c = 0 has no solution in the set of rational numbers when a, b, and c are odd integers, we can use proof by contradiction.

1. Assume that the equation has a solution in the set of rational numbers. This means that the roots can be expressed as x = p/q, where p and q are integers.
2. Substituting x = p/q into the equation, we get ap²/q² + bp/q + c = 0.
3. Since a, b, and c are odd integers, ap²/q² + bp/q + c is also an odd integer.
4. However, in order for an odd integer to equal zero, its numerator must be zero, which means ap² + bpq + cq² = 0.
5. This implies that ap² + bpq + cq² is an even integer, which contradicts our assumption that it is odd.
6. Therefore, our initial assumption that the equation has a solution in the set of rational numbers is false.
7. Thus, the equation ax² + bx + c = 0 has no solution in the set of rational numbers when a, b, and c are odd integers.