Answer
Claim: For all odd integers a, b, and c, if z is a solution of ax² + bx + c = 0 then z is irrational.
Negation: There exists odd integers a, b, and c such that z is a solution of ax² + bx + c = 0 and z is rational.
Proof: Suppose that there exists odd integers a, b, and c such that z is a solution of ax² + bx + c = 0 and z is rational.
By definition of rational, z = p/q and q ≠ 0 (by the zero sum property).
Assume that p and q have no common factor (since if a and b do have a common factor, we can divide a and b
by their greatest common factor to obtain a’ and b’, redefine a=a’ and b=b’, and continue this process
until a and b have no common factor).
By substitution, a(p/q)² + b(p/q) + c.
By laws of algebra, q² * [a(p/q)² + b(p/q) + c] = q² * [ap²/q² + bp/q + c] = ap² + bpq + cq² = 0.
(1) Assuming that p is odd and q is even, it follows that a(odd)² + b(odd)(even) + c(even)².
By substituting odd values for a, b, and c, it follows that:
a(odd)² + b(odd)(even) + c(even)²
= [odd(odd)²] + [odd(odd)(even)] + [odd(even)²]
= odd + even + even
= odd
Therefore, ax² + bx + c both equals 0 and is odd, which is a contradiction because 0 is even.
Hence, p is not odd and q is not even.
(2) Assuming that p is even and q is odd, it follows that a(even)² + b(even)(odd) + c(odd)².
By substituting odd values for a, b, and c, it follows that:
a(even)² + b(even)(odd) + c(odd)²
= [odd(even)²] + [odd(even)(odd)] + [odd(odd)²]
= even + even + odd
= odd
Therefore, ax² + bx + c both equals 0 and is odd, which is a contradiction because 0 is even.
Hence, p is not even and q is not odd.
(3) Assuming that p and q is odd, it follows that a(odd)² + b(odd)(odd) + c(odd)².
By substituting odd values for a, b, and c, it follows that:
a(odd)² + b(odd)(odd) + c(odd)²
= [odd(odd)²] + [odd(odd)(odd)] + [odd(odd)²]
= odd + odd + odd
= odd
Therefore, ax² + bx + c both equals 0 and is odd, which is a contradiction because 0 is even.
Hence, p and q are not odd.
(4) Assuming that p and q are even, it follows that:
a(even)² + b(even)(even) + c(even)²
= odd(even)² + odd(even)(even) + odd(even)²
= even + even + even
= even
Therefore, ax² + bx + c both equals 0 and is even.
Hence, p and q must be even.
Since p and q are even, then they have a common factor (since all even numbers are divisible by 2).
However, this contradicts the supposition that p and q have no common factors.
Therefore, the supposition is false and z is irrational.
Work Step by Step
Claim: For all odd integers a, b, and c, if z is a solution of ax² + bx + c = 0 then z is irrational.
Negation: There exists odd integers a, b, and c such that z is a solution of ax² + bx + c = 0 and z is rational.
Proof: Suppose that there exists odd integers a, b, and c such that z is a solution of ax² + bx + c = 0 and z is rational.
By definition of rational, z = p/q and q ≠ 0 (by the zero sum property).
Assume that p and q have no common factor (since if a and b do have a common factor, we can divide a and b
by their greatest common factor to obtain a’ and b’, redefine a=a’ and b=b’, and continue this process
until a and b have no common factor).
By substitution, a(p/q)² + b(p/q) + c.
By laws of algebra, q² * [a(p/q)² + b(p/q) + c] = q² * [ap²/q² + bp/q + c] = ap² + bpq + cq² = 0.
(1) Assuming that p is odd and q is even, it follows that a(odd)² + b(odd)(even) + c(even)².
By substituting odd values for a, b, and c, it follows that:
a(odd)² + b(odd)(even) + c(even)²
= [odd(odd)²] + [odd(odd)(even)] + [odd(even)²]
= odd + even + even
= odd
Therefore, ax² + bx + c both equals 0 and is odd, which is a contradiction because 0 is even.
Hence, p is not odd and q is not even.
(2) Assuming that p is even and q is odd, it follows that a(even)² + b(even)(odd) + c(odd)².
By substituting odd values for a, b, and c, it follows that:
a(even)² + b(even)(odd) + c(odd)²
= [odd(even)²] + [odd(even)(odd)] + [odd(odd)²]
= even + even + odd
= odd
Therefore, ax² + bx + c both equals 0 and is odd, which is a contradiction because 0 is even.
Hence, p is not even and q is not odd.
(3) Assuming that p and q is odd, it follows that a(odd)² + b(odd)(odd) + c(odd)².
By substituting odd values for a, b, and c, it follows that:
a(odd)² + b(odd)(odd) + c(odd)²
= [odd(odd)²] + [odd(odd)(odd)] + [odd(odd)²]
= odd + odd + odd
= odd
Therefore, ax² + bx + c both equals 0 and is odd, which is a contradiction because 0 is even.
Hence, p and q are not odd.
(4) Assuming that p and q are even, it follows that:
a(even)² + b(even)(even) + c(even)²
= odd(even)² + odd(even)(even) + odd(even)²
= even + even + even
= even
Therefore, ax² + bx + c both equals 0 and is even.
Hence, p and q must be even.
Since p and q are even, then they have a common factor (since all even numbers are divisible by 2).
However, this contradicts the supposition that p and q have no common factors.
Therefore, the supposition is false and z is irrational.
