For all integers, b,c,d, if x is a rational number such that $x^2+bx+c=d$, than x is an integer.
a) express above statment in the form, $Q_1 b,c,d\in U_1 ( Q_2 x\in U_2(p(x)\rightarrow Q(x)))$ , specifying the universes, $U_1, U_2$, and the quantifiers, $Q_1, Q_2$, and the statement $p(x), q(x)$.
b)Prove the statement by contradiction. Use the fact that if a rational number $x$ is not an integer than there are integers $m,n$ such that $x=m/n$ with $n>1$.
I answered for both and got 1.5 point out of 6 points.. Please help!
My answer for a) was that
U_1: All Integer
U_2: All Rational #
Q_2: there is an $x$ such that p(x)
p(x): x is rational
q(x): rational #
I don't know what I've done wrong. For b),
Since I have to prove it by contradiction, I know I should start with 'not p' than 'not $p\Rightarrow\ not q$, and we derive a contradiction, and we find out 'not p' is false, thus p is true.
Not p(x) would be $x$ is irrational, than how should I apply this 'irrational number' to the equation, $x^2+bx+c=d$?
Please help, any hint would be appreciated.