Going through a book here. Stumbled on something confusing.
Goes like this:
Consider the following statement about a polynomial $f(x)$ with real coefficients, such as $x^2+3$ or $x^3-x^2-x$.
(i) For real numbers $a$, if $f(a) = 0$ then $a$ is positive (i.e. $a > 0$).
What is the negation of this statement? The following possibilities spring to mind form the everyday use of 'negation'.
(ii) For real numbers $a$, if $f(a) = 0$ then $a$ is negative.
…
(viii) For some non-positive real number $a$, $f(a) = 0$.
I have omitted a bunch of options because they are irrelevant for this question. Negation of a statement, is explained, to be false when the original statement is true, and is true when the original statement is false.
Now in my head that means that the negation of statement (i) should be (ii). But that is not what the book says is the correct answer. According to the book, the correct answer, is (viii). That is because you can find an example: $x^3-x = x(x+1)(x-1)$ where for a real number $0$, you can have $f(0) = 0$ (as a counter example to picking (ii) as the negation to (i)). But I mean their original statement (i) has this same issue, this example also applies to the original statement (i).
Best Answer
The original statement is of the form "for all $a$, if $p(a)$ then $q(a)$." So it's saying that all elements $a$ (here they are real numbers) satisfy some property (the conditional statement above).
First we have to negate the "for all" part. The negation of the statement "all things satisfy property P" is the statement "at least one thing does not satisfy property $P$." In our case, property $P$ is a conditional so we need to know how to negate conditionals.
The conditional "if $p$, then $q$" is false whenever the hypothesis $p$ is true but the conclusion $q$ is false and it's true in all other cases. So the negation of the conditional should be true when $p$ is true but $q$ is false and should be false in all other cases. This is the statement "$p$ and not $q$."
Putting this all together, we see that the negation of "for all $a$, if $p(a)$ then $q(a)$" is "there is at least one $a$ such that $p(a)$ is true is but $q(a)$ is false."
In your specific example, $a$ is a number, $p(a)$ is the statement $f(a) = 0$, and $q(a)$ is the statement $a > 0$. So the negation is there is at least one real number $a$ such that $f(a) = 0$ but it's not true that $a > 0$ (so $a$ is non-positive).