Negation of a statement about polynomials

Question

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).

Answer 1

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).

Answer 2

Let $A$ denote some assertion that applies to real numbers. Here, it is assumed that for each $x \in \Bbb{R}$, the assertion $A$ with respect to the value $x$ is either true or false.

For any specific value $x \in \Bbb{R}$, let the statement $[A::x]$ denote that the assertion $A$ is (supposedly) true, with respect to the real number $x$.

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Consider the following two statements:

• Statement S-1: For all $x\in \Bbb{R}, [A::x]$.
• Statement S-2: There exists at least one $x \in \Bbb{R}$ such that $[A::x]$.

Now consider how to negate those statements.

For illustrative purposes, let $[A::x]$ denote the specific assertion that $x$ is positive.

Then S-1 is asserting that all real numbers are positive, while S-2 is asserting that there exists at least one real number $x$ such that $x$ is positive.

To negate S-1, it is sufficient to provide a counter-example. One counter-example is $x = -1$, because $(-1)$ is not positive.

Note
In negating S-1, it is not necessary to show that S-1 is false for every real number $(x)$. Instead, it is sufficient to show that there exists at least one real number $(x)$ such that S-1 is false, with respect to $(x)$.

However, the counter-example of $x = -1$ does not negate S-2, because S-2 is not asserting that the assertion $[A::x]$ is true for all real numbers. Instead, S-2 is only asserting that there exists at least one real number $x$ such that the $[A::x]$ is true.

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Now consider the question that you posted.

(i) For real numbers $a$, if $f(a) = 0$ then $a$ is positive (i.e. $a > 0$).

The excerpted assertion is making an assertion that is supposed to apply for all real numbers $(a)$. Therefore, the excerpted assertion analogizes to a statement similar to S-1 above, rather than being similar to S-2 above.

By similar I mean that statement S-1 represents an assertion that is also supposed to apply to all real numbers, while statement S-2 represents an assertion that is only supposed to be true for at least one real number.

Therefore, the way to negate the excerpted assertion is to show that there exists at least one real number $(a)$ such that the assertion is false, for that particular value of $(a)$.

Now consider the two options that you mentioned:

(ii) For real numbers $a$, if $f(a) = 0$ then $a$ is negative.

(viii) For some non-positive real number $a$, $f(a) = 0$.

Option (viii) is the one that focuses on only finding at least one counter-example. Therefore option (viii) represents the negation of (i).

Answer 3

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. $a > 0$).

(ii) For real numbers $a$, if $f(a) = 0$ then $a$ is negative.

(viii) For some non-positive real number $a$, $f(a) = 0$.

Now in my head that means that the negation of statement (i) should be (ii).

But According to the book, $[x^3-x = x(x+1)(x-1)$ where for a real number $0$, you can have $f(0) = 0]$ is 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).

Yes, indeed, (i) and (ii) are both false regarding the given counterexample $x^3-x = x(x+1)(x-1);$ the point of this counterexample is that since (i) and (ii) have the same truth value here, by definition of negation, they cannot possibly be negations of each other.

In any case, do note that the complement of positive is non-positive, rather than negative.

According to the book, the correct answer, is (viii).

This is correct, and here's an analogous example:

(I) For each cat $j,$ if $j$ barks, then $j$ is not a lion. (vacuously TRUE)

(II) For each cat $j,$ if $j$ barks, then $j$ is a lion. (vacuously TRUE)

(IV) For some cat $j$ that is a lion, $j$ barks. (FALSE)

In particular:

1. the negation of the implication $$\text{if }P \text{ then }Q$$ is $$P \text{ yet not } Q.$$
2. the negation of $$\text{for each } x, \:P(x)$$ is $$\text{for some } x, \:\big(\text{not }P(x)\big).$$