I am trying to solve an exercise in which I must decide whether the following sentences are true:

  • $\forall b \; \exists a \; \forall x \; (x^2 + ax + b > 0)$

  • $\exists b \; \forall a \; \exists x \; (x^2 + ax + b = 0)$

  • $\exists a \; \forall b \; \exists x \; (x^2 + ax + b = 0)$

But I don't know how to parse them to check if they are indeed true. I guess I know how to parse two quantifiers easily but this is kinda confusing to me.

  • 1
    $\begingroup$ As for how to read them, try reading the qualifiers aloud in order. For example, the first one is of the form "For every person $b$, there exists some city $a$ such that for every shop $x$, they have bought something there". Specifically, the first statement says that if you pick any old $b$, you will be able to then find some $a$ (possibly depending on $b$) that makes the statement $x^2 + ax + b > 0$ true for all $x$. Can informally judge whether or not that statement is true? $\endgroup$
    – Brian
    Jan 26 '20 at 23:00

For the first one, it reads as follows:

For any $b$, there exist $a$ such that for any $x$, $$x^2 + ax + b > 0.$$

Perhaps by graphing this you can see this statement is most certainly false for $b \le 0$. However, to show the statement is false, we can show that its negation is true.

Negation: There exists $b$ such that for any $a$, there exist an $x$ such that $$x^2 + ax + b \le 0.$$

So, it suffices to show that the negation is true. Take $b=0$ and for any $a$ choose $x=0$. Then $x^2 + ax = x(x+a) = 0 \le 0$.

Hopefully you can see now how to do the other two.


I often find it helpful to mentally insert "such that" after $\exists$.

The first example you gave I would read as "for all $a$, there exists a $b$ such that for all $x$ ...."

Another way to think you this, is that I am giving you an $a$, and you have to find a $b$ such that if I then give you any $x$, the rest of the statement will hold.

Example 2 would read "there exists a $b$ such that for every $a$, there exists an $x$ such that..."

And example three would be the same, but with $a$ and $b$ swapped around.

Let me know if this helps!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.