2
$\begingroup$

a,b $\in$ $\mathbb{R}$

Original statement: $\exists a$ such that $\forall b$, $a+b>0$

My negation: $\forall a$, $\exists b$ such that $a+b \leq 0$

Is my negation correct? If it is, is the negation true whereas the original statement is false?

I drew this conclusion because my interpretation of the original statement was that there was one $a$ that would satisfy the inequality regardless of what value of $b$ was chosen (obviously not true). My understanding of the negation statement was that you could choose different values of $b$ to satisfy the inequality based on the value of $a$ you are dealing with (so $b$ is not fixed unlike $a$ in the original statement). Or am I wrong and in fact $b$ is fixed like $a$ was before? If I am wrong then I am stuck on trying figure out which statement is true. Any help is appreciated.

$\endgroup$
1
  • 3
    $\begingroup$ You negated correctly and the negation is indeed true, making the starting statement false. For an informal proof of the negation, let $a$ be an arbitrary real number. We wish to prove that $\exists b\in \Bbb R(a+b\leq 0)$, so just let $b=-a$. $\endgroup$
    – Git Gud
    Nov 3 '13 at 16:34
1
$\begingroup$

Your negation is correct, and your understanding as to why the original statement is false, and the negated sentence true is "spot on!"

$\endgroup$
1
$\begingroup$

A basic principle worth remembering is this, in headline terms

When you push a negation sign past a quantifier, the quantifier "flips" into its dual.

So $\neg\forall x \varphi \Leftrightarrow \exists x\neg \varphi$, and $\neg\exists x \varphi \Leftrightarrow \forall x\neg \varphi$. [Before reading on make you understand why that has to be right!] And moreover, you can apply this equivalence inside a wff. [Why?]

Applied to this case, the negation of

$\exists a \forall b\, a+b > 0$

is, of course

$\neg\exists a \forall b\, a+b > 0$

Which applying the principle is equivalent to

$\forall a \neg\forall b\, a+b > 0$

which is equivalent to

$\forall a \exists b\neg\, a+b > 0$

which is equivalent to

$\forall a \exists b \, a+b \leq 0$.

As you rightly said!

$\endgroup$

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.