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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.

rattatttta
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    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 $b=-a$. – Git Gud Nov 03 '13 at 16:34

2 Answers2

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

amWhy
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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!

Peter Smith
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