How to write negation of statements? How to write negation of following statements in words?
1. Any integer is either positive or negative.
2. There is a child who is loved by everyone.
3. The connector is loose or the machine is unplugged.
4. No politician cheat voters.

And how to write negation of following statement in symbols?
5. (∀x)(∃y)(x^2 <y)

For 3rd one my answer is "The connector is not loose and the machine is not unplugged" and for 4th one my answer is "politician cheat voters". I am not sure...Are they correct?
 A: Let me give this a go. The first one is trickiest because of the "either-or" construction.


*

*There is an integer that is both positive and negative, or neither positive nor negative.

*(a) There is no child who is loved by everyone. (b) For each child, there is someone who does not love the child.

*The connector is not loose and the machine is not unplugged. (You already said it.)

*There is a politician who cheats voters.

*$(\exists x)(\forall y)(x^2 \geq y)$


Indeed, it is a rule that $(\neg \forall x)(\phi) = (\exists x)(\neg \phi)$ where $\phi$ is a proposition. This should be intuitively clear: if $\phi$ holds for not all $x$, then there must be an $x$ such that $\phi$ does not hold. It is a good exercise to write your original statements in formal symbols and then negate them. For example:


*

*$(\forall x \in \mathbb{Z})((x > 0 \wedge x \nless 0) \vee (x < 0 \wedge x \ngtr 0))$


This seems a bit silly, but your either-or construction forces me to write it like this. If the original statement were
"Any integer is positive or negative",
then I could have written
$(\forall x \in \mathbb{Z})(x > 0 \vee x <0),$
which is equivalent in this case because being positive and being negative are already mutually exclusive. The negation of 1. in formal symbols is
\begin{eqnarray*} \neg(\forall x \in \mathbb{Z})((x > 0 \wedge x \nless 0) \vee (x < 0 \wedge x \ngtr 0)) & = & \\
(\neg \forall x \in \mathbb{Z})((x > 0 \wedge x \nless 0) \vee (x < 0 \wedge x \ngtr 0)) & = & \\
(\exists x \in \mathbb{Z})\neg((x > 0 \wedge x \nless 0) \vee (x < 0 \wedge x \ngtr 0)) & = & \\
(\exists x \in \mathbb{Z})(\neg(x > 0 \wedge x \nless 0) \wedge \neg(x < 0 \wedge x \ngtr 0)) & = & \\
(\exists x \in \mathbb{Z})((\neg(x > 0) \vee \neg(x \nless 0)) \wedge (\neg(x < 0) \vee \neg(x \ngtr 0))) & = & \\
(\exists x \in \mathbb{Z})(x \ngtr 0 \vee x < 0) \wedge (x \nless 0 \vee x > 0)
\end{eqnarray*}
With one more step you can that this is the negation of statement 1. in words. Use the rules of passage.
A: *

*It is not the case that any integer is either positive or negative.

*It is not the case that there is a child who is loved by everyone.

*It is not the case that the connector is loose or the machine is unplugged.

*It is not the case that no politicians cheat voters.

*$\lnot$[(∀x)(∃y)(x^2 < y)]

A: To negate a collection of statements simultaneously, turn  each statement into symbolical form and then negate the whole on the basis of de Morgan's law. That is a very easy way to negate a collection of statements simultaneously.
A: Write the negation of each statement
Some telephone can take photographs
All houses have two stories
Some cars are hybrids
All golf course are green
Some drivers are not safe
