Finding the negation of a statement 
What is the negation of the following statement:
  There is $u \in \mathbb R$ such that for all $v \in \mathbb R$ there exists $w \in \mathbb R$ such that $u + v < w.$

I know the statement begins with "For all." But i'm not sure
Also, is the contrapositive of "If $m + n$ is odd then $m$ is odd or $n$ is even" "If $m, n$ are even, then $m + n$ is even."
 A: (A note on notation:  "$\forall$" = "for all" and "$\exists$" = "there exists".)
The negation of $\forall x, P(x)$ is 
$$  \lnot \forall x, P(x) = \exists x, \lnot P(x)  \text{.}  $$
As an example in words: "it is not the case that all $x$ are people" is the same as "there exists some $x$ such that $x$ is not a person".
The negation of $\exists x, P(x)$ is
$$  \lnot \exists x, P(x) = \forall x, \lnot P(x)  \text{.}  $$
Example: "there does not exist an $x$ such that $x$ is a person" is the same as "for all $x$, it is not the case that $x$ is a person".
To summarize, the negation of a negated quantified statement can be pushed in towards the predicate by reversing the sense of each quantifier that you pass through.
$$  \lnot \exists u, \forall v, \exists w, P(u,v,w) = \forall u, \exists v, \forall w, \lnot P(u,v,w)  \text{.}  $$
The contrapositive of "$a \implies b$" is "$\lnot b \implies \lnot a$".  So the contrapositive of "if $m+n$ is odd then $m$ is odd or $n$ is even" is 


*

*"if not ( $m$ is odd or $n$ is even ) then not( $m+n$ is odd )"

*= "if neither $m$ is odd nor $n$ is even then not( $m+n$ is odd )"

*= "if neither $m$ is odd nor $n$ is even then $m+n$ is not odd"

*(and if you know that not odd is even) "if neither $m$ is odd nor $n$ is even then $m+n$ is even".

A: There are two basic rules for negating quantified statements:

(NOT (for all $x \in A$, $P(x)$) is equivalent to (there exists $x \in A$ such that (NOT $P(x)$)). 

More collquially, to disprove a "for all" statement, it is equivalent to prove that a counterexample exists.

(NOT (there exists $x \in A$ such that $P(x)$) is equivalent to (for all $x \in A$ (NOT $P(x)$)). 

More colloquially, to disprove an existence statement, it is equivalent to prove that the opposite is always true.
These are applied from the outside in, so when negating a statement with nested quantifiers it is particularly important that you parenthesize the statement first. Your statement should first be parenthesized as follows:


*

*There is $u \in \mathbb R$ such that (for all $v \in \mathbb R$ (there exists $w \in \mathbb R$ such that ($u+v<w$))).


It's negation is:


*

*NOT (There is $u \in \mathbb R$ such that (for all $v \in \mathbb R$ (there exists $w \in \mathbb R$ such that ($u+v<w$))))


And now you simply apply the negation rule one parenthesis at a time. The first step is


*

*For all $u \in \mathbb R$, NOT (for all $v \in \mathbb R$ (there exists $w \in \mathbb R$ such that ($u+v<w$))).


I'm sure you can finish this now: it's just an automatic, step-by-step process, one parenthesis at a time.
A: It should be
$$\forall u ∈ R \quad \exists v ∈ R\quad \forall w ∈ R \quad u + v \ge w$$
and for 
$$m+n =2k+1 \implies m=2h+1\quad \lor \quad n=2i$$
the contrapositive is
$$ m\neq2h+1\quad \land \quad n\neq2i \implies m+n \neq2k $$
A: I find it the easiest to write it down rigorously.
$$
(\exists u \in \mathbb{R})(\forall v \in \mathbb{R}) (\exists w \in \mathbb{R}) u+v<w 
$$
Then do the negation part by part using the fact that $\neg \exists \iff \forall$. So your expression would be
$$
(\forall u\in \mathbb{R}) (\exists v \in \mathbb{R}) (\forall w \in \mathbb{R}) u+v\geq w
$$
You can learn these elementary logical expressions easily if you don't know them, and the negation happens bit by bit. You can figure out that negation of forall is there exists because for the claim to be false one counterexample is enough, meaning there would need to be at least one. 
