Negating statements help I need to negate (move the negation inside) or write the contrapositive statements of the following statements:


*

*Write the contrapositive of: If everyone is here, then someone will leave.

*Write the negation of: If Alice and Bob go, then Carol or Dave will come.

*Write the negation of Every integer is even or odd, but no integer is even and odd.

*rewrite as if then statement, then write its contrapositive: Every integer bigger than $1$ is divisible by some prime.

*rewrite as if then statement, then negate statement:  Every Integer that is divisible by $2$ and $3$ is divisible by $6$.
This is what I’ve done:


*

*If someone didn’t leave, then not everyone was here. 

*Alice and Bob went; however, neither Carol or Dave came along.

*There is an integer that is neither even or odd, or there is an integer that is even or odd.

*If $x$ is an integer bigger than $1$, then it is divisble by some prime. Negation: 
$x$ is an integer bigger than $1$, however $x$ is not divisible by any prime. 

*If $x$ is an integer divisible by $2$ and $3$ then $x$ is also divisible by $6$. 
contrapositive: If $x$ is an integer not divisible by $6$, then $x$ is not divisble by $2$ and $3$.
These are not entirely correct, so I was hoping you guys would tell me what's wrong with them and how to fix them.
 A: Comment
Regarding 

3) Write the negation of "Every integer is even or odd, but no integer is even and odd", 

we have that it is a conjunction ("but"). 
Thus, its negation will be a disjunction : "Either (there is an integer that is neither even nor odd), or (...)". Up to now, Ok.
But the second disjunct must be the negation of "no integer is even and odd".
In formula, this sentence is $\lnot \ \exists n ( \text E (n) \land \text O(n))$.
If we negate it, we have only to remove the leading negation sign : $\exists n \ ( \text E (n) \land \text O(n))$.
In conclusion, the correct negation of 3) will be :

"Either there is an integer that is neither even nor odd, or there is an integer that is both even and odd". 

A: You can always use the following rules. $A$ and $B$ will be statements in the following.
To negate something an easy way to start is to write "it is not true that..." in the beginning of the statement and then try to rearrange the sentence to a more sensible one.
If you would like to negate something like "$A$ and $B$" you will get "not $A$ or not $B$".
If you would like to negate "$A$ or $B$" you will get "not $A$ and $B$". 
If you would like to negate something of the form "if $A$ is true then $B$ is true" you will get "not $B$ then not $A$".
Negating "every" could be thought of as "there exist at least one which is not..."
With this in mind lets look at your question, 
1) looks ok to me
2) let $A,B$ stand for "Alice goes",  "Bob goes" respectively and $C,D$ stand for "Carol came", "Dave came" respectively then we have to negate
$$
A \land B\Rightarrow C\land D
$$
$$
\neg (A \land B\Rightarrow C\land D)
$$
so
$$
\neg (C \land D)\Rightarrow \neg(A\land B)
$$
which means
$$
\neg C \lor \neg D \Rightarrow \neg A \lor \neg B.
$$
That is "if neither Carol or Dave come then neither of Alice or Bob goes". (with some eventual tweaks about the past tense)
3) look at Mauros answer
4) I would say that the negation looks like "if $x$ is not divisible by any prime then $x$ is not an integer greater than $1$"
5) looks fine to me
I hope I could help
A: 


*

*Write the contrapositive of: If everyone is here, then someone will leave.



*

*If someone didn’t leave, then not everyone was here.


"Did not" is not the complement for "will".  "Was not" is not the complement for "is".  Don't change the tenses.




*Write the negation of: If Alice and Bob go, then Carol or Dave will come.




*Alice and Bob went; however, neither Carol or Dave came along.


Don't tamper with the verbs; just leave them as they are, or negate them if required.   "Go" either remains "go" or negates to "did not go". So on.




*Write the negation of Every integer is even or odd, but no integer is even and odd.




*There is an integer that is neither even or odd, or there is an integer that is even or odd.


The negation of "all are" is "some are not", but the negation of "not some are" is just "some are".
Also the negation of "are even or odd" is "are not-even and not-odd", which you may say as "are odd and even".




*rewrite as if then statement, then write its contrapositive: Every integer bigger than $1$ is divisible by some prime.




*If $x$ is an integer bigger than $1$, then it is divisble by some prime.

Negation: $x$ is an integer bigger than $1$, however $x$ is not divisible by any prime.

Where did $x$ come from?  Don't introduce terms not used in the original statement.
Start with "If any integer is blah blah blah, then it is rhubarb rhubarb," then negate that.
Hint: That is still a universal quantified statement.




*rewrite as if then statement, then negate statement:  Every Integer that is divisible by $2$ and $3$ is divisible by $6$.




*If $x$ is an integer divisible by $2$ and $3$ then $x$ is also divisible by $6$.

contrapositive: If $x$ is an integer not divisible by $6$, then $x$ is not divisble by $2$ and $3$.

Same. Try again
A: You need to break the statements into smaller pieces. I will give you an example with the first statement but you will have to do the rest by yourself:
If everyone is here, then someone will leave.
There are two statements here: "$A = \text{ everyone is here }$" and "$B = \text{ someone will leave }$". Then the statement you're given is "If $A$ then $B$" which means that whenever $A$ happens $B$ must happen. Therefore the negation would be that $A$ happens and $B$ does not happen. Formally the negation is "$A$ and not $B$".
Negating $B$ is easy $\text{not } B = \text{nobody will leave}$. Therefore $A$ and not $B$ means:
$\text{everyone is here and nobody will leave.}$

Note that I'm not a native English speaker. 
