# Negating implications

• Mar 23rd 2009, 08:50 PM
jusstjoe
Negating implications
Hello all,

1)If a>2 and b<-5, then |a|+ |b| >=7

So, I think this can be rewritten as A= a>2 and b<-5; B=|a|+ |b| >=7
and thus, a>2 and b<-5 and |a| + |b|<7

Though, I'm also confused cause it might be:

If a<2 and b>5 OR |a| + |b| >=7.....so as you guys can tell i'm throughly confused.

2) If the numbers a and b are primes, then the number a+b is composite.

Sort of having the same problem as above.

• Mar 23rd 2009, 09:32 PM
n0083
Quote:

Originally Posted by jusstjoe
Hello all,

1)If a>2 and b<-5, then |a|+ |b| >=7

Couple of things to lay the ground work.

• $\displaystyle A \implies B \iff \neg A \lor B$
• $\displaystyle \neg (A \land B) \iff \neg A \lor \neg B$
• $\displaystyle \neg (A \lor B) \iff \neg A \land \neg B$
• $\displaystyle \neg (\neg A) \iff A$

Your statement is of the form

$\displaystyle A \land B \implies C$

You want to find,

$\displaystyle \neg (A \land B \implies C)$

Apply the rules above.

Quote:

2) If the numbers a and b are primes, then the number a+b is composite.
Suppose $\displaystyle a \neq b \neq 2$, then a and b are odd.

For some $\displaystyle n,m \in \mathbb{N}$ then,

$\displaystyle a = 2n-1$

$\displaystyle b = 2m-1$

This implies,
$\displaystyle a+b = 2(n+m) - 2 = 2 (n+m-1)$

Therefore a+b is even and not prime.

Now, if a or b is 2 then...this is not true. Consider the counter example with
$\displaystyle a=2$
$\displaystyle b=3$
then $\displaystyle a+b=5$ is prime (not composite)