# My last 2 what should be easy problems are crushing my soul

• Jan 29th 2009, 09:34 AM
thecleric
My last 2 what should be easy problems are crushing my soul
Let x, y, z, be real numbers and let n be an integer. Negate this statement:

For all of x, For all of y [(x > 0 and y > 0) --> there exists n(nx > y)]

AND...

Use the properties of logical equivalence to show that:
~((~p and Q) or (~p and ~q)) or (p and q) logically equivalent p

maybe its because i have been up all night but these are killing me.
• Jan 29th 2009, 10:02 AM
Plato
Quote:

Originally Posted by thecleric
Let x, y, z, be real numbers and let n be an integer. Negate this statement:

For all of x, For all of y [(x > 0 and y > 0) --> there exists n(nx > y)]

Try to read this negation: $\left( {\exists x} \right)\left( {\exists y} \right)\left[ {\left( {x > 0 \wedge y > 0} \right) \wedge \left( {\forall n} \right)\left[ {nx \leqslant y} \right]} \right]$
• Jan 29th 2009, 10:04 AM
clic-clac
Your formula $F$ is $F:=\ \forall x\forall y((x>0\wedge y>0)\Rightarrow \exists n(nx>y))$

$\neg F\equiv \exists x\exists y ((x>0\wedge y>0)\wedge \forall n(nx\leq y))$

which can be written $\exists x\exists y (x>0\wedge y>0\wedge \forall n(nx\leq y))$

$\neg((\neg p\wedge q)\vee(\neg p\wedge\neg q))\vee(p\wedge q)$
$\equiv (\neg(\neg p\wedge q)\wedge\neg(\neg p\wedge\neg q))\vee(p\wedge q)$
$\equiv ((p\vee\neg q)\wedge(p\vee q))\vee(p\wedge q)$
$\equiv ((p\wedge(\neg q\vee q))\vee(p\wedge q)$
$\equiv p\vee(p\wedge q)$
$\equiv p$
• Jan 29th 2009, 10:32 AM
thecleric
ok i didn't know how to do the symbols but here it is:

$
\forall x\forall y [(x>0\wedge y>0) \rightarrow \exists n(nx > y)]
$

where $\rightarrow$ is the symbol for implies

i have to find and simplify the negation of this statement
• Jan 29th 2009, 12:13 PM
Plato
Did you see my reply above?