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1) $(\exists x \in R)[(x^2 = (x+1)^2 ∧ (x^3 \in Z))]$

ATTEMPT : $((∀ x \in R)[(x^2 \not= (x+1)^2 ∧ (x^3 \notin Z))])$

2)$(∀x \in R)(x>0) ⇒ (\exists n \in N)(n . x >1)$ Note: the (n.x) is multiplication.

ATTEMPT: $(\exists x \in R)(x>0)∧(∀n \in N)(n . x <=1)$

Would be grateful if anyone could look my answers over.

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Shouldn't the $\wedge$ in the first line change to something else in the second line? –  Jay Dec 13 '13 at 21:46
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1 Answer 1

For the first, you need DeMorgan's to get $$(∀ x \in R)[(x^2 \not= (x+1)^2 \lor (x^3 \notin Z))]$$

For the second, we have an implication of the form $p \rightarrow q\equiv \lnot p \lor q$.

To negate it gives us, again, with DeMorgan's, $\lnot(\lnot p \lor q) \equiv p \land \lnot q$. So your quantifier in p should be $\forall$:

$$(\forall x \in R)\Big((x>0)∧(∀n \in N)(nx \leq 1)\Big)$$

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+1 thanks, I knew that, just forgot to change it to or when I copy/pasted –  FA1om Dec 13 '13 at 21:18
    
You're welcome! Note also the first quantifier in the second problem. (Unless that's what you forgot to change.) –  amWhy Dec 13 '13 at 21:38
    
Great answer though eh, this is exactly a question, I am studying for from a past exam, this guy is probably getting ready for the same exam tomorrow :) –  Mac Dec 13 '13 at 21:43
    
@amWhy I can't accept it, but I can upvote for sure :)... So if I have two sets A,B. ∀x(x∈A) → x∉B, would I get A^c ⊆ B, or would I negate ¬P and get A ⊆ B^c. Kind of confused when you say ¬p∨q –  Mac Dec 13 '13 at 21:53
    
Thanks, @Mac! ;-) –  amWhy Dec 13 '13 at 21:53
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