My question is that when I negate the statement $$(\forall x\in \mathbb{R})( \exists n \in \mathbb{N})(x < 1/n),$$ do I negate all of the statement or just the first part $(\forall x \in \mathbb{R})$?
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1"$(\forall x\in R)$" is not a complete formula and therefore cannot be negated alone. It only becomes a complete formula when you attach a formula to the right of it. – hmakholm left over Monica Jul 06 '15 at 15:38
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¬(∀x∈ℝ)(∃n∈ℕ)(x<1/n) in this case I negate the whole thing ? – Arjun Dhiman Jul 06 '15 at 15:39
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@ZainPatel I think removing the parens makes the formula even harder to read and less clear (also takes away from what seems to be bothering OP, namely how a negation is to be applied in the midst of parens). – Daniel W. Farlow Jul 06 '15 at 16:43
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@DanielW.Farlow Fair enough, I submit to your experience. I personally think that he expression without the brackets looks more elegant, but I agree with your point about how it might confuse the OP. :-) – Zain Patel Jul 06 '15 at 16:44
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1@ZainPatel I've definitely seen both styles so I won't say that one is "objectively better" than the other, but I realized shortly after I wrote my answer that the parentheses themselves seem to be at the root of OP's confusion. If not, then the notation really doesn't matter I don't think. :) – Daniel W. Farlow Jul 06 '15 at 16:46
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If I understand correctly, you want to negate the statement $$\forall x\in \mathbb{R} ~ \exists n \in \mathbb{N} : x < 1/n$$ To do this, you swap every quantifier and then negate the actual formula, i.e. you get $$\exists x\in \mathbb{R} ~ \forall n \in \mathbb{N} : x \geq 1/n$$
j4GGy
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∀x∈ℝ)¬(∃n∈ℕ)(x<1/n) does the same apply for this one too? so i would get ∀x∈ℝ)(∀n∈ℕ)(x>=1/n) – Arjun Dhiman Jul 06 '15 at 15:41
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1@Zero Looking back over your original question and your first comment on this answer, it seems like part of the problem (and maybe the problem) is understanding that the negation $\neg$ may be thought of as being sort of pushed through whatever formula you are dealing with, changing all of the quantifiers along the way and finally concluding by negating the actual statement. Does that make sense? – Daniel W. Farlow Jul 06 '15 at 16:41
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@Zero I notice you have been a user for quite some time with $10$ questions but zero accepted answers given--I'd recommend accepting an answer, whichever one best meets your needs (or modifying your question to seek further clarification, or make additional comments, etc.). This 1) acknowledges the answerer and thanks the person in a way for their help and 2) lets other users in the MSE community know that your question is effectively "closed" (i.e., you are satisfied with the answer(s) you have received). I'd recommend doing this for your other questions as well ($10$ isn't so bad). – Daniel W. Farlow Jul 06 '15 at 16:50
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You need to negate every block, which gives $$ (\exists x \in \mathbb{R})(\forall n \in \mathbb{N})(x \geqslant 1/n) $$
J.-E. Pin
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It appears you have still not quite "gotten there" just yet--perhaps the following will make the other answers easier to understand and will help as a reference in the future. (More info here if desired)
To negate a statement of the form $$ Q_1x_1 Q_2x_2 \ldots Q_nx_n\; P(x_1,x_2,\ldots,x_n), $$ where $Q_i$ is $\forall$ or $\exists$ for $1 \leq i \leq n$, we do the following:
(i) Change every $\forall$ to $\exists$ and every $\exists$ into $\forall$.
(ii) Replace $P$ by its negation.
In your case, consider the positive "linguistic translation" of what you are dealing with symbolically (of which you are trying to find the negation):
For every real number $x$ there exists a natural number $n$ such that $x<1/n$.
Expressing the above in its more formal symbolic form: $$ (\forall x\in\mathbb{R})(\exists n\in\mathbb{N})(x<1/n).\tag{1} $$ How would you negate $(1)$ in light of rules (i) and (ii) specified at the beginning of this answer? First, change every $\forall$ to $\exists$ and every $\exists$ into $\forall$:
- $(\forall x\in\mathbb{R})$ becomes $(\exists x\in\mathbb{R})$
- $(\exists n\in\mathbb{N})$ becomes $(\forall n\in\mathbb{N})$
Then replace $P$ by its negation (in this case, we have $P : x<1/n$): $$ \neg P \equiv x\geq1/n. $$ Thus, we can finally see that $$ \neg[(\forall x\in\mathbb{R})(\exists n\in\mathbb{N})(x<1/n)]\equiv(\exists x\in\mathbb{R})(\forall n\in\mathbb{N})(x\geq1/n), $$ confirming the two other answers you have already received.
Daniel W. Farlow
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