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I have to check which of these propositions are false, i tried to solve this but i was able to find 3 alternatives resulting in false.

I don't know which symbols to use in these 2 situations:

  • For any
  • For some

Well, from that:

Which one is the only false, or is there more than one false and my solution is right?

$\exists (x) \in \mathbb{Z}$ | $ x+4 = -4$

$\forall(x) \in \mathbb{N}$ | $ x>15$

For any X $\in \mathbb{R}$ | $ x < -5$

$\exists x \in \mathbb{R} $ | $ 3x + 5 = 15$

For some $x \in \mathbb{N} $ | $ x^2 + 15 = 0$

PlayMa256
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    What have you tried? Have you, say, determined which of these are true and which are false? – lulu Dec 15 '17 at 14:03
  • You can find there:https://math.stackexchange.com/questions/402020/whats-the-difference-between-any-all-and-some – 1ENİGMA1 Dec 15 '17 at 14:09
  • @lulu yeah, i did that. I was able to determine which ones are false, but as i said, i found 3 falses, but my professors is asking for only one. So i assumed that my answers are wrong. – PlayMa256 Dec 15 '17 at 14:12
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    Maybe the proposition "my professor is asking for only one..." is false? – Paul Dec 15 '17 at 14:16
  • well, could you share your work with us? Which do you consider false? – lulu Dec 15 '17 at 14:19

1 Answers1

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You are in fact correct, three of these statements are false.

The first, $\exists x \in \mathbb{Z}[x+4=-4]$ is true as it suffices to simply choose $x=-8$ to prove existence.

The second, $\forall x \in \mathbb{N}[x>15]$ is false as $x=1$ is certainly less than $15$, so the statement does not hold true for all naturals, the same reasoning can be applied to the third one as $\forall x \in \mathbb{R}[x<-5]$ as again there exist real numbers such as $1$ which are greater than $-5$.

The fourth $\exists x \in \mathbb{R}[3x+5=15]$ is true as such a number does indeed exist and it is $x=\frac{10}{3}$

Finally, the last statement is false as it is equivalent to $\exists x \in \mathbb{N}[x=\sqrt{-15}]$ which is certainly absurd as you cannot square a natural and get any sort of negative number.

Michael Connor
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