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$
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.