Can i say that $(\forall x \in \mathbb{N}:x^2=x) \vee(\forall x \in \mathbb{N}:x>1)$
is the same statement as:$\forall x \in \mathbb{N}:(x^2=x)\vee(x>1)$ ?
If not, why?
Thanks.
Can i say that $(\forall x \in \mathbb{N}:x^2=x) \vee(\forall x \in \mathbb{N}:x>1)$
is the same statement as:$\forall x \in \mathbb{N}:(x^2=x)\vee(x>1)$ ?
If not, why?
Thanks.
On
The first is false, the second is true, regardless of whether $0 \in \Bbb N$.
$(\forall x \in \mathbb{N}:x^2=x) \vee(\forall x \in \mathbb{N}:x>1)$ is a distjunction of two universally qualified sentences. The first is false for $x=2$, the second is false for $x=1$, so both are false. it is a bit clearer is you write the equivalent $(\forall x \in \mathbb{N}:x^2=x) \vee(\forall y \in \mathbb{N}:y>1)$ where it is easier to see that $x$ need not equal $y$
In $\forall x \in \mathbb{N}:(x^2=x)\vee(x>1)$ you only get to choose $x$ once and for all $x$ one of the two disjuncts is true.
The first sentences says 'every natural number is equal to its square or every natural number is greater than one'. This is false since both sentences on the sides of the 'or' are false.
The other sentence is true. It says 'Every natural number is equal to its square or is greater than one'. Since the only natural number that is equal to its square is 1, and every other natural number is greater than 1, this statement holds for every natural number.