State whether the following statements are true or false and justify your answers
Emily Wong 
(a) ∃x ∈ Z ∃y ∈ Z (x + y = 2) ∧ (x − y = 3)
(b) ∀x ∈ R ∀y ∈ R x > y −→ x^2 < y^2
(c) ∀x ∈ R ∃y ∈ R x^2 + y^2 > 1
my answers: A. False, let’s assume x=1 and y=1 1+1=2 (correct) for some x and y values 1-1≠2 (wrong)
B. False, let’s assume x=2 and y=1 2>1 (correct) for all x values 2^2<1^2 (wrong)
C. True, let’s assume x=2 and y=1 2^2+1^2>1 (correct) for all x values and y values
are they correct and well justified?
$\endgroup$ 21 Answer
$\begingroup$Let's see these statements one by one:
a) If we operate in the second condition we get $x=y+3$; which, substituting in the first one, gives us: $(3+y)+y=2$. Or what is the same $y=-1/2$, so we can not choose $y$ to be an integer.
Just as a remark, pay attention to the fact that in this case, a counterexample would not be valid as the statement doesn't talk about a property that is true for "every" integer but for a particular one ("exist $x$ and $y$ such that...").
b) Just take $x$ and $y$ bigger or equal than $1$, e.g. $x=2$ and $y=1$ would do the trick.
c) Take $x\in\mathbb{R}$. If $x>1$ or $x<-1$ then $x^2>1$, and we can take $y=0\in\mathbb{R}$. If $-1\leq x\leq 1$, in particular $x^2<2$, so $2-x^2>0$. We can then take $y=\sqrt{2-x^2}\in\mathbb{R}$ and get $y^2+x^2=2>1$.
We see that in this case, the statement is true, unlike the other ones.
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