List members of the set: $\{x: (x \in\mathbb{Z})\land(x^2<8) \}$ I have this problem:
$$\{x: (x \in\mathbb{Z})\land(x^2<8) \}$$
Originally I thought the solution was $\{-4,-1,0,1,4\}$ but $-2^2$ is $-4$ not $4$ so the answer would be infinite. (my original assumption was that a negative number squared was positive because $-2 \cdot -2=4$, but I've since found out this is not the case).
This is a new subject in my discrete maths exam that our lecturer threw at us today. I've never worked with set notation before. The exam is tomorrow.
Also... just to confirm I'm actually working out the answers correctly...
$$\{x: (x \in\mathbb{N})\land(3x<10) \}$$
is
$$\{0,3,6,9\}?$$
 A: You are looking for all integers $x$ such that $x^2<8.$ This is the set consisting of $-2,-1,0,1,$ and $2.$
A: *

*$$\{x: (x \in\mathbb{Z})\land(x^2<8) \}= \{ -2, -1, 0, 1,2\}$$

*Squaring of negative number does give you positive number.
$$-2^2=-4$$
but we have $$(-2)^2=4$$
If $x=-2$, $x^2$ means $(-2)^2$.


*

*Note that $4^2=16 \ge 8$, hence $4 \not \in \{x: (x \in\mathbb{Z})\land(x^2<8) \}$

*Note that $3x < 10$ is equivalent to $x < \frac{10}3$.

*Depending on your definition of $\mathbb{N}$,
If you consider $0 \in \mathbb{N}$
$$\{x: (x \in\mathbb{N})\land(3x<10) \}=\{0,1,2,3\}$$ 
if you consider $0 \not \in \mathbb{N}$
$$\{x: (x \in\mathbb{N})\land(3x<10) \}=\{1,2,3\}$$ 
A: Your original thought process is correct. 
The only thing to look out for is order of operations: You're right that the square of a negative number is a positive number. After all, $(-2)^2 = (-2)\times (-2) = 4$.  But if you type $-2^2$ into your calculator, then due to order of operations, it will be interpreted as $-(2^2)=-4$, where the squaring unexpectedly happens before the minus sign is applied.

If set notation is confusing, you can put it into words:

$$\{x : (x \in \mathbb{Z}) \wedge (x^2 < 8)\}$$
means "Find every possible $x$, where $x$ is a positive-or-negative integer and $x^2 < 8$". The answer is $\{-2,-1,0,+1,+2\}$.
Similarly,
$$\{x : (x \in \mathbb{N})\wedge 3x < 10\}$$
means "Find every possible $x$, where $x$ is a whole number, and $3x < 10$." Your process looks exactly right: $\{0,1,2,3\}$. 

Just make sure in each case that you're listing the values of $x$ itself,  rather than listing $x^2$ or $3x$. To check your answer, like when you say the answer is $\{0,3,6,9\}$, take each of those values and plug them in to the equation $3x < 10$. For example, you can plug in $x=6$ and you find that $3x = 3(6) = 18 \not < 10$. So $x=6$ can't be an answer.
A: 
but $-2^2$ is $-4$ and not $4$

And $-2^2$ is not $(-2)^2$.
$-2^2$ is not a square number at all.
If $x = -2$ then $x^2 = (-2)^2 \ne -2^2$.
$(-2)^2 = 4$.  There is NO possible number so that $anything^2 = -2^2$.  That can NEVER happen.
(Err... that can never happen if we are working with real numbers.  If we are working with complex or imaginary numbers than... well, then we are taking a different class....)

(my original assumption was that a negative number squared was positive because −2∗−2=4, but I've since found out this is not the case).

I sincerely hope you haven't!  Go back!  You were right the first time.
$(-2)^2 = -2*-2 = 4$
$x^2 \ge 0$. 
And $(-3)^2 = (-3)*(-3) = 9 > 8$.  
So no, it is not infinite as all $x < -3$ are such that $x^2 > 9$.
.....

Originally I thought the solution was {−4,−1,0,1,4} 

The set is a set of the $x$s.  Not of the $x^2$s.
So since $(-2)^2 = 4 < 8; (-1)^2 = 1 < 8; 0^2 = 0 < 8; 1^2 = 1 < 8; 2^2 = 4 < 8$ whereas $x \ge 3 \implies x^2 \ge 9 > 8$ and $x \le -3 \implies x^2 \ge 9 > 8$ it follows that acceptable values for $x$ are $-2,-1,0,1,2$ and all others are not acceptable.
.....
$\{x: (x \in\mathbb{Z})\land(x^2<8)\}$
means
All $x$ so that 1) $x$ is an integer and 2) $x^2 < 8$
It should be clear that that is $\{-2,-1,0,1,2\}$.
To be formal and pedantic $x^2 < 8 \iff -\sqrt 8 < x < \sqrt 8$ so $-3 < \sqrt 8 < x < \sqrt 8 < 3$ and as $x\in \mathbb Z$ then the set is $\mathbb Z \cap (-\sqrt 8, \sqrt 8)= \{-2,-1,0, 1,2\}$.
