1) For every shape A, there is a circle D, such that D surrounds A
2) There is a circle C, such that for every line l, l intersects C
This is what I got are my answers correct
1) There is at least one shape of A that no circle D surroundings. 2) There is at least one line I which does not intersect this circle C.
On
Your negation of (1) is fine, though you may want to end the statements:
(2) There is a circle $C$, such that for every line $\mathcal{l},\;\,\mathcal{ l}$ intersects $C$
Let the domain of concern be all geometric objects (lines, shapes, points, etc.).
Let $C(x)$ denote "x is a circle"
Let $L(y)$ denote "y is a line"
Let $I(x, y)$ denote "x intersects y".
Then we have, essentially, "there exists an x such that for all y, [(x is a circle and y is a line) and y intersects x", which we can write:
$$\exists x \forall y [(C(x) \land L(y) \land I(y, x)]\tag{1}$$
Now we negate $(1)$: $$\lnot\exists x \forall y [(C(x) \land L(y) \land I(y, x)]\tag{2}$$
$$ \equiv \forall x \lnot\forall y[(C(x) \land L(y) \land I(y, x)]\tag{3}$$ $$\equiv \forall x \exists y \lnot[(C(x) \land L(y) \land I(y, x)]\tag{4}$$ $$\equiv \forall x \exists y \left[\lnot[C(x)\land L(y)] \lor \lnot I(y,x)\right]\tag{5}$$ $$\equiv \forall x \exists y \left[(C(x) \land L(y)) \rightarrow \lnot I(y, x)\right]\tag{6}$$
That is, we end with a statement such as the following: For every x there exists a y such that (x is a circle and y is a line) and y does not intersect x:
Less formally: For every circle $C$, there exits some line $L$ such that $L$ does not intersect $C$.
The negation of 1) is correct. The negation of 2) is: For every circle C, there exists a line L which does not intersect C.
One might take this occasion to mention that there is an automatic procedure to negate logical sentences. Let me give an example. Let $f:\mathbb R\to\mathbb R$ denote a function. The assertion that $f$ is continuous can be written as: $$ \forall\varepsilon\in\mathbb R_+^*,\ \forall x\in\mathbb R,\ \exists\delta\in\mathbb R_+^*,\ \forall y\in\mathbb R,\ |x-y|\leqslant\delta\implies|f(x)-f(y)|\leqslant\varepsilon. $$ How to negate this? One can work sequentially, starting from the leftmost end of the formula, and making its logical structure apparent. The original formula is: $$ \color{red}{\forall}\varepsilon\in\mathbb R_+^*,\ \color{red}{\forall} x\in\mathbb R,\ \color{red}{\exists}\delta\in\mathbb R_+^*,\ \color{red}{\forall} y\in\mathbb R,\ P(x,y,\delta,\varepsilon), $$ for some logical assertion $P(x,y,\delta,\varepsilon)$. Thus its negation is: $$ \color{red}{\exists}\varepsilon\in\mathbb R_+^*,\ \color{red}{\exists} x\in\mathbb R,\ \color{red}{\forall}\delta\in\mathbb R_+^*,\ \color{red}{\exists} y\in\mathbb R,\ \color{red}{\lnot} P(x,y,\delta,\varepsilon), $$ That is, starting from the left, one replaces each $\forall$ by $\exists$, and vice versa. Finally, one should negate $P(x,y,\delta,\varepsilon)$, but perhaps you already know how to do this?