Stavros
Reputation
370
Next privilege 500 Rep.
Access review queues
 Aug6 comment Set Theory Laws We say that $A$ is a subset of $B$ if every element of $A$ is also an element of $B$. The proper subset is a different thing. If $A$ is a subset of $B$, but $A$ is not equal to $B$, then $A$ is a proper subset of $B$. To prove $(A \cup B \cup C) - (A \cup B) \subseteq (C - A) - B$, let $x \in (A \cup B \cup C) - (A \cup B)$ be arbitrary and then prove $x \in (C - A) - B$. Aug6 comment Set Theory Laws I've only proved that $(C - A) - B \subseteq (A \cup B \cup C) - (A \cup B)$ and left the proof of $(A \cup B \cup C) - (A \cup B) \subseteq (C - A) - B$ to you. To prove $(C - A) - B = (A \cup B \cup C) - (A \cup B)$ you must prove both statements. Aug6 comment Set Theory Laws What are $F$ and $G$? Elaborate on this. Jul30 comment prove that : $\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c}+ \frac{c^2+a^2}{c+a} \geq 3$ Check this answer. Jul27 comment If the same message is sent to Alice and Bob who are using different public keys, how can somoene following the channel find $m$ Are you sure that the only given is $m < N_1$? Beware of the fact that if $m > N_2$, you won't get the exact $m$ back when you decrypt $c_2$. Jul26 comment If the same message is sent to Alice and Bob who are using different public keys, how can somoene following the channel find $m$ It's just an application of CRT. We know that $N_1$ and $N_2$ are relative prime. Then, for the given sequence of integers $c_1, c_2$, there exists an integer $x$ solving the system of the following simultaneous congruences: $x \equiv c_1\pmod {N_1}$ and $x \equiv c_2\pmod {N_2}$. Sep1 comment Prove that \$D_r \circ D_s = \{ (x,y) \in \mathbb{R}^2 \ | \ |x-y|