Need help with LaTex code. I am going to copy and paste it because it will not let me attach the tex so can someone help. Just recopy it without the errors as an answer or attack compilable copy! Thank you so much.

\documentclass[12pt]{article} \usepackage{amsmath, amssymb}

\begin{document}

\thispagestyle{empty}

{\bf \centerline{Writing Assignment 1}}

\vskip 0.5cm

{\bf 1.} For any set A, B, and C, prove that: \vskip 0.5cm

$(A\cap (B\cup C)=(A\cap B) \cup (A\cap C)$ \vskip 0.5cm

a) Prove: $(A\cap(B\cap C)\subset(A\cap B)\cup C)$

$(\forall x\in A\cap(B\cap C) so x\in A and x\in (B\cup C))$ \vskip 0.5cm

x is always belongs to set A and x is also a set to B or C. \vskip 0.5cm

$(x\in A and (x\in B)\bigvee (x\in C))$

Case 1: $(x\in B)$ \vskip 0.5cm

x is always belongs in A and in this case in B. So therefore, it is also in either A and B or the set is in C. Since x is in B in case one, it does not need to be in . \vskip 0.5cm

$(x\in(A\cap B)\cup C)$ \vskip 0.5cm

Case 2: $(x\in C)$ \vskip 0.5cm

$(x\in C)$ in case two, so the statement still holds true. \vskip 0.5cm

$(x\in(A\cap B)\cup C)$ \vskip 0.5cm

Therefore, $(A\cap(B\cup C)\subset (A\cap B)\cup C)$ \vskip 0.5cm

b) Prove: $((A\cap B)\cup(A\cap C)\subset A\cap (B\cup C))$ \vskip 0.5cm

$(\forall x\in(A\cap B)\cup(A\cap C) so x\in A and x\in (A\cap B)\bigvee(A\cap C))$ \vskip 0.5cm

case 1: $(x\in B)$ \vskip 0.5cm

since x always belongs to A, then $(A\cap(B\cup C))$ is true because it is in B and only needs to belong to B or C to be true. \vskip 0.5cm

$(x\in A\cap(B\cup C))$ \vskip 0.5cm

case 2: $(x\in C)$ \vskip 0.5cm

x always belongs to A so $(x\in A\cap(B\cupC))$ is true because x only needs to also belong in B and C and in case two, it belongs in C. \vskip 0.5cm

Therefore, $((A\cap B)\cup(A\cap C)\subset A\cap(B\cup C))$ and $(A\cap(B\cup C)\subset (A\cap B)\cup(A\cap C))$ so it follows $(A\cap(B\cup C)=(A\cap B)\cup(A\cap C))$ \vskip 0.5cm

{\bf 2.} using principle of mathematical induction, prove: $(1\cdot 1!+2\cdot 2!+3\cdot 3!+ \dots + n\cdot n!=(n+1)!-1$$ for all $n \in \mathbb{N})$ \vskip 0.5cm

a) base of induction: P(1)-T \vskip 0.5cm

Let $(n=1)$ $(1\cdot 1! = (1+1)!-1)$ $(1\cdot 1! = 1$ and $(1+1)!-1=1)$ $(1=1)$ so P(1) is True \vskip 0.5cm

b) Induction Step: $(P(n)\Longrightarrow P(n+1))$ \vskip 0.5cm

Assume: P(m) = $(1\cdot 1! + 2\cdot 2! + \dots + m\cdots m! = (m+1)!-1)$ \vskip 0.5cm

Want: $((m+1)!-1+(m+1)\cdot (m+1)!=(m+2)!-1)$ \vskip 0.5cm

$(1\cdot 1! + 2\cdot 2!+\dots+m\cdot m!+ (m+1)\cdot (m+1)! = (m+1)!-1+(m+1)\cdot (m+1)!)$ \vskip 0.5cm

$(=(m+1)!\cdot (1+m+1)-1)$ $(=(m+1)!\cdot (m+2)-1)$ $((m+2)!-1)$

\vskip 0.5cm

a) base of induction: P(1)-T \vskip 0.5cm

Let $(n=1)$ $(1\cdot 1! = (1+1)!-1)$ $(1\cdot 1! = 1$ and $(1+1)!-1=1)$ $(1=1)$ so P(1) is True \vskip 0.5cm

b) Induction Step: $(P(n)\Longrightarrow P(n+1))$ \vskip 0.5cm

Assume: P(m) = $(1\cdot 1! + 2\cdot 2! + \dots + m\cdots m! = (m+1)!-1)$ \vskip 0.5cm

Want: $((m+1)!-1+(m+1)\cdot (m+1)!=(m+2)!-1)$ \vskip 0.5cm

$(1\cdot 1! + 2\cdot 2!+\dots+m\cdot m!+ (m+1)\cdot (m+1)! = (m+1)!-1+(m+1)\cdot (m+1)!)$ \vskip 0.5cm

$(=(m+1)!\cdot (1+m+1)-1)$ $(=(m+1)!\cdot (m+2)-1)$ $((m+2)!-1)$

\end{document}