# flapjackery

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 4h comment Prove $f(A \cup B) = f(A) \cup f(B)$ where $f: X \rightarrow Y$ DanZimm, your comments on double inclusion are very helpful in my journey. Along those lines, would this version of a proof of this claim work as well? mathb.in/20048 4h comment Prove $f(A \cup B) = f(A) \cup f(B)$ where $f: X \rightarrow Y$ Very insightful comments. Thanks very much 4h comment Prove $f(A \cup B) = f(A) \cup f(B)$ where $f: X \rightarrow Y$ This is incredibly helpful. Thank you 4h accepted Prove $f(A \cup B) = f(A) \cup f(B)$ where $f: X \rightarrow Y$ 5h asked Prove $f(A \cup B) = f(A) \cup f(B)$ where $f: X \rightarrow Y$ 5h accepted If $f$ and $g$ are bijective functions, prove that $(g \circ f)' = f' \circ g'$ 8h comment If $f$ and $g$ are bijective functions, prove that $(g \circ f)' = f' \circ g'$ Yes, that's what I intended. My apologies. 8h asked If $f$ and $g$ are bijective functions, prove that $(g \circ f)' = f' \circ g'$ Sep7 comment Zero divisor in $R[x]$ Why does this imply that F(fG)=0? I understand that FG = 0 but why does this say something about the effect of F on the polynomial fG with deg(fG) < deg(G)? Sep6 asked Prove that for any $f,g$ polynomials in a polynomial ring, there are no zero divisors Jul22 accepted Prove: If $A \subseteq B$ and $B \subseteq C$, then $A \subseteq C$. Jul21 asked Prove: If $A \subseteq B$ and $B \subseteq C$, then $A \subseteq C$. Jul20 comment Prove $A \subset \emptyset \iff A = \emptyset$ I gave your answer the points because it was the most thorough and points out the issues in my proof. Although, what does Git Gud mean by his comment? What is the prejudice against the statement? Jul18 accepted Prove $A \subset \emptyset \iff A = \emptyset$ Jul17 accepted Why does direct substitution work for limits? Jul17 asked Prove $A \subset \emptyset \iff A = \emptyset$ Jul2 awarded Curious May23 asked How do I work out the last sentence in this section of a proof of the Unique Factorization Theorem? Mar19 asked How can I show that $u=e^{\sigma\sqrt{\Delta t}}$ in the binomial option pricing model Dec20 comment Please show $|\sin(n+1)x| = |\sin(nx+x)|$ Thanks. That makes sense. Is this typically how arguments for sine are written? I feel like the way you have written in in your answer makes much more sense, in fact, aren't the extra set of brackets necessary? Otherwise what I wrote in my question should be interpreted as (sin(n+1))*x, right?