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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 $(g \circ f)' = f' \circ g'$
Sep
7
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)?
Sep
6
asked Prove that for any $f,g$ polynomials in a polynomial ring, there are no zero divisors
Jul
22
accepted Prove: If $A \subseteq B$ and $B \subseteq C$, then $A \subseteq C$.
Jul
21
asked Prove: If $A \subseteq B$ and $B \subseteq C$, then $A \subseteq C$.
Jul
20
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?
Jul
18
accepted Prove $A \subset \emptyset \iff A = \emptyset$
Jul
17
accepted Why does direct substitution work for limits?
Jul
17
asked Prove $A \subset \emptyset \iff A = \emptyset$
Jul
2
awarded  Curious
May
23
asked How do I work out the last sentence in this section of a proof of the Unique Factorization Theorem?
Mar
19
asked How can I show that $u=e^{\sigma\sqrt{\Delta t}}$ in the binomial option pricing model
Dec
20
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?