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I am a graduate student at UCLA.


1d
comment Boolean algebra proof (a+b) (a+c)' = a'bc'
Yes, that's right.
1d
comment Boolean algebra proof (a+b) (a+c)' = a'bc'
Yes, you probably have some identity that tells you $aa'$ is false, and then you need to know how false and's and or's with boolean variables.
1d
comment Boolean algebra proof (a+b) (a+c)' = a'bc'
Yes. Make sure you justify why you can remove the $aa'c'$ term.
1d
answered Boolean algebra proof (a+b) (a+c)' = a'bc'
2d
comment Prove that if A and B are both invertible nxn matrices then A and B are row equivalent to each other.
What do you mean by column b? If both matrices can be row-reduced to $I$, then they are each row equivalent to $I$, and therefore to each other.
Sep
17
answered Let $f:(\mathbb{R}\setminus\mathbb{Q})\cap [0,1]\to \mathbb{Q}\cap [0,1]$. Prove there exists a continuous$f$.
Sep
17
comment Let $f:(\mathbb{R}\setminus\mathbb{Q})\cap [0,1]\to \mathbb{Q}\cap [0,1]$. Prove there exists a continuous$f$.
@Winther better yet, one which is not locally constant, as it's easy to concoct a function which "jumps" on the missing rational points in the domain.
Sep
17
comment Let $f:(\mathbb{R}\setminus\mathbb{Q})\cap [0,1]\to \mathbb{Q}\cap [0,1]$. Prove there exists a continuous$f$.
@SujaanKunalan well as Arthur mentioned, constant functions are always continuous...
Sep
17
comment Let $f:(\mathbb{R}\setminus\mathbb{Q})\cap [0,1]\to \mathbb{Q}\cap [0,1]$. Prove there exists a continuous$f$.
@Arthur lol I meant I really doubt that all such functions are continuous. Ambiguities arising in the English language. :P
Sep
17
comment Let $f:(\mathbb{R}\setminus\mathbb{Q})\cap [0,1]\to \mathbb{Q}\cap [0,1]$. Prove there exists a continuous$f$.
I really doubt that any such function is continuous...
Sep
15
comment Order of a permutation
The order of a permutation $\sigma$ is the order of the subgroup $\langle \sigma \rangle$ generated by $\sigma$.
Sep
13
revised Given S, prove that C is an open cover, but there is no finite subcover of C that covers S.
added 8 characters in body
Sep
13
reviewed Approve suggested edit on Given S, prove that C is an open cover, but there is no finite subcover of C that covers S.
Sep
13
comment Given S, prove that C is an open cover, but there is no finite subcover of C that covers S.
I've edited the math formatting. Let me know if there are any mistakes.
Sep
13
revised Given S, prove that C is an open cover, but there is no finite subcover of C that covers S.
added 69 characters in body
Sep
13
comment Show that there exists an uncountable family of sets
Let $T$ be an uncountable set, for instance $T = \mathbb R$. No choice needed.
Sep
13
comment Show that there exists an uncountable family of sets
Yes. The element has, in itself, labelled which two $X_t$'s it's supposed to come from.
Sep
13
comment Show that there exists an uncountable family of sets
The subscript doesn't mean anything unless you define it to mean something. How about $\{t, t'\}$?
Sep
13
answered Show that there exists an uncountable family of sets
Sep
10
comment How to check which axioms hold for models in set theory?
I believe the arrows are intended to signify the $\in$ relation, the only relation defined in the basic language of set theory.