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I'm an undercover :) computer science teacher


1h
comment Decidability of given languages
When I teach the Rice's theorem, I usually explain it as : if a set S is defined by $\{w\;|\;\phi(M_w)\}$, where $\phi$ is any formula then either $S$ is trivial (empty or all) or $S\ge_m K$ (with $\ge_m$ the many one reduction and $K$ the halting set). So $L_2$, $L_3$ and $L_4$ all applies, it easy to see they are not trivial, so they are not decidable directly by Rice's theorem.
2h
comment Decidability of given languages
And what is $f$ in your Rice's theorem ?
20h
comment Prove that $a+b$ cant divide $a^a+b^b$ nor $a^b+b^a$
New try, I simplified using the fact that $a=-b\mod p$
20h
revised Prove that $a+b$ cant divide $a^a+b^b$ nor $a^b+b^a$
added 167 characters in body
22h
comment Prove that $a+b$ cant divide $a^a+b^b$ nor $a^b+b^a$
@Assaultous2 Sorry I fixed my proof
22h
revised Prove that $a+b$ cant divide $a^a+b^b$ nor $a^b+b^a$
Fixed
23h
answered Prove that $a+b$ cant divide $a^a+b^b$ nor $a^b+b^a$
2d
answered Find $f''(x)$ if $f\circ f'(x) = 4x^2 + 3$
Jan
21
answered how do I parenthesize the product of abcdef??
Jan
21
comment how do I parenthesize the product of abcdef??
Words must contain the same number of 1's and 0's. The answers to a.1 and a.3 doesn't. They are obviously false.
Jan
20
revised Number of ways to arrange $n$ items in $m$ positions having exactly $k$ items adjacent to each other
added 2 characters in body
Jan
20
answered Number of ways to arrange $n$ items in $m$ positions having exactly $k$ items adjacent to each other
Jan
15
reviewed Approve $(3/4) \cdot (8/9) \cdot (15/16) \cdot \ldots$ How to prove that limit is $1/2$
Jan
14
awarded  Enlightened
Jan
14
awarded  Nice Answer
Jan
12
revised probability divisible by 11
added 59 characters in body
Jan
12
comment probability divisible by 11
@barakmanos no because all permutations have a $0$, in the 17 part or in the 28 part. So you have to remove it from all permutations. What you did with your computation is to suppose that all sets have a $0$, but only one of the two sets (either the 17 or the 28, not both).
Jan
12
comment probability divisible by 11
What is really fun, is that if you include numbers with a leading 0, you get the same probability. This is due to the fact that you can choose the first number and it doesn't change the problem...
Jan
12
comment probability divisible by 11
It seems right to me... @barakmanos. The "attempt" is ok.
Jan
12
comment probability divisible by 11
@PM2Ring you're right he missed also (0,1,3,6,7) and (0,1,2,6,8) is shown twice !