331 reputation
111
bio website gabi.purcaru.com
location
age 21
visits member for 2 years, 7 months
seen Aug 15 at 22:02

Jul
2
awarded  Curious
Dec
10
comment $1/i=i$. I must be wrong but why?
@arsmath that's why I said an not the
Dec
10
comment $1/i=i$. I must be wrong but why?
my highschool math teacher told me to never use $\sqrt{-1}$; $i$ is just an imaginary number with the property that $i^2 = -1$. There's no such thing as the square root of a negative number.
Aug
28
accepted How to prove that $(X^a -1, X^b - 1) = X^{(a,b)}-1$
Aug
28
comment How to prove that $(X^a -1, X^b - 1) = X^{(a,b)}-1$
this is so elegant, thank you!
Aug
28
asked How to prove that $(X^a -1, X^b - 1) = X^{(a,b)}-1$
Jun
29
awarded  Nice Question
Mar
18
awarded  Yearling
Dec
16
comment How to find all morphisms from $(\mathbb{N}, \mid)$ to $(\mathbb{N}, \mid)$?
thanks for your input. I selected Hagen's answer because it describes all solutions though
Dec
16
accepted How to find all morphisms from $(\mathbb{N}, \mid)$ to $(\mathbb{N}, \mid)$?
Dec
15
comment How to find all morphisms from $(\mathbb{N}, \mid)$ to $(\mathbb{N}, \mid)$?
$0 \in \mathbb{N}$, yes. Sorry for the confusion... This being my homework, I figured I was missing something, but now I'm starting to think the teacher was messing with us. Thanks!
Dec
15
awarded  Commentator
Dec
15
comment How to find all morphisms from $(\mathbb{N}, \mid)$ to $(\mathbb{N}, \mid)$?
@Isomorphism I don't think that's the case, because $f(1^n) = f(1)$, not $f(1)^n$
Dec
15
comment How to find all morphisms from $(\mathbb{N}, \mid)$ to $(\mathbb{N}, \mid)$?
@Isomorphism but isn't this just $f(x) = cx^0$ ? Anyway, I'm not convinced that these are the only morphisms either, because, for example, $f(0)=0$ and $\forall x > 0$, $f(x) = c$, or $f(0) = 8, f(1) = 2, f(2) = 4, f(x) = 8 $ for any other $x$, etc. are valid morphisms.
Dec
15
comment How to find all morphisms from $(\mathbb{N}, \mid)$ to $(\mathbb{N}, \mid)$?
@Isomorphism can you give me an example for that? My reasoning is that if $x \mid a$ and $x \mid b$, then $x \mid a + b$, which means that $a+b = kx (k \in \mathbb{N})$. But then $(a+b)^r = (kx)^r = k^rx^r$ and $x^r \mid k^rx^r$. (I removed the other constant to make this simpler, it doesn't change anything) Is this wrong?
Dec
15
revised How to find all morphisms from $(\mathbb{N}, \mid)$ to $(\mathbb{N}, \mid)$?
edited body
Dec
15
asked How to find all morphisms from $(\mathbb{N}, \mid)$ to $(\mathbb{N}, \mid)$?
Sep
21
awarded  Custodian
Jul
19
accepted Help me evaluate $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$
Jul
18
revised Help me evaluate $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$
added 9 characters in body