Of course, one can make a transformation between the two, but solving the transformed one in integers isn't necessarily equivalent to solving the original in integers.
it has been ages. i don't have much spare time anymore. i work nights and am taking a full load at school while also exploring who i am and all that growing up crap
In topology, a clopen set (a portmanteau of closed-open set) in a topological space is a set which is both open and closed. That this is possible may seem counter-intuitive, as the common meanings of open and closed are antonyms. But their mathematical definitions are not mutually exclusive. A set is closed if its complement is open, which leaves the possibility of an open set whose complement is itself also open, making both sets both open and closed, and therefore clopen.
== Examples ==
In any topological space X, the empty set and the whole space X are both clopen.
Now consider the space X which...
i feel your pain. i had to do a chemistry presentation over the effects of gold on rheumatoid arthritis. i didn't understand a single word in my presentation.
Could you glance over a test paper for me and tell me if you could have done 4 questions in 3 hours at the end of your first year of college, @meer2kat?
i wish this could have been an exam i'd taken. this doesn't look hard at all. 3 hours you said? i think i could finish in an hour or hour and a half at the end of my first year in college. a lot faster now that i'm in my third year.
yes and yes. i could have done this at the end of my third year of high school no problem. i may have been able to figure it out at my second, but it would have taken a long time since i have very limited knowledge of calc back then
so i could do it when i was 15/16 easily. 14/15 possibly
has anyone ever had galub jamun? it's so yummy!!!!!
Off topic If $f:\mathbb{R} \to \mathbb{R}$ satisfies $|(f(x)-f(y) / (x-y) | \ge 1$, then I can conclude $|f(x+r)|\le |f(x) + f(r)|$ Can this be generalized to the case that $f$ is a function from $\mathbb{R^n}$ onto itself if I use the standard metric?
Well, the linear system which at least has one solution is called "consistent" linear system.
Find an equation involving g, h, and k that makes
this augmented matrix correspond to a consistent system:
$$\begin{bmatrix}
1 &-4 &7 &g \\
0 & 3 & -5 &h \\
-2 & 5 & -9 &k
\end{bmatrix}$$
Alright, ...