I was thinking about the question above and can't find an easy way of proving it. Any suggestions?
53351
Prove: Let $R$ a Artinian commutative ring with $1 \neq 0$. If $I$ is prime, then $I$ is maximal.     I got stuck on this. I understand that every ideal is generated by finitely many elements. Here is my approach:  It's enough to prove the following implication: $$ R/I \ \text{is a domain} \quad \Rightarrow \quad R/I \ \text{is a field}$$ Then only finding an inverse is left. So let $x \in R$ in arbitrary. We have to find an inverse element of $x+I \in R/I$, that as an element y so that $x \cdot y \in I+1$. We denote $I = (x_1, x_2, \cdots, x_n)$.  Now I have to find y so that $xy -1 = r_1 x_1+ r_2 x_2 + \cdots + r_n x_n$. This is where I no more knew what to do.   Your advise will be appreciated.
35166
There are several proofs I have seen of this, but they all seem to use choice subtely at some point. Is there any way to prove this without choice, or is it possibly unproveable?
36899
how is this possible? I know there is some trick, should someone please explain?!
12472
As the question states - Is it possible to prove the null hypothesis? From my (limited) understanding of hypothesis, the answer is no but I can't come up with a rigorous explanation for it. Does the question have a definitive answer?
27829
I've been messing around with modular arithmetic recently, and stumbled across this, but couldn't find a proof for it anywhere. I hate taking things as truth without knowing why, so could anyone provide a (fairly simple) proof?
120379
I've scoured the math stackexchange and the philosophy one for some guidance on how to go about this while using the Fitch System.  Anyone can attempt it here;    I can reach the conclusion, but it's not in the correct "layer" of the proof. I'm stuck with the answer as a subproof.  (Note that the fitch system lacks contradiction elimination (RAA) as a method for getting out of assumptions).
12307
I've tried deriving a proof for this using the handshaking lemma, but that was unsuccessful. Could someone please explain to me how they would do this proof?
54519
I am not being able to understand the graphical method of solving this, any simple explanation will be appreciated.  A non-graphical calculation will be very helpful too.  Thank you so much in advance!
81041
How can I prove that if $ac&gt;bc$ and $c&gt;0$, then $a&gt;b$ without division? Or Should I prove it by contrapositive?
43327
I'm trying to see why this is the case. I have taken  the LHS and added $X\hat{\beta}$ and subtracted $X\hat{\beta}$.  From this I can get both terms on the right but I end up with a cross product term  $2(y-X\hat{\beta})^T(X\hat{\beta} - X\beta)$  that I can't prove equals to zero. I think there is some trick I'm missing, anybody remember it?
25366
One can obtain the solution to a $2$-Body problem analytically. However, I understand that obtaining a general solution to a  is impossible.  Is there a proof somewhere that shows this possibility/impossibility?  Edit: I am looking to prove or disprove the below statement:     there exists a power series that that   solve this problem, for all the terms   in the series and the summation of the   series must converge.
37580
It's a question of a some time ago test, I've found a way to solve the problem using calculus, but always I've thought that exist a solution with algebra and geometry.  Thank you for your time.
63022
Given that $a \equiv b \pmod n$ and $c\equiv d \pmod n$, I need to prove that $ac \equiv bd \pmod n$  So far, I've only managed to deduce that $a+b \equiv c+d \pmod n$. I don't know if this is usable, but it's there, at least.  Any hint would be greatly appreciated!
113749
This is an exercise from Kelley's book. Could someone help to show me a proof?   It seems very natural, and it is easy to prove by utilizing the arctan function in $\mathbb{R}^1$.  Thanks a lot.
15643
Is it possible to prove De Morgan’s Laws without showing a contradiction otherwise. I’m actually asking because I’m wondering if it’s possible to prove De Morgan’s law without assuming the Law of Non-Contradiction
32207
The only solution that I have been able to come up with is $d=1$, but because $d&gt;2$, I can't seem to figure out how to find a solution.
85075
A friend and I are going back and forth with brain-teasers and I have no idea how to solve this one.  My assumption is that it's possible with some bitwise operators, but not sure.