Encountered difficulties in proving the attached image. Greatly appreciate for the help!
A counter example is $\displaystyle \begin{align*} 2\,\textrm{mod}\,4 \times 2\,\textrm{mod}\,4 = 0\,\textrm{mod}\,4 \end{align*}$.
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Prove It
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Alexis87 said:
Encountered difficulties in proving the attached image. Greatly appreciate for the help!
As for the proof: Since $\displaystyle \begin{align*} p \end{align*}$ is prime, it is a positive integer.
$\displaystyle \begin{align*} 0\,\textrm{mod}\,p = k \, p \end{align*}$, where $\displaystyle \begin{align*} k \end{align*}$ is some integer.
As $\displaystyle \begin{align*} p \end{align*}$ is prime, the only way to get a multiple of $\displaystyle \begin{align*} p \end{align*}$ is if that number has $\displaystyle \begin{align*} p \end{align*}$ as a factor. But any multiple of $\displaystyle \begin{align*} p \end{align*}$ is itself $\displaystyle \begin{align*} 0\,\textrm{mod}\,p \end{align*}$, and thus the only way to get $\displaystyle \begin{align*} 0\,\textrm{mod}\,p \end{align*}$ through multiplication in $\displaystyle \begin{align*} \mathbf{Z}_p \end{align*}$ is to multiply by $\displaystyle \begin{align*} 0\,\textrm{mod}\,p \end{align*}$.
Are there known conditions under which a Markov Chain is also a Martingale? I know only that the only Random Walk that is a Martingale is the symmetric one, i.e., p= 1-p =1/2.
Hello !
I derived equations of stress tensor 2D transformation.
Some details: I have plane ABCD in two cases (see top on the pic) and I know tensor components for case 1 only. Only plane ABCD rotate in two cases (top of the picture) but not coordinate system. Coordinate system rotates only on the bottom of picture.
I want to obtain expression that connects tensor for case 1 and tensor for case 2.
My attempt:
Are these equations correct? Is there more easier expression for stress tensor...