Going into dual space for a vector product I am a bid confused regarding the notation for tensor products when going into dual-space  If $\left| \Psi \rangle \right. = \left| A \rangle \right. \left| B \rangle \right.$ is $ \left. \langle \Psi \right| = \left. \langle A \right| \left. \langle B \right|$ or $\left. \langle B  \right| \left. \langle A \right|$?  My guess is the last choice, since operators should act on   $\left| 0 \rangle \right. = \left( \begin{matrix}   1 \\   0  \end{matrix} \right) $  $\left| 1 \rangle \right. = \left( \begin{matrix}   0 \\   1  \end{matrix} \right) $  so  $\left| 01 \rangle \right. = \left| 0 \rangle \right. \left| 1 \rangle \right. = \left( \begin{matrix}   0 \\   1 \\   0 \\   0  \end{matrix} \right) $  But then, for me, it should be  $\left( \begin{matrix}   0 &amp; 1 &amp; 0 &amp; 0  \end{matrix} \right) = \left \langle 10 \right|$ But this is not correct if i calculate the tensor product, since   $\left \langle 1 \right| \left \langle 0 \right| = \left( \begin{matrix}   0 &amp; 0 &amp; 1 &amp; 0  \end{matrix} \right)$  For instance, shoulden't it be  $\left| 01 \rangle \right. \left \langle 10 \right| = \left( \begin{matrix}   0 &amp; 0 &amp; 0 &amp; 0\\   0 &amp; 1 &amp; 0 &amp; 0\\   0 &amp; 0 &amp; 0 &amp; 0\\   0 &amp; 0 &amp; 0 &amp; 0\\  \end{matrix} \right) $ ?  I hope I have made my confusion clear
Tensor product in dual-space I am a bid confused regarding the notation for tensor products when going into dual-space  If $\left| \Psi \rangle \right. = \left| A \rangle \right. \left| B \rangle \right.$ is $ \left. \langle \Psi \right| = \left. \langle A \right| \left. \langle B \right|$ or $\left. \langle B  \right| \left. \langle A \right|$?  My guess is the last choice, since operators should act on   $\left| 0 \rangle \right. = \left( \begin{matrix}   1 \\   0  \end{matrix} \right) $  $\left| 1 \rangle \right. = \left( \begin{matrix}   0 \\   1  \end{matrix} \right) $  so  $\left| 01 \rangle \right. = \left| 0 \rangle \right. \left| 1 \rangle \right. = \left( \begin{matrix}   0 \\   1 \\   0 \\   0  \end{matrix} \right) $  But then, for me, it should be  $\left( \begin{matrix}   0 &amp; 1 &amp; 0 &amp; 0  \end{matrix} \right) = \left \langle 10 \right|$ But this is not correct if i calculate the tensor product, since   $\left \langle 1 \right| \left \langle 0 \right| = \left( \begin{matrix}   0 &amp; 0 &amp; 1 &amp; 0  \end{matrix} \right)$  For instance, shoulden't it be  $\left| 01 \rangle \right. \left \langle 10 \right| = \left( \begin{matrix}   0 &amp; 0 &amp; 0 &amp; 0\\   0 &amp; 1 &amp; 0 &amp; 0\\   0 &amp; 0 &amp; 0 &amp; 0\\   0 &amp; 0 &amp; 0 &amp; 0\\  \end{matrix} \right) $ ?  I hope I have made my confusion clear
$ac\equiv bc \pmod{\!m}\!\!\iff\!\! a\equiv b \pmod{\!\!\frac{m}{(c,m)}}$ [Euclid's Lemma, congruence form] I came across this theorem:   For all integers a,b,c and m&gt;0, if d = GCD(c,m) then a*c ≡ b*c (mod m) iff a ≡ b (mod m/d).  The =&gt;proof states this:  a*c ≡ b*c (mod m) =&gt; m|(a-b)*c                    =&gt; m/d | (a-b)*c/d                   =&gt; m/d | (a-b)                   =&gt; a ≡ b (mod m/d)   I don't understand how to go from step 2 to step 3. The proof states that this is possible because m/d and c/d are coprime. To me this means GCD(m/d, c/d) = 1 but how does this cancel out the whole term? How do comment @replies work? How do the comment replies (aka &quot;pinging users&quot;) work?  Who can I send an @reply to? How do I respond to a specific user when entering my comment? Will they be notified? What do people mean when they talk about &quot;pinging&quot; another user?  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The old advice used to be to double-quote any expression involving a $VARIABLE, at least if one wanted it to be interpreted by the shell as one single item, otherwise, any spaces in the content of $VARIABLE would throw off the shell.  I understand, however, that in more recent versions of shells, double-quoting is no longer always needed (at least for the purpose described above).  For instance, in bash:  % FOO='bar baz' % [ $FOO = 'bar baz' ] &amp;&amp; echo OK bash: [: too many arguments % [[ $FOO = 'bar baz' ]] &amp;&amp; echo OK OK % touch 'bar baz' % ls $FOO ls: cannot access bar: No such file or directory ls: cannot access baz: No such file or directory   In zsh, on the other hand, the same three commands succeed.  Therefore, based on this experiment, it seems that, in bash, one can omit the double quotes inside [[ ... ]], but not inside [ ... ] nor in command-line arguments, whereas, in zsh, the double quotes may be omitted in all these cases.  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