- #1
kostoglotov
- 234
- 6
I feel like I almost understand the solution I've come up with, but a step in the logic is missing. I'll post the question and my solution in LaTeX form.
Paraphrasing of text question below in LaTeX. Text question can be seen in its entirety via this imgur link: http://i.imgur.com/41fvDRN.jpg
[tex]
\ if \begin{pmatrix}
a\\
b
\end{pmatrix} \ is \ a \ multiple \ of \begin{pmatrix}
c\\
d
\end{pmatrix} \ with \ abcd \neq 0, show \ that \begin{pmatrix}
a\\
c
\end{pmatrix} \ is \ a \ multiple \ of \begin{pmatrix}
b\\
d
\end{pmatrix}
[/tex]
My solution so far
[tex]
assume \ \lambda \begin{pmatrix}
c\\
d
\end{pmatrix} = \begin{pmatrix}
a\\
b
\end{pmatrix} \rightarrow \begin{matrix}
a = \lambda c\\
b = \lambda d
\end{matrix}
[/tex]
[tex]
now \ assume \ \gamma \begin{pmatrix}
b\\
d
\end{pmatrix} = \begin{pmatrix}
a\\
c
\end{pmatrix} \rightarrow \begin{matrix}
a = \gamma b\\
c = \gamma d
\end{matrix}
[/tex]
So I'm making two assumptions
Let's take the assumptions and put them into a system of four equations
[tex]1: \ a = \lambda c \ \ \ \ 2: \ b = \lambda d \\ 3: \ a = \gamma b \ \ \ \ 4: \ c = \gamma d[/tex]
Now if we sub 3 into 1 to get A, and sub 2 into A to get B and then sub 4 into B to get C
[tex]C \rightarrow \lambda \gamma d = \lambda \gamma d[/tex]
Similarly if we, sub 2 into 3 to get A, and 1 into A to get B, and 4 into B to get C
[tex]C \rightarrow \lambda \gamma d = \lambda \gamma d[/tex]
I want to stop trying all the possible ways to get C now, because I want to look for a generalized way to show that they will all end up at the same point.
But more than this...what is the step of logic that connects the final equation C to proving the first two assumptions. I feel like this should prove the assumptions, but I don't know how exactly, or how exactly to express it.
Thanks :)
Paraphrasing of text question below in LaTeX. Text question can be seen in its entirety via this imgur link: http://i.imgur.com/41fvDRN.jpg
[tex]
\ if \begin{pmatrix}
a\\
b
\end{pmatrix} \ is \ a \ multiple \ of \begin{pmatrix}
c\\
d
\end{pmatrix} \ with \ abcd \neq 0, show \ that \begin{pmatrix}
a\\
c
\end{pmatrix} \ is \ a \ multiple \ of \begin{pmatrix}
b\\
d
\end{pmatrix}
[/tex]
My solution so far
[tex]
assume \ \lambda \begin{pmatrix}
c\\
d
\end{pmatrix} = \begin{pmatrix}
a\\
b
\end{pmatrix} \rightarrow \begin{matrix}
a = \lambda c\\
b = \lambda d
\end{matrix}
[/tex]
[tex]
now \ assume \ \gamma \begin{pmatrix}
b\\
d
\end{pmatrix} = \begin{pmatrix}
a\\
c
\end{pmatrix} \rightarrow \begin{matrix}
a = \gamma b\\
c = \gamma d
\end{matrix}
[/tex]
So I'm making two assumptions
Let's take the assumptions and put them into a system of four equations
[tex]1: \ a = \lambda c \ \ \ \ 2: \ b = \lambda d \\ 3: \ a = \gamma b \ \ \ \ 4: \ c = \gamma d[/tex]
Now if we sub 3 into 1 to get A, and sub 2 into A to get B and then sub 4 into B to get C
[tex]C \rightarrow \lambda \gamma d = \lambda \gamma d[/tex]
Similarly if we, sub 2 into 3 to get A, and 1 into A to get B, and 4 into B to get C
[tex]C \rightarrow \lambda \gamma d = \lambda \gamma d[/tex]
I want to stop trying all the possible ways to get C now, because I want to look for a generalized way to show that they will all end up at the same point.
But more than this...what is the step of logic that connects the final equation C to proving the first two assumptions. I feel like this should prove the assumptions, but I don't know how exactly, or how exactly to express it.
Thanks :)