Yes. Conceptually, all the matrices in the group have the same structure, except for the variable component 
. So, each matrix is identified by its top-right coefficient, since the other three entries remain constant.
However, let's prove in a more formal way that
![\phi:\ \mathbb{R} \to G,\quad \phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cphi%3A%5C%20%5Cmathbb%7BR%7D%20%5Cto%20G%2C%5Cquad%20%5Cphi%28x%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20)
is an isomorphism.
First of all, it is injective: suppose 
. Then, you trivially have 
, because they are two different matrices:
![\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right],\quad \phi(y) = \left[\begin{array}{cc}1&y\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cphi%28x%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%2C%5Cquad%20%5Cphi%28y%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26y%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20)
Secondly, it is trivially surjective: the matrix
![\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cphi%28x%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20)
is clearly the image of the real number x.
Finally, 
and its inverse are both homomorphisms: if we consider the usual product between matrices to be the operation for the group G and the real numbers to be an additive group, we have
![\phi (x+y) = \left[\begin{array}{cc}1&x+y\\0&1\end{array}\right] = \left[\begin{array}{cc}1&x\\0&1\end{array}\right] \cdot \left[\begin{array}{cc}1&y\\0&1\end{array}\right] = \phi(x) \cdot \phi(y)](https://tex.z-dn.net/?f=%20%5Cphi%20%28x%2By%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%2By%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26y%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cphi%28x%29%20%5Ccdot%20%5Cphi%28y%29)
A) 1/6
B) 5/6
C) 1
D) 20
Explanation:
A) There is one 6 on a 6-sided die, out of 6 numbers.
B) There are 5 numbers that are not 6 on a 6 sided die, out of 6 numbers.
C) P( 6 or ~6) = P(6) + P(~6) = 1/6 + 5/6 = 6/6 = 1
D) 1/6(120) = 120/6 = 20
Answer:
D 2
Step-by-step explanation:
hope this helps you
The change in elevation is -15 meters
Answer:
4 blu-rays
Step-by-step explanation:
*Note: I'm rounding up to full dollars to make this easier. You will get the same answer, whether you round up or not.
First, get rid of the shipping cost from the budget.
$150 - $10 = <u>$140 left for blu-rays</u>
Then divide $140 by $30, which is the price for a single blu-ray to get the number of blu-rays Chris can get.
$140 ÷ $30 ≈ 4.6 repeating. You can't have a fraction of a movie, so Chris can get 4 blu-rays
