This test is gonna be the end of me lol
1 answer:
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Inequation 1:
to plot the pairs (x, y) for which the inequation holds, draw the line
then pick a point in either side of the line. If that point is a solution of the inequation, than color that region of the line, if that point is not a solution, then color the other part of the line.
we do the same for the second inequation. Then the solution, is the region of the x-y axes colored in both cases.
inequation 2:
draw the lines
i) use points (0, -3), (3, -1)
ii)
use points ( 0, 4), (3, 2)
let's use the point P(3, 3) to see what region of the lines need to be coloured:
;
, not true so we color the region not containing this point
not true, so we color the region not containing the point (3, 3)
The graph representing the system of inequalities is the region colored both red and blue, with the blue line not dashed, and the red line dashed.
to plot the pairs (x, y) for which the inequation holds, draw the line
then pick a point in either side of the line. If that point is a solution of the inequation, than color that region of the line, if that point is not a solution, then color the other part of the line.
we do the same for the second inequation. Then the solution, is the region of the x-y axes colored in both cases.
inequation 2:
draw the lines
i)
ii)
let's use the point P(3, 3) to see what region of the lines need to be coloured:
The graph representing the system of inequalities is the region colored both red and blue, with the blue line not dashed, and the red line dashed.
Part A:
Given
defined by 
but
Since, f(xy) ≠ f(x)f(y)
Therefore, the function is not a homomorphism.
Part B:
Given
defined by
Note that in
, -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular 
and
Therefore, the function is a homomorphism.
Part C:
Given
, defined by 
Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.
Part D:
Given
, defined by 
but
Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.
Part E:
Given
, defined by ![\left([x_{12}]\right)=[x_4]](https://tex.z-dn.net/?f=%5Cleft%28%5Bx_%7B12%7D%5D%5Cright%29%3D%5Bx_4%5D)
, where ![[u_n]](https://tex.z-dn.net/?f=%5Bu_n%5D)
denotes the lass of the integer 
in 
.
Then, for any![[a_{12}],[b_{12}]\in Z_{12}](https://tex.z-dn.net/?f=%5Ba_%7B12%7D%5D%2C%5Bb_%7B12%7D%5D%5Cin%20Z_%7B12%7D)
, we have
![f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)](https://tex.z-dn.net/?f=f%5Cleft%28%5Ba_%7B12%7D%5D%2B%5Bb_%7B12%7D%5D%5Cright%29%3Df%5Cleft%28%5Ba%2Bb%5D_%7B12%7D%5Cright%29%20%5C%5C%20%20%5C%5C%20%3D%5Ba%2Bb%5D_4%3D%5Ba%5D_4%2B%5Bb%5D_4%3Df%5Cleft%28%5Ba%5D_%7B12%7D%5Cright%29%2Bf%5Cleft%28%5Bb%5D_%7B12%7D%5Cright%29)
and
![f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)](https://tex.z-dn.net/?f=f%5Cleft%28%5Ba_%7B12%7D%5D%5Bb_%7B12%7D%5D%5Cright%29%3Df%5Cleft%28%5Bab%5D_%7B12%7D%5Cright%29%20%5C%5C%20%5C%5C%20%3D%5Bab%5D_4%3D%5Ba%5D_4%5Bb%5D_4%3Df%5Cleft%28%5Ba%5D_%7B12%7D%5Cright%29f%5Cleft%28%5Bb%5D_%7B12%7D%5Cright%29)
Therefore, the function is a homomorphism.
Given
but
Since, f(xy) ≠ f(x)f(y)
Therefore, the function is not a homomorphism.
Part B:
Given
Note that in
and
Therefore, the function is a homomorphism.
Part C:
Given
Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.
Part D:
Given
but
Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.
Part E:
Given
Then, for any
and
Therefore, the function is a homomorphism.