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For part a) I got phi(x)=3x. For the kernel (part c)) I got {0,5,10,15,20,25,30,35,40,45}. I wasn't really sure how to do the image and the inverse though. Does anyone have any ideas? I thought that the inverse might be 3*ker(phi) by theorem 10.6.1, but I wasnt sure. --[[User:Clwarner|Clwarner]] 14:44, 18 February 2009 (UTC) | For part a) I got phi(x)=3x. For the kernel (part c)) I got {0,5,10,15,20,25,30,35,40,45}. I wasn't really sure how to do the image and the inverse though. Does anyone have any ideas? I thought that the inverse might be 3*ker(phi) by theorem 10.6.1, but I wasnt sure. --[[User:Clwarner|Clwarner]] 14:44, 18 February 2009 (UTC) | ||
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| + | ---- | ||
| + | Yea I got the same thing in part a by showing:<br> | ||
| + | 7^2 = 49 = -1 mod 50<br> | ||
| + | φ(-1) = φ(49) = φ(7*7) = 7*φ(7) = 7*6 = 42 = 12 mod 15<br> | ||
| + | φ(1) = -φ(-1) = -12 = 3 mod 15<br> | ||
| + | Therefore, φ(x) = 3x. The answer checks out, but I don't know if this is the right way to get it. I'm also not certain on how to get the image and the inverse. | ||
Revision as of 11:15, 18 February 2009
For part a) I got phi(x)=3x. For the kernel (part c)) I got {0,5,10,15,20,25,30,35,40,45}. I wasn't really sure how to do the image and the inverse though. Does anyone have any ideas? I thought that the inverse might be 3*ker(phi) by theorem 10.6.1, but I wasnt sure. --Clwarner 14:44, 18 February 2009 (UTC)
Yea I got the same thing in part a by showing:
7^2 = 49 = -1 mod 50
φ(-1) = φ(49) = φ(7*7) = 7*φ(7) = 7*6 = 42 = 12 mod 15
φ(1) = -φ(-1) = -12 = 3 mod 15
Therefore, φ(x) = 3x. The answer checks out, but I don't know if this is the right way to get it. I'm also not certain on how to get the image and the inverse.
