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φ(-1) = φ(49) = φ(7*7) = 7*φ(7) = 7*6 = 42 = 12 mod 15<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> | 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. | + | Therefore, φ(x) = 3x. The answer checks out, but I don't know if this is the right way to get it.<br><br> You can get the inverse by looking at all the possible x values and then determining which of these values of x give us φ(x) = 3. We know x = {0,1,2,...,49}. To find which values of x give us 3*x mod 15 = 3, we could iterate all of the 3*x values as {0,3,6,9,12,...,147} and then determining which ones are equivalent to 3 in Z_15. It's easier however to use property 6 of Theorem 10.1 in the book which says that φ^-1(g') = gKer(φ). This means all we need is one value x such that 3x mod 15 = 3 and then add each value of Kerφ to that x to determine the inverse set. The simplest value for x is 1. Since our kernel is just the multiples of 5 in [0,45] we just get the inverse as {1+0, 1+5, 1+10,..., 1+45} = {1,6,11,16,...,46}. <br><br> |
| + | The image I believe is just {0,3,6,9,12} since φ(x) = 3x. This is the multiples of 3 from 0 to 14.<br> | ||
--[[User:Jniederh|Jniederh]] 15:15, 18 February 2009 (UTC) | --[[User:Jniederh|Jniederh]] 15:15, 18 February 2009 (UTC) | ||
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| + | I think the image of φ is just the elements in Z mod 15Z that are being mapped to. So {0, 3, 6, 9, 12}.--[[User:Jcromer|Jcromer]] 15:58, 18 February 2009 (UTC) | ||
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| + | Yeah I was having trouble with that question too, but thanks for the explanation! I just have one question though, do you want to pick an x so when it gets multipled by 3 and then divided by 3 it becomes a whole number? | ||
| + | --[[User:Lchinn|Lchinn]] 18:51, 18 February 2009 (UTC) | ||
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| + | But why can you simply take out that 7 inside "φ(7*7)" to get "7*φ(7)"? That's my main question!! | ||
| + | -K. Brumbaugh | ||
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| + | "I just have one question though, do you want to pick an x so when it gets multiplied by 3 and then divided by 3 it becomes a whole number?" | ||
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| + | I'm not sure what you meant by this question. If you pick any x, multiply it by 3, and then divide it by 3, you will always get back x. Therefore, if x is a whole number, then the result of this process is also a whole number. If we are mapping Z50 to Z15, then we know, by definition, that all possible values of x are already whole numbers because Z50 is the set of integers {0,1,2,...,49}. | ||
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| + | --[[Ysuo|Ysuo]] | ||
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| + | I don't understand how you can just φ(7*7) = 7*φ(7) shouldn't that be φ(7*7) = φ(7)*φ(7) | ||
| + | --[[User:Rpkersey|Rpkersey]] | ||
Latest revision as of 03:01, 19 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.
You can get the inverse by looking at all the possible x values and then determining which of these values of x give us φ(x) = 3. We know x = {0,1,2,...,49}. To find which values of x give us 3*x mod 15 = 3, we could iterate all of the 3*x values as {0,3,6,9,12,...,147} and then determining which ones are equivalent to 3 in Z_15. It's easier however to use property 6 of Theorem 10.1 in the book which says that φ^-1(g') = gKer(φ). This means all we need is one value x such that 3x mod 15 = 3 and then add each value of Kerφ to that x to determine the inverse set. The simplest value for x is 1. Since our kernel is just the multiples of 5 in [0,45] we just get the inverse as {1+0, 1+5, 1+10,..., 1+45} = {1,6,11,16,...,46}.
The image I believe is just {0,3,6,9,12} since φ(x) = 3x. This is the multiples of 3 from 0 to 14.
--Jniederh 15:15, 18 February 2009 (UTC)
I think the image of φ is just the elements in Z mod 15Z that are being mapped to. So {0, 3, 6, 9, 12}.--Jcromer 15:58, 18 February 2009 (UTC)
Yeah I was having trouble with that question too, but thanks for the explanation! I just have one question though, do you want to pick an x so when it gets multipled by 3 and then divided by 3 it becomes a whole number? --Lchinn 18:51, 18 February 2009 (UTC)
But why can you simply take out that 7 inside "φ(7*7)" to get "7*φ(7)"? That's my main question!! -K. Brumbaugh
"I just have one question though, do you want to pick an x so when it gets multiplied by 3 and then divided by 3 it becomes a whole number?"
I'm not sure what you meant by this question. If you pick any x, multiply it by 3, and then divide it by 3, you will always get back x. Therefore, if x is a whole number, then the result of this process is also a whole number. If we are mapping Z50 to Z15, then we know, by definition, that all possible values of x are already whole numbers because Z50 is the set of integers {0,1,2,...,49}.
--Ysuo
I don't understand how you can just φ(7*7) = 7*φ(7) shouldn't that be φ(7*7) = φ(7)*φ(7) --Rpkersey
