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For this problem you can consider things such as 1/2 as the multiplicative inverse of 2. Then you have to ask, what element in Z7 is an inverse of 2? 2*4=8 and 8=1 in Z7 so a reasonable interpretation for 1/2 is 4. You can find reasonable interpretations for the other expressions in a similar fashion.<br> | For this problem you can consider things such as 1/2 as the multiplicative inverse of 2. Then you have to ask, what element in Z7 is an inverse of 2? 2*4=8 and 8=1 in Z7 so a reasonable interpretation for 1/2 is 4. You can find reasonable interpretations for the other expressions in a similar fashion.<br> | ||
--[[User:Jniederh|Jniederh]] 03:05, 11 March 2009 (UTC) | --[[User:Jniederh|Jniederh]] 03:05, 11 March 2009 (UTC) | ||
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| + | I understand what we are doing here, but what are we supposed to do for something like <math>\sqrt{-3}</math>? --[[User:Podarcze|Podarcze]] 11:34, 11 March 2009 (UTC) | ||
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| + | For <math>\sqrt{-3}</math> you use the same approach. First you observe -3 mod 7 = 4. Then you ask, what element(s) in Z_7 that have been squared give you 4? If you test all the elements in Z_7 you'll find 2^2 = 4 and 5^2 = 25 = 4 so both 2 and 5 are reasonable interpretations for <math>\sqrt{-3}</math>. Hope this helps.<br> | ||
| + | --[[User:Jniederh|Jniederh]] 14:00, 11 March 2009 (UTC) | ||
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| + | i'm still having a bit of problems. can someone explain with either the other ones or made up examples | ||
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| + | Take -2/3 as an example. First we want to find a value x such that -2/3*x = 1 in Z_7. A good candidate for this is x = -12. This gives us the equation -2/3*-12 = 8 = 1 in Z_7. I prefer to work with positive numbers so we can have x = -12 mod 7 instead. We know -12 = -2*7 + 2 so in Z_7 -12 = 2. Now that we have a value for x, namely x = -12 = 2, we must find a value y in Z_7 that is a multiplicative inverse of 2 just as -2/3 was a multiplicative inverse of -12. To do this we try to satisfy the equation 2*y = 1. Just as above in the example with 1/2 we get 2*4 = 8 = 1, therefore 4 is a multiplicative inverse of 2 in Z_7 and consequently 4 is a decent representation of -2/3 in Z_7. Hope this helps.<br> | ||
| + | --[[User:Jniederh|Jniederh]] 00:05, 12 March 2009 (UTC) | ||
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| + | Thanks so much Jared! This really helped! | ||
| + | --[[User:Eraymond|Eraymond]] 09:15, 12 March 2009 (UTC) | ||
Latest revision as of 03:11, 12 March 2009
For this problem you can consider things such as 1/2 as the multiplicative inverse of 2. Then you have to ask, what element in Z7 is an inverse of 2? 2*4=8 and 8=1 in Z7 so a reasonable interpretation for 1/2 is 4. You can find reasonable interpretations for the other expressions in a similar fashion.
--Jniederh 03:05, 11 March 2009 (UTC)
I understand what we are doing here, but what are we supposed to do for something like $ \sqrt{-3} $? --Podarcze 11:34, 11 March 2009 (UTC)
For $ \sqrt{-3} $ you use the same approach. First you observe -3 mod 7 = 4. Then you ask, what element(s) in Z_7 that have been squared give you 4? If you test all the elements in Z_7 you'll find 2^2 = 4 and 5^2 = 25 = 4 so both 2 and 5 are reasonable interpretations for $ \sqrt{-3} $. Hope this helps.
--Jniederh 14:00, 11 March 2009 (UTC)
i'm still having a bit of problems. can someone explain with either the other ones or made up examples
Take -2/3 as an example. First we want to find a value x such that -2/3*x = 1 in Z_7. A good candidate for this is x = -12. This gives us the equation -2/3*-12 = 8 = 1 in Z_7. I prefer to work with positive numbers so we can have x = -12 mod 7 instead. We know -12 = -2*7 + 2 so in Z_7 -12 = 2. Now that we have a value for x, namely x = -12 = 2, we must find a value y in Z_7 that is a multiplicative inverse of 2 just as -2/3 was a multiplicative inverse of -12. To do this we try to satisfy the equation 2*y = 1. Just as above in the example with 1/2 we get 2*4 = 8 = 1, therefore 4 is a multiplicative inverse of 2 in Z_7 and consequently 4 is a decent representation of -2/3 in Z_7. Hope this helps.
--Jniederh 00:05, 12 March 2009 (UTC)
Thanks so much Jared! This really helped! --Eraymond 09:15, 12 March 2009 (UTC)
