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You are not logged in. #1 20071107 12:01:35
Things they never teach you in algebraProve that the following polynomial has no integer roots: "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #2 20071107 15:54:56
Re: Things they never teach you in algebraRicky, Character is who you are when no one is looking. #3 20071108 01:36:14
Re: Things they never teach you in algebraWe can immediately rule out all integers from 0 to positive infinity. 0 can be ruled out by trial, and it can't be any positive integer since every coefficient is positive. Using this new equation we will consider only positive integers for x, which is exactly equivalent to considering only negative integers of x for the original equation. Now, since the highest power is even we know that the equation will tend to positive infinity as x tends to positive infinity. What we do now is try, by trial and error, all integers from 1 up to the point that the equation is strictly increasing, at which point we'll be done. Start with x = 1: 1  3 + 15  6 + 33  45 + 42  24 + 3  9 + 6 = 13 x = 2: 1024  1536 + 3840  768 + 2112  1440 + 672  192 + 12  18 + 6 = 3712 For x = 3, we'll look at prime factorizations. We'll further rewrite the equation as follows: When x = 3 we can see that the first two terms will cancel out. The next two terms give us , which is greater than 0. Combining the next two terms gives us , again greater than 0. The next two terms clearly give us a term greater than 0, since the positive term has both a higher power of 3 and a larger coefficient. The final 3 terms give us 27  27 + 6 = 6. Since we were able to combine all of the terms to give us either 0 or a positive number we know that the equation is greater than 0 for x = 3. We will use this same trick to show that all integer values of x greater than 3 result in a positive answer and thus are not roots. Combine the first two terms to give us . Note that this is equivalent to . For all x > 3 this is clearly greater than 0, since both x^9 and x  3 are greater than 0. Combine the next two terms to give us . This term is clearly greater than 0 since the positive term has both a larger coefficient and a larger power than the negative term. Combine the next 2 terms for . Again, this is positive for all x > 3 since 3x^5 and 11x  15 will both be positive. Combining the next 2 terms gives us , which again has a positive term with both a larger power and a larger coefficient, so this is positive. Finally, combine the final 3 terms for . The 6 is obviously positive, and for all x > 3 so will 3x and x  3. Therefore, this final term is positive. Add it all up and we know that the equation is positive for all x > 0, which is equivalent to saying that the original equation is greater than 0 for all x < 0. Since we also know that 0 and x < 0 are not roots where x is an integer, we know that the equation has no integer roots. Edit: Holy cow, your answer is way easier than mine. I don't quite understand how it works, though. Do you have a proof, or is this a well known theorem? Last edited by TheDude (20071108 01:45:57) Wrap it in bacon #4 20071108 04:33:03
Re: Things they never teach you in algebraThe result assumes that the polynomial is of degree at least 2. (Linear polynomial equations always have solutions.) It also assumes that none of the coefficients (except possibly the leading one) is ±1. Last edited by JaneFairfax (20071108 04:43:43) #5 20071108 04:37:46
Re: Things they never teach you in algebraGanesh, be very careful. A polynomial of degree greater than 5 can not be solved in genera. However, many of them can be factored. For example:
This is known as Eisenstein's criterion. It is proved in general for polynomial rings, and can even be applied to a ring's field of fractions. In this case, we may not only say that my given polynomial has no integer roots and is not reducible in the integers, but also that it is not reducible in the rationals as well (and hence, no rational roots either). A simple proof can be found on wikipedia.
1 is not prime, so the preconditions can't be met if a coefficient is +/ 1 "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #6 20071108 04:57:47
Re: Things they never teach you in algebraVery cool, thanks Jane and Ricky Wrap it in bacon #7 20071108 14:44:15
Re: Things they never teach you in algebraThis is great! Ricky (or anyone else) do you have any more "things they never teach you in algebra." I'm looking forward to learning new things like number theory and what not, but I love to expand on my preexisting knowledge (like Algebra). There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction. 