BMAD 64 Report post Posted July 15, 2015 Prove that if a, b, and c are odd integers, then a x^2 + b x + c = 0 has no rational roots. Quote Share this post Link to post Share on other sites
0 Joyandwarmfuzzies 2 Report post Posted July 16, 2015 For the roots to be irrational, the discriminant must not be a square number. Using a proof by contradiction, assume that the roots can be rational. i.e. assume that the discriminant is a square number, k^{2} b^{2} - 4ac = k^{2}.b^{2} = 4ac + k^{2} Per the Pythagoren Theorem, 4ac must also be a square, which only occurs when a = c. b^{2} - 4ac = k^{2}b^{2} - 4a^{2} = k^{2} Note that because b is odd, b^{2} is odd. And because 4a^{2} is even, k^{2}, and therefore k, must also be odd. Let b = 2m + 1 and k = 2n+1 b^{2} - 4a^{2} = k^{2}(2m+1)^{2} - 4a^{2} = (2n+1)^{2}4m^{2} + 4m + 1 - 4a^{2} = 4n^{2} + 4n + 14m^{2} + 4m - 4a^{2} = 4n^{2} + 4nm(m+1) - a^{2} = n(n+1) Note that m(m+1) is even, a^{2} is odd, and n(n+1) is even. An odd number subtracted from an even number is always odd. We have reached a contradiction. Therefore, the discriminant can never be a squared number, and the roots can never be rational. Quote Share this post Link to post Share on other sites
0 DejMar 9 Report post Posted July 16, 2015 Joyandwarmfuzzies wrote:"Per the Pythagoren Theorem, 4ac must also be a square, which only occurs when a = c."4ac may be a square other than when a = c. It is a square when all the factors of 4, a and c can be paired up (4 is is paired up with factors 2*2). As each "a" and "c" are given as odd, neither will have the prime factor 2 as a factor. Yet, "a" may have, for example, the factors 3*3*3 and "c" may simply have the factored 3. This would allow 4ac in this given example to be the square of 18 (=2*3*3). Quote Share this post Link to post Share on other sites
0 Joyandwarmfuzzies 2 Report post Posted July 16, 2015 Turns out my note about a = c doesn't matter. For the roots to be irrational, the discriminant must not be a square number. Using a proof by contradiction, assume that the roots can be rational. i.e. assume that the discriminant is a square number, k^{2} b^{2} - 4ac = k^{2}. Per the Pythagorean Theorem, 4ac must also be a square. Note that because b is odd, b^{2} is odd. 4ac must be even. Therefore k^{2}, and therefore k, must also be odd. Let b = 2m + 1 and k = 2n+1 b^{2} - 4ac = k^{2}(2m+1)^{2} - 4ac = (2n+1)^{2}4m^{2} + 4m + 1 - 4ac = 4n^{2} + 4n + 14m^{2} + 4m - 4ac = 4n^{2} + 4nm(m+1) - ac = n(n+1) Note that m(m+1) is even, ac is odd, and n(n+1) is even. An odd number subtracted from an even number is always odd. We have reached a contradiction. Therefore, the discriminant can never be a squared number, and the roots can never be rational. Quote Share this post Link to post Share on other sites
0 BMAD 64 Report post Posted July 17, 2015 Turns out my note about a = c doesn't matter. For the roots to be irrational, the discriminant must not be a square number. Using a proof by contradiction, assume that the roots can be rational. i.e. assume that the discriminant is a square number, k^{2} b^{2} - 4ac = k^{2}. Per the Pythagorean Theorem, 4ac must also be a square. Note that because b is odd, b^{2} is odd. 4ac must be even. Therefore k^{2}, and therefore k, must also be odd. Let b = 2m + 1 and k = 2n+1 b^{2} - 4ac = k^{2}(2m+1)^{2} - 4ac = (2n+1)^{2}4m^{2} + 4m + 1 - 4ac = 4n^{2} + 4n + 14m^{2} + 4m - 4ac = 4n^{2} + 4nm(m+1) - ac = n(n+1) Note that m(m+1) is even, ac is odd, and n(n+1) is even. An odd number subtracted from an even number is always odd. We have reached a contradiction. Therefore, the discriminant can never be a squared number, and the roots can never be rational. 1 Quote Share this post Link to post Share on other sites
0 Joyandwarmfuzzies 2 Report post Posted July 17, 2015 Oops... the reference to the Pythagorean Theorem is now irrelevant after making the change that DejMar noted, since 4ac being a square doesn't actually matter. I should have removed that, my mistake. For the roots to be irrational, the discriminant must not be a square number. Using a proof by contradiction, assume that the roots can be rational. i.e. assume that the discriminant is a square number, k^{2} b^{2} - 4ac = k^{2}. Note that because b is odd, b^{2} is odd. 4ac must be even. Therefore k^{2}, and therefore k, must also be odd. Let b = 2m + 1 and k = 2n+1 b^{2} - 4ac = k^{2}(2m+1)^{2} - 4ac = (2n+1)^{2}4m^{2} + 4m + 1 - 4ac = 4n^{2} + 4n + 14m^{2} + 4m - 4ac = 4n^{2} + 4nm(m+1) - ac = n(n+1) Note that m(m+1) is even, ac is odd, and n(n+1) is even. An odd number subtracted from an even number is always odd. We have reached a contradiction. Therefore, the discriminant can never be a squared number, and the roots can never be rational. Quote Share this post Link to post Share on other sites
0 BMAD 64 Report post Posted July 19, 2015 Proof by contradictionGiven ax^2 + bx +c = 0 and a,b,c are oddAssume that x has a rational solutionThen there exist a p and q such that x = p/q where p/q is a reduced fraction and p,q are integers.Then a(p/q)^2 + b(p/q) + c =0Times everything by q^2Ap^2 +bq +cq^2 = 0Now we have three cases, both p and q are odd or one of them is odd.1. Both are odd...If both p and q are odd then ap^2, bq, and cq^2 are odd, meaning their sum is odd not zero.2. p is odd ...Then ap^2 is odd, while bq and cq^2 are even. And odd summed with two evens is odd not zero.3. Q is odd...The same reasoning follows if Q is odd, except you end up with an even plus and even plus and odd, which results in an odd not zero. Quote Share this post Link to post Share on other sites
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