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Prove by Contradiction: For all Prime Numbers a, b, and c .. |
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| Aug1-12, 11:38 AM | #1 |
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Prove by Contradiction: For all Prime Numbers a, b, and c ..
1. The problem statement, all variables and given/known data
Prove by Contradiction: For all Prime Numbers a, b, and c, a^2 + b^2 =/= c^2 2. Relevant equations Prime number is a number whose only factors are one and itself. Proof by contradiction means that you take a statement's negation as a starting point, and find a contradiction. 3. The attempt at a solution The statement's negation is: There exists prime numbers a, b, and c, such that a^2 + b^2 = c^2 Rearrange it: a^2 = c^2 - b^2 a^2 = (c - b) (c + b) a = √(c-b)(c+b) I'm stuck here. To show that it's a contradiction I would have to show that it's factors are not equal to 1 or a, and I've been staring at this a little too long, my head just keeps going in circles.. some help would be appreciated! |
| Aug1-12, 12:25 PM | #2 |
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So you must have c- b= 1 and c+ b= a^2 or c- b= a and c+ b= a. Now if all of a, b, and c are odd that is impossible. |
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