1873_solutions

and 3 3 2 2 2 2 2 2 2

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Unformatted text preview: number 12 m n is an integer that has a factor 3. But the integer 2 n cannot have a factor 3 and we have contradicted the fact that 12 m n  2 n . c. 3 4 To obtain a contradiction, suppose that the number 3 4 is rational. Choose positive integers m and n that have no common factor such that 34  m. n Since 21 4n 3  m 3 and since n is a factor of the number we conclude that n is a factor of m 3 . Therefore, since m and n have no common factor we must have n  1 and the equation 4n 3  m 3 becomes m3  4 which is impossible since there is no integer whose cube is 4. 4n 3 d. Any solution of the equation 8x 3 6x 1  0. Solution: Suppose that x is a solution of the equation 8x 3 6x 1  0 and, to obtain a contradiction, assume that x is rational. Choose integers m and n such that n  0 and m and n have no common factor and x  m . Since n 3 8m 6m 10 n n we have 8m 3 6mn 2 n 3  0. From the fact that 8m 3  n 6mn  n 2 we deduce that n is a factor of 8m 3 . Therefore, since m and n have no common factor, the number n must be 1 or 2 or 4 or 8. Returning to the equation 8m 3 6mn 2 n 3  0. we deduce that n 3  m 8m 2 6n 2 and so n is a factor of Therefore, since m and n have no common factor we must have m  1. We deduce that x must be one of the numbers 1, 1 , 1 or 1 and, by trying each of these numbers in 2 4 8 the equation 8x 3 6x 1  0, we can verify that none of them are solutions. Therefore no rational number can be a solution of this equation. n3. A quick way to verify that none of the numers 1, Note: 1 2 , 1 4 or 1 8 can be solutions of the 1  0 is to supply the equation f x  8x 3 6x 1 as a definition to Scientific Notebook by pointing at this equation and clicking on the new definition equation 8x 3 6x button and then evaluating the expression 1 f 1 2 1 4 1 8 1 1 2 1 4 1 8 1 3  19 8 111 64 3 1 3 8 17 64 which shows us at once that f x is not zero when x is one of the eight rational numbers that might have been solutions of the equation 8x 3 6x 1  0. 22 2. Given that m and n are integers and that mn does not have a factor 3, prove that neither m nor n can have a factor 3. What we want to prove is that if at least one of the integers m and n has a factor 3 then mn must have a factor 3. Strictly speaking, there are two cases to consider because the statement that at least one of the integers m and n has a factor 3 means that either m has a factor 3 or n has a factor 3. However, these two cases are analogous. We can turn one into the other by renaming the numbers. We can therefore restrict ourselves safely to just one of the two cases. Our proof can begin: Without loss of generality, we assume that m has a factor 3. Since m/3 is an integer and since mn  3 m/3 n we see that mn has a factor 3, which is what we wanted to prove. 3. Suppose that we know that x 2 2x  0 and that we wish to prove that 0  x  2. Write down the first line of a proof of this assertion that uses the method of proof by contradiction. Do so in such a way that your proof split...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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