Adv Alegbra HW Solutions 27

Adv Alegbra HW Solutions 27 - g i are nonnega-tive Solution...

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27 Solution. Suppose there are q and p for ( 9 , 12 , 15 ) . Then 2 qp = 12 and qp = 6. Since q > p are positive integers, the only possibilities are q = 6 and p = 1o r q = 3 and p = 2. The f rst possibility gives the Pythagorean triple ( 12 , 35 , 37 ) while the second gives the Pythagorean triple ( 5 , 12 , 13 ) . (iii) Using a calculator which can f nd square roots but which can dis- play only 8 digits, show that ( 19597501 , 28397460 , 34503301 ) is a Pythagorean triple by f nding q and p . Solution. If q and p exist, then we have q 2 + p 2 = 34503301 q 2 p 2 = 19597501 . Therefore, 2 p 2 = 14905800 and p 2 = 7452900. Hence, p = 2730. Finally, 2 qp = 28397460, and so q = 5201. Since we were able to f nd q and p , the original trio does form a Pythagorean triple. 1.68 True or false with reasons. (i) | 2 19 3 12 | < 1 2 . Solution. False. (ii) If r = p g 1 1 ··· p g n n , where the p i are distinct primes and the g i are integers, then r
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Unformatted text preview: g i are nonnega-tive. Solution. True. (iii) The least common multiple [ 2 3 · 3 2 · 5 · 7 2 , 3 3 · 5 · 13 ] = 2 3 · 3 5 · 5 2 · 7 2 · 13 / 45. Solution. True. (iv) If a and b are positive integers which are not relatively prime, then there is a prime p with p | a and p | b . Solution. True. (v) If a and b are relatively prime, then ( a 2 , b 2 ) = 1. Solution. True. 1.69 (i) Find gcd ( 210 , 48 ) using factorizations into primes. Solution. 210 = 2 1 · 3 1 · 5 1 · 7 1 and 48 = 2 4 · 3 1 · 5 · 7 , so that ( 210 , 48 ) = 2 1 · 3 1 · 5 · 7 = 6. (ii) Find gcd ( 1234 , 5678 ) ....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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