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Unformatted text preview: MATH 74 HOMEWORK 10 (DUE WEDNESDAY NOVEMBER 28) 1(a). We first prove that if 5  a , then 5  a 2 . Suppose 5  a . This means that there is q ∈ Z with a = 5 q . Squaring both sides we see a 2 = 25 q 2 = 5( q 2 ), so that 5  a 2 . We now prove that if 5  a 2 , then 5  a . Suppose that 5  a 2 . From the division theorem we know there are q ∈ Z and 0 ≤ r < 5 satisfying a = 5 q + r . If 5 6  a then r 6 = 0, so that r ∈ { 1 , 2 , 3 , 4 } . But (5 q + 1) 2 = 25 q 2 + 10 q + 1 = 5(5 q 2 + 2 q ) + 1 (5 q + 2) 2 = 25 q 2 + 20 q + 4 = 5(5 q 2 + 4 q ) + 4 (5 q + 3) 2 = 25 q 2 + 30 q + 9 = 5(5 q 2 + 6 q + 1) + 4 (5 q + 4) 2 = 25 q 2 + 40 q + 16 = 5(5 q 2 + 8 q + 3) + 1 , so no matter what we see that 5 6  a 2 , a contradiction. 1(b). The statement of 1(a) is false if 5 is replaced with 4; e.g. 4  2 2 but 4 6  2. 1(c). Numbers like 6 , 7 , 11 , 15 are like 5. Numbers like 9 , 12 , 18 , 25 are like 4. 2. We first prove that if 9  a 2 then 3  a 2 . To see this, note that if 9  a 2 there is q ∈ Z with a 2 = 9 q . But then a 2 = 3(3 q ), showing that 3  a 2 . We now show that if 3  a 2 , then 9  a 2 . Suppose 3  a 2 . By the division theorem we have q ∈ Z and 0 ≤ r < 3 with a = 3 q + r . We cannot have r = 1 or r = 2, because then we would have either a 2 = (3 q + 1) 2 = 9 q 2 + 6 q + 1 = 3(3 q 2 + 2 q ) + 1 or a 2 = (3 q + 2) 2 = 9 q 2 + 12 q + 4 = 3(3 q 2 + 4 q + 1) + 1; in either case we’d conclude that 3 6  a 2 , a contradiction. So, a contradiction....
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This note was uploaded on 02/10/2010 for the course MATH 74 taught by Professor Courtney during the Fall '07 term at University of California, Berkeley.
 Fall '07
 COURTNEY
 Math

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