319HW3Correction

319HW3Correction - MAT 319/320 Correction of HW3 Exercise...

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Unformatted text preview: MAT 319/320 Correction of HW3 Exercise 1. Page 50, #2. Proof. 1. If S is bounded then there exists a lower bound m and an upper bound M . By defini- tion, they are such that any x in S satisfies m lessorequalslant x lessorequalslant M . But this means x [ m, M ] .Therefore S [ m,M ] . 2. Conversely, S [ m, M ] exactly means that any x in S is bounded above by M , and below by m . square Exercise 2. Page 50, #9. Proof. By contradiction: assume that the intersection is non empty, and therefore contains some real number x . Pick any integer K strictly larger than x (for example 1 + E ( x ) , where E ( x ) is the integral part of x ): then clearly x ( K, ) and thus x intersectiontext n =1 ( n, ) , a contradic- tion. square Exercise 3. Page 50, #13. Proof. Since 1/3 is strictly less than 1, the binary representation starts with 0. We want to find a, b, c, d { , 1 } such that the binary representation of 1/3 starts with (0 . abcd ) 2 . We notice that 1 2 > 1 3 , so the first digit a must be 0 (not one). Then 1 4 < 1 3 , so the next digit is 1. Then 1 4 + 1 8 is too large so the following digit must be 0. Similarly the fourth digit is 1 because 2 + 1 2 2 + 2 3 + 1 2 4 < 1/3 . It seems that there is a pattern: so lets prove that the binary expansion of 1/3 is 0.010101 Call x ( 0.0101010101 ) 2 Then notice that 2 2 .x = ( 1.01010101 ) 2 , so by subtraction one has that (2 2- 1) .x = 1 which means exactly that x = 1/3 ....
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319HW3Correction - MAT 319/320 Correction of HW3 Exercise...

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