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Unformatted text preview: a h4 ig6 e k# 8 i b iC a aha 3.1 Problem 8: We start with x 2 5 y 2 = 2 and then take the remainder of both sides when divided by 5: x 2 5 y 2 x 2 2 (mod 5) . However, if x mod 5 0 then x 2 mod 5 1 then x 2 mod 5 1 2 then x 2 mod 5 4 3 then x 2 mod 5 4 4 then x 2 mod 5 1 . Thus it is impossible to have x 2 2 (mod 5), so x 2 5 y 2 = 2 has no solutions. 3.1 Problem 12: Lets try a = 2 and b = 4. Then radicalbigg a 2 + b 2 2 = radicalbigg 4 + 16 2 = 10 3 . 16 > a + b 2 = 3 . Lets try a = 1 and b = 7. Then radicalbigg a 2 + b 2 2 = radicalbigg 1 + 49 2 = 25 = 5 > a + b 2 = 4 . Lets try a = 3 and b = 3. Then radicalbigg a 2 + b 2 2 = radicalbigg 9 + 9 2 = 9 = 3 = a + b 2 = 3 . We conjecture that the quadratic mean is always larger than or equal to the arithmetic mean. First, if ( a + b ) / 2 < 0 then we already have radicalbig ( a 2 + b 2 ) / 2 > > ( a + b ) / 2. Otherwise, we work this one backwards: radicalbigg a 2 + b 2 2 a + b 2 a 2 + b 2 2 parenleftbigg a + b 2 parenrightbigg 2 = a 2 + 2 ab + b 2 4 2 a 2 + 2 b 2 a 2 + 2 ab + b 2 a 2 2 ab + b 2 ( a b ) 2 . The last line is true, with equality only when a = b . To turn this into a proof we run it the right way around: ( a b ) 2 a 2 2 ab + b 2 a 2 + b 2 2 a 2 + 2 ab + b 2 4 = parenleftbigg a + b 2 parenrightbigg 2 2 a 2 + 2 b 2 a 2 + 2 ab + b 2 radicalbigg a 2 + b 2 2 vextendsingle vextendsingle vextendsingle...
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This note was uploaded on 07/12/2008 for the course MAT 243 taught by Professor Callahan during the Spring '06 term at ASU.
 Spring '06
 CALLAHAN
 Remainder

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