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Unformatted text preview: Math 131A Analysis Summer Session A Homework 1 Solutions 1. Prove by induction that 2 n n ! for n 4. Proof. Let S = { n N : n 4 and 2 n n ! } . Since 2 4 = 16 < 24 = 4!, we have 4 S . Now suppose n S . Since 2 n n !, we have 2 n +1 = 2(2 n ) 2( n !) ( n + 1)( n !) = ( n + 1)! So n + 1 S . By the Principle of Math Induction I, S = { n N : n 4 } . 2. Prove by induction that ( n + 1) n 1 n n for n 1. Proof. Let S = { n N : ( n + 1) n 1 n n } . When n = 1, both sides of the inequality are 1 so 1 S . Now assume that n S so that ( n + 1) n 1 n n . Multiply both sides of the inequality by ( n +1) n n to obtain n + 1 n n n + 1 But then we have ( n + 1) + 1 n + 1 n = 1 + 1 n + 1 n 1 + 1 n n = n + 1 n n n + 1 . Now multiply by ( n + 1) n to obtain (( n + 1) + 1) n ( n + 1) n +1 . This shows that n +1 S whenever n S . By the Principle of Math Induction I, we have S = N . 3. Prove that if r 6 = 0 is a rational number and x is an irrational number, then rx and r + x are also irrational numbers. Proof. Suppose not. Then rx Q and r 1 Q imply x = r 1 rx Q , a contradiction. Similarly, r + x Q and r Q imply x = x +0 = x +( r r ) = ( x + r ) r = ( r + x ) + ( r ) Q , another contradiction. Thus rx and r + x are both irrational. 1 4. Prove that (5 3) 1 3 is not a rational number. Proof. If a = (5 3) 1 3 , then a 3 = 5 3. Thus a 3 5 = 3. Squaring both sides, we have 3 = ( 3) 2 = ( a 3 5) 2 = a 6 10 a 3 + 25. Therefore, a is a root of the polynomial...
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This note was uploaded on 09/04/2008 for the course MATH 131a taught by Professor Hitrik during the Summer '08 term at UCLA.
 Summer '08
 hitrik
 Math

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