Rudin_1

# Rudin_1 - The Real and Complex Number Systems Written by Men-Gen Tsai email [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ 1 2 3 4 5 6 Fix b > 1(a If m n p q are integers n

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The Real and Complex Number Systems Written by Men-Gen Tsai email: [email protected] 1. 2. 3. 4. 5. 6. Fix b> 1. (a) If m, n, p, q are integers, n> 0, q> 0, and r = m/n = p/q , prove that ( b m ) 1 /n =( b p ) 1 /q . Hence it makes sense to de±ne b r =( b m ) 1 /n . (b) Prove that b r + s = b r b s if r and s are rational. (c) If x is real, de±ne B ( x ) to be the set of all numbers b t , where t is rational and t x . Prove that b r = sup B ( r ) where r is rational. Hence it makes sense to de±ne b x = sup B ( x ) for every real x . (d) Prove that b x + y = b x b y for all real x and y . 1

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Proof: For (a): mq = np since m/n = p/q . Thus b mq = b np . By Theorem 1.21 we know that ( b mq ) 1 / ( mn ) =( b np ) 1 / ( mn ) , that is, ( b m ) 1 /n =( b p ) 1 /q , that is, b r is well-defned. For (b): Let r = m/n and s = p/q where m, n, p, q are integers, and n> 0 , q > 0. Hence ( b r + s ) nq =( b m/n + p/q ) nq =( b ( mq + np ) / ( nq ) ) nq = b mq + np = b mq b np =( b m/n
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## This note was uploaded on 07/15/2008 for the course MATH 118 taught by Professor Ponce during the Spring '08 term at UCSB.

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Rudin_1 - The Real and Complex Number Systems Written by Men-Gen Tsai email [email protected] 1 2 3 4 5 6 Fix b > 1(a If m n p q are integers n

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