hw5solutions23b

# hw5solutions23b - Anna Medvedovsky [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){}}()/* ]]> */ Math...

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Anna Medvedovsky [email protected] Math 23b / Spring 2009 HW #3 solutions 1. For polynomial functions f ( x ) = x - 1 and g ( x ) = x 2 - 1 ﬁnd f g and g f . Computing, f g ( x ) = x 2 - 2 and g f ( x ) = x 2 - 2 x. 2. Explain why multiplication by 2 deﬁnes a bijection from R to R but not from Z to Z . Multiplication by 2 is not a surjective function from Z to Z . For example, there’s no element of Z that you can multiply by 2 to get 1. So multiplication by 2 is not bijective from Z to Z . But multiplication by 2 is bijective from R to R because dividing by 2 deﬁnes an inverse function. (Note that dividing by 2 does not deﬁne a function from Z to Z .) 3. Let f : R R be a bijective increasing function. Prove that f - 1 : R R is an increasing function as well. Given x > y in R , we want to show that f - 1 ( x ) > f - 1 ( y ). So suppose not. Then f - 1 ( x ) f - 1 ( y ), so that either f - 1 ( x ) = f - 1 ( y ) or f - 1 ( x ) < f - 1 ( y ). If f - 1 ( x ) = f - 1 ( y ), then x = f ( f - 1 ( x ) ) = f ( f - 1 ( y ) ) = y. And if f - 1 ( x ) < f - 1 ( y ), then, since f is an increasing function, we must have x = f ( f - 1 ( x ) ) < f ( f - 1 ( y ) ) = y. In either case,

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## This note was uploaded on 10/15/2011 for the course MATH 23b taught by Professor Bonglian during the Spring '09 term at Brandeis.

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hw5solutions23b - Anna Medvedovsky [email protected] Math...

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