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Precalculus Mathematics Homework Help
Prove that ## \sum_{d\mid n} f(d)=\sum_{d\mid n} f(\frac{n}{d}) ##.
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[QUOTE="fresh_42, post: 6803229, member: 572553"] Yes, but why? More detailed: ##n## is even, i.e. not odd. So ##n\neq 2k+1##. Since ##2(k-1)+1=2k-1## and ##2(k+1)+1=2k+3## are all odd per definition, we can only have ##2(k-1)+0,2k+0,2(k+1)+0## within the gaps of odd numbers for ##n.## Thus ##n=2k## for some ##k## and so ##2\,|\,n.## The other direction? So far, we have shown ## \{n\, : \,n \text{ is even }\}\subseteq \{n\, : \, 2\,|\,n\}. ## Now assume ##2\,|\,n.## Then ##n=2\cdot e## for some ##e.## If ##n## was odd, then ##2e=2k+1## for some ##k.## Therefore ##2(e-k)=1## and ##2\,|\,1## which is impossible. Hence, ##n## cannot be odd, which is the definition of even, i.e. ## \{n\, : \,n \text{ is even }\}\supseteq \{n\, : \, 2\,|\,n\}. ## Both inclusions ## \{n\, : \,n \text{ is even }\}\subseteq \{n\, : \, 2\,|\,n\} \subseteq \{n\, : \,n \text{ is even }\} ## imply equality. You cannot just jump to what you want to show. Read the definition! I defined 'even' as 'not odd', and 'odd' as ##2k+1.## I did not define 'even' as ##2k.## This had to be proven. This is an almost trivial example, admitted, but it should show you the steps from what we know to what we want to prove. [/QUOTE]
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Precalculus Mathematics Homework Help
Prove that ## \sum_{d\mid n} f(d)=\sum_{d\mid n} f(\frac{n}{d}) ##.
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