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MCIII Number Theory 2.16

 
 
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MCIII Number Theory 2.16
by Tina Jin - Thursday, 4 January 2024, 9:19 PM
 

Hello,

From the solutions at back of book, 

floor(199/97)+floor(199*96/97)=198 because their sum is equal to 199. But it is still possible that the fractional(199/97)+fractional(199(96/97) to be greater than 1, which would make the sum 197 instead.

Thanks,

Tina Jin


 
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Re: MCIII Number Theory 2.16
by Dr. Kevin Wang - Saturday, 6 January 2024, 1:20 AM
 

"But it is still possible that the fractional(199/97)+fractional(199(96/97) to be greater than 1".  Think very carefully about this.

 
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Re: MCIII Number Theory 2.16
by Tina Jin - Sunday, 14 January 2024, 9:43 AM
 
Oh, is it because fractional(199/97)+fractional(199(96)/97) is equal to 1? I tried to prove it works for all k, but I need some help. Here is the algebraic proof:

{199*k/97}+{199*(97-k)/97}={199*k/97}+{199(-k)/97}={199*k/97}+{-(199*k/97)} and {x}+{-x}=1 so {199*k/97}+{-(199*k/97)}=1. But why is {x}+{-x}=1 true? 

Thanks,

Tina Jin

 
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