Question #185757

i) Let S={2,4,7} and T={1,3,5}. Find f(S×T) if

f(x,y)=⌊14x/3y⌋


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f(x,y)=x^2+y^3


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1
Expert's answer
2021-05-07T09:25:46-0400

S={2,4,7}, T={1,3,5}


S×\times T={(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(7,1),(7,3),(7,5)}


(i) f(x,y)=14x3yf(x,y)=\dfrac{14x}{3y}


So,

f(S×T)=f(S\times T)=

14(2)3(1)+14(2)3(3)+14(2)3(5)+14(4)3(1)+14(4)3(3)+14(4)3(5)+14(7)3(1)+14(7)3(3)+14(7)3(5)\dfrac{14(2)}{3(1)}+\dfrac{14(2)}{3(3)}+\dfrac{14(2)}{3(5)}+\dfrac{14(4)}{3(1)}+\dfrac{14(4)}{3(3)}+\dfrac{14(4)}{3(5)}+\dfrac{14(7)}{3(1)}+\dfrac{14(7)}{3(3)}+\dfrac{14(7)}{3(5)}


Hence f(S×T)f(S\times T) ={283,289,2815,563,569,5615,983,989,9815\dfrac{28}{3},\dfrac{28}{9},\dfrac{28}{15},\dfrac{56}{3},\dfrac{56}{9},\dfrac{56}{15},\dfrac{98}{3},\dfrac{98}{9},\dfrac{98}{15} }


(ii) f(x,y)=x2+y2f(x,y)=x^2+y^2


f(S×T)f(S\times T) ={

(22+12),(22+32),(22+52),(42+12),(42+32),(42+52),(72+12),(72+32),(72+52)(2^2+1^2),(2^2+3^2),(2^2+5^2),(4^2+1^2),(4^2+3^2),(4^2+5^2),\\(7^2+1^2),(7^2+3^2),(7^2+5^2) }


F(S×T)=F(S\times T)= {5,13,29,17,25,41,50,58,74{5,13,29,17,25,41,50,58,74} }


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