Discrete Mathematics with Applications 4th Edition Chapter 7 - Functions - Exercise Set 7.1 - Page 394 13

Answer

$ J_{5} = \left \{ 0, 1, 2, 3, 4 \right \}, f :J_{5}\rightarrow J_{5} \,\,and\,\, g :J_{5}\rightarrow J_{5}\\ f (x)=(x + 4)^{2}\,\,mod\,\,5\,\,and\,\,g(x)=(x^{2} + 3x + 1)\,\,mod\,\,5.\\ f(0)=(0+4)^2\,\,mod\,\,5=16\,mod\,5=1\\ f(1)=(1+4)^2\,\,mod\,\,5=25\,mod\,5=0\\ f(2)=(2+4)^2\,\,mod\,\,5=36\,mod\,5=1\\ f(3)=(3+4)^2\,\,mod\,\,5=49\,mod\,5=4\\ f(4)=(4+4)^2\,\,mod\,\,5=64\,mod\,5=4\\ g(0)=(0^{2} + 3(0) + 1)\,\,mod\,\,5=1\,mod\,5=1\\ g(1)=(1^{2} + 3(1) + 1)\,\,mod\,\,5=5\,mod\,5=0\\ g(2)=(2^{2} + 3(2) + 1)\,\,mod\,\,5=11\,mod\,5=1\\ g(3)=(3^{2} + 3(3) + 1)\,\,mod\,\,5=19\,mod\,5=4\\ g(4)=(4^{2} + 3(4) + 1)\,\,mod\,\,5=29\,mod\,5=4\\ so\,\,\\ f(0)=g(0)=1 \\ f(1)=g(1)=0 \\ f(2)=g(2)=1 \\ f(3)=g(3)=4 \\ f(4)=g(4)=4 \\ \therefore f=g $

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