Explicație pas cu pas:
[tex]x * y = x + y + xy[/tex]
a)
[tex]x*(y*z) = x + (y*z) + x(y*z) = x + (y + z + yz) + x(y + z + yz) = x + y + z + yz + xy + xz + xyz = x + y + z + xy + yz + xz + xyz[/tex]
[tex](x*y)*z = (x*y) + z + (x*y)z = (x + y + xy) + z + (x + y + xy)z = x + y + xy + z + xz + yz + xyz = x + y + z + xy + yz + xz + xyz[/tex]
[tex] = > (x*y)*z = x*(y*z)[/tex]
=> legea * este asociativă
b)
[tex]f(x) = x + 1[/tex]
[tex]f(x * y) = (x * y) + 1 \\ = x + y + xy + 1 = x(y + 1) + (y + 1) \\ = (x + 1)(y + 1) = f(x) \times f(y)[/tex]
[tex] = > f(x * y) = f(x) \times f(y)[/tex]
c)
[tex]S_{1001} = 1 * \frac{1}{2} * \frac{1}{3} *...* \frac{1}{1000} * \frac{1}{1001} \\ [/tex]
legea * este asociativă
observăm că:
[tex]S_{2} =1* \frac{1}{2} = 1 + \frac{1}{2} + \frac{1}{2} = 1 + 1 = 2 \\ [/tex]
[tex]S_{3} =1 * \frac{1}{2} * \frac{1}{3} = \left(1 * \frac{1}{2} \right) * \frac{1}{3} \\ = S_{2} * \frac{1}{3} = 2 * \frac{1}{3} = 2 + \frac{1}{3} + \frac{2}{3} = 2 + 1 = 3 \\ [/tex]
[tex]S_{4} = 1 * \frac{1}{2} * \frac{1}{3}* \frac{1}{4} = \left(1 * \frac{1}{2}* \frac{1}{3} \right) * \frac{1}{4} \\ = S_{3} * \frac{1}{4} = 3* \frac{1}{4} = 3 + \frac{1}{4} + \frac{3}{4} = 3 + 1 = 4 \\ [/tex]
presupunem că este adevărat pentru k:
[tex]S_{k} =1* \frac{1}{2} * \frac{1}{3} *...* \frac{1}{k - 1} * \frac{1}{k} = k \\ [/tex]
și demonstrăm pentru k + 1:
[tex]S_{k + 1} = 1* \frac{1}{2} * \frac{1}{3} *...* \frac{1}{k} * \frac{1}{k + 1} \\ = \left(1* \frac{1}{2} * \frac{1}{3} *...* \frac{1}{k} \right) * \frac{1}{k + 1}= \\ = k * \frac{1}{k + 1} = \\ = k + \frac{1}{k + 1} + \frac{k}{k + 1} = k + 1\\ [/tex]
=>
[tex]S_{1001} = 1* \frac{1}{2} * \frac{1}{3} *...* \frac{1}{1000} * \frac{1}{1001} = \\ = 1001 \\ [/tex]
