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## Here's the Solution to this Question

Let $X = \{a, b, c\}$ defined by $f : X \to X$ such that $f = \{(a, b), (b, a), (c,c)\}$ .

a) Find the values of $f^{–1}, f^2$ and $f^4$. Since $f^{-1}(y)=x$ iff $f(x)=y,$ we concluse that $f^{-1} = \{(b, a), (a, b), (c,c)\}.$ Taking itno aaccount that $f^2(x)=f(f(x))$ and $f^4(x)=f^2(f^2(x)),$ we conclude that $f^2 = \{(a, a), (b, b), (c,c)\}$ and $f^4 = \{(a, a), (b, b), (c,c)\}.$

b) Let $L = \{3, 4, 12, 24, 48, 82\}$ and the relation < be defined on L such that x < y if x divides y. Draw the Hasse diagram.

The Hasse diagram is a graphical rendering of a partially ordered set displayed via the cover relation of the partially ordered set with an implied upward orientation. A point is drawn for each element of the poset, and line segments are drawn between these points according to the following two rules:

1. If $x  in the poset, then the point corresponding to $x$ appears lower in the drawing than the point corresponding to $y$.

2. The line segment between the points corresponding to any two elements  $x$ and  $y$ of the poset is included in the drawing iff  $x$ covers $y$  or  $y$  covers $x$.

In our case, $x if and only if $x|y.$ Therefore, the Hasse diagram is the following:

c) The functions $f:X\to Y$ and $g:Y\to X$ are inverse of one another if $f\circ g=id_Y$ and $g\circ f=id_X,$ that is $f(g(y))=y$ and $g(f(x))=x$ for any $x\in X$ and $y\in Y.$