group

Sampras · #1 $Old$ November 4th, 2009, 01:51 AM

Suppose $K = \{r \in \mathbb{R}: r \neq 0, r \neq 1 \}$ . Suppose G consists of the following functions: $f(x) = \frac{1}{1-x}, \ g(x) = \frac{x-1}{x}, \ h(x) = \frac{1}{x}, \ i(x) = x, j(x) = 1-x, \ k(x) = \frac{x}{x-1}$ . Is G a group under function composition?

No because $k(x)$ doesn't have an inverse?

Is this correct?

Opalg · #2 $Old$ November 4th, 2009, 02:42 AM

Quote:

Originally Posted by Sampras $View Post$

Suppose $K = \{r \in \mathbb{R}: r \neq 0, r \neq 1 \}$ . Suppose G consists of the following functions: $f(x) = \frac{1}{1-x}, \ g(x) = \frac{x-1}{x}, \ h(x) = \frac{1}{x}, \ i(x) = x, j(x) = 1-x, \ k(x) = \frac{x}{x-1}$ . Is G a group under function composition?

No because $k(x)$ doesn't have an inverse?

Is this correct?

$(k\circ k)(x) = k(k(x)) = \frac{\frac x{x-1}}{\frac x{x-1}-1} = \frac x1 = i(x)$ . So $k\circ k = i$ , and therefore k is its own inverse (under function composition), i being the identity element for this operation.