01_hwc1SolnsODDA

01_hwc1SolnsODDA - for f ◦ g . In the same way we find...

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Unformatted text preview: for f ◦ g . In the same way we find 12 ↓↓ 24 3 ↓ 4 4 ↓ 2 5 ↓ 2 1 ↓ 4 2 ↓ 2 3 ↓ 2 4 ↓ 4 5 ↓ 4 1 ↓ 1 2 ↓ 2 3 ↓ 3 4 ↓ 4 5 ↓ 5 g◦f : h◦g : h◦h : Note that h ◦ h is the identity map. (b) We find 12 ↓↓ 42 3 ↓ 2 4 ↓ 4 5 ↓ 4 1 ↓ 4 2 ↓ 2 3 ↓ 2 4 ↓ 4 5 ↓ 4 h ◦ (g ◦ f ) : and (h ◦ g ) ◦ f : These are the same, which is consistent with the associativity of composition, which asserts that h ◦ (g ◦ f ) = (h ◦ g ) ◦ f . (c) Only f and h are one to one, onto and invertible, while g is none of these things. We have seen above that h ◦ h is the identity map, so h−1 = h. To find f −1 , just run it backwards: since f (5) = 1, f −1 (1) = 5. since f (1) = 2, f −1 (2) = 1, and so on, with the result that 12345 f −1 : ↓ ↓ ↓ ↓ ↓ 51324 1.5 Consider the function from IR2 to IR2 defined by f (a) Is f one to one? explain why not. (b) Is f onto? x y = x−y . 2x + y (c) Is f invertible? If so find a formula for the inverse. If not, SOLUTION The three questions are closely related, and the best way to approach them u all at once is to try and compute a formula for an inverse. Given in IR2 , we seek to v find x y in IR2 so that f By the definition of f , this amounts to x−y =u 2x + y = v 21/september/2005; 22:09 x y = u v . (∗) 2 ...
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