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Assignment 11

Assignment 11 - Homework#11 Section 5.2 3a 3b 3c 3d 3e 3f...

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Homework #11 *********** Section 5.2 ********** 3a) false 3b) true 3c) false 3d) true 3e) true 3f) false 4b) Define h:N->E+ by h(n)= { 16 if n=1, 12 if n=2, 2 if n=3, 4 if n=8, 6 if n=6, 2n else 5b) c 5d) Xo 5f) c 5g) Xo 6a) A=N, B=N-{1} 6b) A=N, B=positive even integers 6c) A=N, B=N-{1} 7a) A=(2,5), B=(3,4) 7b) A=R, B=R-{0} 7c) A=(1,0), B=(3,5) 10.) This is clearly a function on (0,1), as it is defined for all numbers in this open interval. If you compute the derivative you will see that it is negative throughout the interval, the function is thus decreasing, and it is thus 1-1. As x gets arbitrarily close to 0 from the right, f gets arbitrarily large. As x gets arbitrarily close to 1 from the left, f gets arbitrarily large negative. We can then use the Intermediate Value Theorem and the fact that f is continuous everywhere to conclude that f takes on any real value. f is thus onto. 11.) [An intuitive note is that both RXR and C are represented graphically with a coordinate system in the plane, so this suggests that there is a 1-1 correspondence. A convenient one would be to map (x,y) to (x,y).] To show this, let f:C -> RXR be given by f(a+bi) = (a,b). This is defined for all of C, and the assignments are unique, so it is a function.

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