Problem Set 4
3 points

The problems are taken from the text. Keep in mind starred problems have solutions in the back!

NOT GRADED (do not turn in - do only if you need more practice)
Chapter 1.4
Exercises #7(d, k, l)

GRADED (please turn in)
Chapter 1.5
Exercises 5 (all), 8, 9

Chapter 1.5
5. Recall that a circle of radius R and center (2,4) is the set of all points of distance R from the point (2,4).
c) We prove the contrapositive, i.e that if (3,1) is inside the circle, then (0,3) is inside the circle. Suppose that (3,1) is inside the circle. Then since the distance between (3,1) and (2,4) is the square root of (3-2)^2 +(1-4)^ = 10, the radius must be greater than the square root of 10. This implies that all points closer than square root of 10 from (2,4) are in the circle. The distance of the point (0,3) to (2,4) is the square root of (0-2)^2+(3-4)^2 = 5. However, the square root of 5 is less than the square root of 10, so (0,3) is inside the circle.

8. We have to show both that m, n different parity implies m^2-n^2 is odd, and that m^^2-n^2 odd implies different parity. There are many ways to sole this question.
First assume that m and n have different parity. Assume first that m is even and n is odd. Then m^2 is even and -n^2 is odd, so that m^2-n^2 is the sum of an odd and an even number, which is odd. Similarly, if m were odd and n were even, m^2 is odd and -n^2 is even, so their sum is odd.
Conversely, suppose m^2-n^2 is odd. Then m^2 and n^2 have different parity, for if they have the same parity their difference would be even. Therefore m and n have different parity.

Here's another way to prove this.
Assume that m and n have different parity. Suppose first that that m = 2k and n = 2l+1 for some integers k, l. Then m^2 - n^2 = (2k)^2-(2l+1)^2 = 4k^2 - [4l^2 + 4l+1], which is odd.  The proof is similar in the case that m = 2l+1 and n=2k for some integers k, l. On the other hand, suppose that m^2-n^2 is odd. Then m^2 -n^2 = 2k+1 for some integer k. We factor the left hand side, to obtain (m+n)(m-n) = 2k+1. Since the right hand side is not divisible by 2, m+n must be odd. Therefore m and n must have differnt parity.