Problem Set 6
8 points

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

NOT GRADED (do not turn in)
Chapter 1.6
Exercise #2(d, e), 4(b,d,f)

GRADED (please turn in)
Chapter 1.6
5(a,b), 6(a,l), 7(c,e,f)
Chapter 1.7
Exercises 1(c,f), 2(a, f), 4(a),
Chapter 2.1
Exercises 2 (all), 3 (all), 4 (d, j), 5(f,h,j), 7, 19(a)


Chapter 1.6
5(b) Since p is prime and larger than 3, p is not 3k for any integer k. Therefore p=3k+1 or p=3k+2. In both cases, a direct computation will prove that p^2+2 is divisible by 3.

6(l) We need to show that for every natural number x there exists an integer k such that 3.3x+k<50. In particular, our choice of k will depend on x. Let k=-4x. Since is an integer, so is k. Note that
3.3(x) +k = 3.3x-4x=-.7x.
Since x is a natural number, -.7x is negative, and in particular it is less than 50.


Chapter 1.7
2 (a) If n is even, then n=2k for some integer k. Then 5n^2+3n+4=5(2k)^2+3(2k)+4=20k^2+6k+4 = 2(10k^2+3k+2) is even. On the other hand, if n is odd, n=2k+1 for some integer k. Then 5n^2+3n+4=5(2k+1)^2+3(2k+1)+4 = 20k^2+20k+5+6k+3+4 = 2(10k^2+13k+6) is even.

(f) We note that (n^3-n)(n+2) = (n-1)*n*(n+1)*(n+2) is the product of four consecutive integers. Therefore, it must be a mutliple of 3 and of 4. Then for some integer (n^3-n)(n+2) =4mm, and 3 divides 4m. Euclid's lemma tells us that if 3 must divide 4 or divide m. Since 3 does not divide 4, it must divide m and hence m=3l for some integer l. Therefore, (n^3-n)(n+2) =4m=4(3l)=12l is a multiple of 12.

4(a) (Proof by contradition.) Suppose that x is rational and y is irrational, but x+y is rational. Then x+y=a/b for some integers a,b. Since x is rational, x = c/d for some integers c,d. Therefore,
y= (a/b) - x = (a/b)-(c/d) = (ad-cb)/bd
is rational, contrary to assumption.

Chapter 2.1
5(f) True.
(h)  False.
(j) False.

7. We prove the contrapositive. Suppose that x is in A. Since A is a subset of B, x is in B.