Problem Set 5
5 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 1.5
Exercise #10, 12 (a,b)

GRADED (please turn in)
Chapter 1.5
Exercises 12 (c,d)
Chapter 1.6
Exercises 1(d,e,g), 2(a,c), 4(a,c,e,g)

Chapter 1.6

2 (a) Since c divdes a, we have a = ck for some integer k. Similarly b=cl fr some integer l. Let x and y be any integer. Then
ax+by = ckx+cly = c(kx+ly) is divisible by c.
(c) If a divides b, then b=ak for some integer k. Therefore, for any natural number n, b^n = (ak)^n = a^n k^n. Therefore a^n divides b^n.

4(a) This is false. A counterexample occurs when x=41, since (41)^2+41+41= (41)^2 + 2(41) = 41(41+2)=41*43 is not prime.
(c) This is false. A counterexample occurs when x=2 and y=1/2. Then x>1 and y>0, but y^x = 1/4 is not greater than 2.  This is a counterexample since the sentence is stated for all x and for all y.
(e) This is true. Suppose a, b,c,d are integers, and a divides b-c and a divides c-d. Then b-c=am for some integer m and c-d=an for some integer n. Therefore
b-d = (b-c)+(c-d) = am + an = a(m+n),
so a divides b-d.
(g) This is false. A counter example can be found with x=.01. (One way to see that it's false is to do a Taylor's series expansion of 2^x near x=0. Keep in mind that d/dx(2^x) = ln 2 * 2^x. When you do the expansion you find that 2^x is approximately 1 + (ln 2)x near 0. Since ln 2 is less than 1 (e is greater than 2), you can conclude that there are values near 0 in which (ln 2)x is less than x, so 1+(ln2)x <1+x, so 2^x<1+x. )