Answer:
See below
Step-by-step explanation:
a) Direct proof: Let m be an odd integer and n be an even integer. Then, there exist integers k,j such that m=2k+1 and n=2j. Then mn=(2k+1)(2j)=2r, where r=j(2k+1) is an integer. Thus, mn is even.
b) Proof by counterpositive: Suppose that m is not even and n is not even. Then m is odd and n is odd, that is, m=2k+1 and n=2j+1 for some integers k,j. Thus, mn=4kj+2k+2j+1=2(kj+k+j)+1=2r+1, where r=kj+k+j is an integer. Hence mn is odd, i.e, mn is not even. We have proven the counterpositive.
c) Proof by contradiction: suppose that rp is NOT irrational, then rp=m/n for some integers m,n, n≠. Since r is a non zero rational number, r=a/b for some non-zero integers a,b. Then p=rp/r=rp(b/a)=(m/n)(b/a)=mb/na. Now n,a are non zero integers, thus na is a non zero integer. Additionally, mb is an integer. Therefore p is rational which is contradicts that p is irrational. Hence np is irrational.
d) Proof by cases: We can verify this directly with all the possible orderings for a,b,c. There are six cases:
a≥b≥c, a≥c≥b, b≥a≥c, b≥c≥a, c≥b≥a, c≥a≥b
Writing the details for each one is a bit long. I will give you an example for one case: suppose that c≥b≥a then max(a, max(b,c))=max(a,c)=c. On the other hand, max(max(a, b),c)=max(b,c)=c, hence the statement is true in this case.
e) Direct proof: write a=m/n and b=p/q, with m,q integers and n,q nonnegative integers. Then ab=mp/nq. mp is an integer, and nq is a non negative integer. Hence ab is rational.
f) Direct proof. By part c), √2/n is irrational for all natural numbers n. Furthermore, a is rational, then a+√2/n is irrational. Take n large enough in such a way that b-a>√2/n (b-a>0 so it is possible). Then a+√2/n is between a and b.
g) Direct proof: write m+n=2k and n+p=2j for some integers k,j. Add these equations to get m+2n+p=2k+2j. Then m+p=2k+2j-2n=2(k+j-n)=2s for some integer s=k+j-n. Thus m+p is even.
