Hello,
Can someone please help me with this home work problem? I am totally lost and stumped. I've read my text and notes several times and I don't get it. The problem is attached.

(a) Assume n is even and 3n+2 is odd. Since any multiple of an even number is even, we know that 3n is even. We also know that the difference between an odd number and an even number is odd. Therefore

2 = (3n+2) - 3n is odd.

But since we know ...

Solution Summary

We prove by contradiction that 3n+2 is even if and only if n is even.

I would love to have a usual word format answer with detailed proofs.
The problems are 6 in total, the numbers are: from attached page #1: #2.82; 2.86; 2.87, from attached page #2: # 2.97; 2.98; 2.99. Thanks in advance.
The student.

Prove by contradiction that there does not exist a largest integer.
Hint: observe that for any integer n there is a greater one, say n+1. So begin the proof
"Suppose for contradiction that there is a largest integer. Let this integer be n...."

Check if the proof is correct. I need help to justify some of my answers by using Theorems, Definitions and etc. You can change my wording but try to stick to my idea. It's really important that you explain your work. Thanks!
Note: R=Real Number and C=Complex Number

Show the error in the following fallacious "proof" that P NP. Proof by contradiction. Assume that P = NP. Then SAT P. So, of some k, SAT TIME(nk). Because every language in NP is polynomial time reducible to SAT, NP TIME(nk). Therefore P TIME(nk). But, by the time hierarchy the

Let (X, E, u) be a measure space with u a non-negative
measure. Suppose that
1. f : X -> R is measurable
2. f (x) >= 0 a.e. with respect to u.
3. integral (over X) f du = 0
Prove that f (x) = 0 a.e. with respect to u.
note: E = Capital Sigma
u = lowercase mu

(a) Let f be analytic in a bounded region D and its boundary C, such that |f(z)| = 1
on C. Show that f has at least one zero inside D, unless f is a constant.
(b) Let f(z) be an analytic function in a region D except for one simple pole and
assume |f(z)| = 1 on the boundary of D. Prove that every value a with |a| > 1 is
take

Here is what I have can you add anything to help this out?
You want to show that if m is a natural number (0, 1, 2, 3, ...), then m x m is not equal to m + m. You can do this simply by showing that it is not in at least one case (because if it's not true for at least one natural number, then it's not true for natural numbers

A) Prove that if H is nontrivial normal subgroup of the solvable group G then there is a nontrivial subgroup A of H with A normal subgroup of G and A abelian.
b)Prove that if there exists a chain of subgroups G1<=G2<=.....<=G such that
G=union(from i=1 to infinity)of Gi and each Gi is simple, then G is simple
Part a of thi

Prove or disprove:
a. There exist a connected n-vertex simple graph with n+1 edges that contains exactly two cycles.
b. There does not exist a connected n-vertex simple graph with n+2 edges that contains four edge-disjoint cycles.