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...."

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

Describe the steps in and/or define how to accomplish the following types of proofs:
a. That two sets are equal.
b. That two sets are disjoint.
c. A proof by contra-positive.
d. A proof by contradiction.
e. A proof by Mathematical Induction.
(Question is repeated in attachment)

There is no bijection between any set A and its power set P(A) of A.
For finite sets, proof is trivial since |A| = n and |P(A)| = 2^n. For finite sets, this is done by contradiction. Suppose there is a bijection $ between a set A and its power set P(A). Consider the set B={x|x is a member A where x is not a member $(x)}For e

I don't understand how you count the degree of the vertices.
(See attached file for full problem description)
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2.- Prove that a tree with Delta(T)=k ( Delta means maximum degree) has at least k vertices of degree 1.
Proof. We prove it by contradiction. Suppose that and there are s vertices of degree 1, where s