 3.4.1: Prove that if n is an integer, then n2 3n + 5 is odd.
 3.4.2: Prove that if n is an integer, then n3 n is even.
 3.4.3: Let n Z. Prove that if n2 + n = 0, then 2n+3n 12n is even.
 3.4.4: Prove that if n is an integer, then 3n + 1 and 5n + 2 are of opposi...
 3.4.5: (a) Let m and n be two integers. Prove that m+ n is even if and onl...
 3.4.6: Let m and n be two integers. Prove that 3mn is even if and only if ...
 3.4.7: Let m and n be two integers. Prove that 7m+3n is odd if and only if...
 3.4.8: Let m and n be two integers. Prove that if m and n are of opposite ...
 3.4.9: Let m and n be two integers. Prove that mn2 is odd if and only if m...
 3.4.10: Let m and n be two integers. Prove that mn and m+ n are both even i...
 3.4.11: Give a proof of Let n Z. If 2n 1 5, then n 3 and n 2. using (a) a...
 3.4.12: In the proof of Result 3.28 it was proved, for integers m and n, th...
 3.4.13: Give a proof of Result 3.29 where without loss of generality is use...
 3.4.14: Use a proof by cases (as in the proof of Result 3.29) to prove the ...
 3.4.15: Prove the following: Let A and B be sets. Then (A B) (B A) = (A B) ...
 3.4.16: Let S = {0, 1, 3, 4}. Prove that if x S, then x2 4x + 3 = 0 or x2 4...
Solutions for Chapter 3.4: Proof by Cases
Full solutions for Discrete Mathematics  1st Edition
ISBN: 9781577667308
Solutions for Chapter 3.4: Proof by Cases
Get Full SolutionsDiscrete Mathematics was written by and is associated to the ISBN: 9781577667308. Chapter 3.4: Proof by Cases includes 16 full stepbystep solutions. This expansive textbook survival guide covers the following chapters and their solutions. Since 16 problems in chapter 3.4: Proof by Cases have been answered, more than 8755 students have viewed full stepbystep solutions from this chapter. This textbook survival guide was created for the textbook: Discrete Mathematics, edition: 1.

Affine transformation
Tv = Av + Vo = linear transformation plus shift.

CayleyHamilton Theorem.
peA) = det(A  AI) has peA) = zero matrix.

Characteristic equation det(A  AI) = O.
The n roots are the eigenvalues of A.

Complex conjugate
z = a  ib for any complex number z = a + ib. Then zz = Iz12.

Echelon matrix U.
The first nonzero entry (the pivot) in each row comes in a later column than the pivot in the previous row. All zero rows come last.

Elimination.
A sequence of row operations that reduces A to an upper triangular U or to the reduced form R = rref(A). Then A = LU with multipliers eO in L, or P A = L U with row exchanges in P, or E A = R with an invertible E.

Fibonacci numbers
0,1,1,2,3,5, ... satisfy Fn = Fnl + Fn 2 = (A7 A~)I()q A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].

Four Fundamental Subspaces C (A), N (A), C (AT), N (AT).
Use AT for complex A.

Full row rank r = m.
Independent rows, at least one solution to Ax = b, column space is all of Rm. Full rank means full column rank or full row rank.

Left nullspace N (AT).
Nullspace of AT = "left nullspace" of A because y T A = OT.

Markov matrix M.
All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.

Multiplicities AM and G M.
The algebraic multiplicity A M of A is the number of times A appears as a root of det(A  AI) = O. The geometric multiplicity GM is the number of independent eigenvectors for A (= dimension of the eigenspace).

Outer product uv T
= column times row = rank one matrix.

Partial pivoting.
In each column, choose the largest available pivot to control roundoff; all multipliers have leij I < 1. See condition number.

Pascal matrix
Ps = pascal(n) = the symmetric matrix with binomial entries (i1~;2). Ps = PL Pu all contain Pascal's triangle with det = 1 (see Pascal in the index).

Semidefinite matrix A.
(Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.

Symmetric factorizations A = LDLT and A = QAQT.
Signs in A = signs in D.

Symmetric matrix A.
The transpose is AT = A, and aU = a ji. AI is also symmetric.

Trace of A
= sum of diagonal entries = sum of eigenvalues of A. Tr AB = Tr BA.

Vandermonde matrix V.
V c = b gives coefficients of p(x) = Co + ... + Cn_IXn 1 with P(Xi) = bi. Vij = (Xi)jI and det V = product of (Xk  Xi) for k > i.