 1.4.1: Consider the statements P: 5 is odd. Q: 23 + 1 is even. (a) Express...
 1.4.2: Determine, with explanation, whether the statement 12 > 0 if and on...
 1.4.3: For an integer n and the open sentences P(n): 3n2 is even. Q(n): n3...
 1.4.4: For an integer n and the open sentences P(n): n is odd. Q(n): n2 is...
 1.4.5: For an integer n, consider the biconditional 2n > n2 if and only if...
 1.4.6: For an integer n, consider the open sentences P(n): 5n + 7 is even....
 1.4.7: For integers a and b, consider the biconditional ab is even if and ...
 1.4.8: For integers a and b, consider the biconditional a + b is even if a...
 1.4.9: For integers a, b and c, consider the biconditional At least two of...
 1.4.10: For an integer n, consider the open sentences P(n) : n(n + 1)(2n + ...
 1.4.11: For two statements P and Q, give an explanation why the bicondition...
 1.4.12: For every two statements P and Q, use truth tables to verify the fo...
Solutions for Chapter 1.4: Biconditionals
Full solutions for Discrete Mathematics  1st Edition
ISBN: 9781577667308
Solutions for Chapter 1.4: Biconditionals
Get Full SolutionsDiscrete Mathematics was written by and is associated to the ISBN: 9781577667308. Since 12 problems in chapter 1.4: Biconditionals have been answered, more than 12901 students have viewed full stepbystep solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Discrete Mathematics, edition: 1. Chapter 1.4: Biconditionals includes 12 full stepbystep solutions.

Associative Law (AB)C = A(BC).
Parentheses can be removed to leave ABC.

Basis for V.
Independent vectors VI, ... , v d whose linear combinations give each vector in V as v = CIVI + ... + CdVd. V has many bases, each basis gives unique c's. A vector space has many bases!

Block matrix.
A matrix can be partitioned into matrix blocks, by cuts between rows and/or between columns. Block multiplication ofAB is allowed if the block shapes permit.

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].

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.

Fundamental Theorem.
The nullspace N (A) and row space C (AT) are orthogonal complements in Rn(perpendicular from Ax = 0 with dimensions rand n  r). Applied to AT, the column space C(A) is the orthogonal complement of N(AT) in Rm.

Hankel matrix H.
Constant along each antidiagonal; hij depends on i + j.

Kirchhoff's Laws.
Current Law: net current (in minus out) is zero at each node. Voltage Law: Potential differences (voltage drops) add to zero around any closed loop.

Kronecker product (tensor product) A ® B.
Blocks aij B, eigenvalues Ap(A)Aq(B).

lAII = l/lAI and IATI = IAI.
The big formula for det(A) has a sum of n! terms, the cofactor formula uses determinants of size n  1, volume of box = I det( A) I.

Multiplier eij.
The pivot row j is multiplied by eij and subtracted from row i to eliminate the i, j entry: eij = (entry to eliminate) / (jth pivot).

Orthogonal subspaces.
Every v in V is orthogonal to every w in W.

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).

Pivot columns of A.
Columns that contain pivots after row reduction. These are not combinations of earlier columns. The pivot columns are a basis for the column space.

Row space C (AT) = all combinations of rows of A.
Column vectors by convention.

Saddle point of I(x}, ... ,xn ).
A point where the first derivatives of I are zero and the second derivative matrix (a2 II aXi ax j = Hessian matrix) is indefinite.

Spanning set.
Combinations of VI, ... ,Vm fill the space. The columns of A span C (A)!

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

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

Tridiagonal matrix T: tij = 0 if Ii  j I > 1.
T 1 has rank 1 above and below diagonal.