(a) Predicate: 6/d is an integer, domain: Z (b) Predicate: 6/d is an integer, domain: Z + (c) Predicate: 1 ≤ x^2 ≤ 4, domain: R (d) Predicate: 1 ≤ x^2 ≤ 4, domain: Z
Identify the explicit formula for the nth term of the sequence written below. Note that the first term is for n = 1. -6, 18, -72,324, - 7776/5 , 7776, …
Let Q(x, y) be the predicate "If x < y then x2 < y2" with domain for both x and y being R the set of real numbers. (a) Explain why Q(x, y) is false if x = -2 and y = 1. . The conclusion is --Select-- , which is --Select--- . Thus Q(-2, 1) is a conditional statement with a ---Select--- hypothesis and a The hypothesis of Q(-2, 1) is --Select-- , which is --Select-- --Select--- conclusion. So Q(-2, 1) is false. (b) Give values different from those in part (a) for which Q(x, y) is false. 1. (x, y) =( (c) Explain why Q(x, y) is true if x = 3 and y = 8. . The conclusion is -Select-- v, which is --Select--- . Thus Q(3, 8) is a conditional statement with a ---Select-- hypothesis and a The hypothesis of Q(3, 8) is --Select-- , which is --Select-- -Select-- conclusion. So Q(3, 8) is true. (d) Give values different from those in part (c) for which Q(x, y) is true. (X,Y)= ( )
Identify the explicit formula for the nth term of the sequence written below. Note that the first term is for n = 1. -12, 72, -576, 5184, -248832/5, 497664, … Select the correct answer below: an = (-1) ⋅ 12^n/n ( ) an = (-1)^n ⋅ 12^n/n an = (-1)^n ⋅ 12^n-1 / n an = (-1)^n ⋅ 12^n / n + 1
An integer n is even if, and only if, n equals twice some integer. An integer n is odd if, and only if, n equals twice some integer plus 1. Symbolically, for any integer, n n is even ⇔ n = 2k for some integer k n is odd ⇔ n = 2k + 1 for some integer k Prove the following statement. The difference of any two odd integers is even. Proof: Let m and n be any odd integers. By definition of odd, there are integers r and s so that m can be expressed in terms of r and n can be expressed in terms of s as follows: m = n = Write m - n in terms of r and s and factor out a 2 to obtain m - n = Now is an integer because --- Select-- v of integers are integers. Therefore, m - n = 2 ⋅ (an integer), and so m - n is … - Select-- by definition of … - Select--.
Write the first four terms of the sequence defined recursively by the equation below. a0 = 1 an+1 = - 3/4an^-1 + 2 You may either enter an exact answer or round to two decimal places.
Assume x is a particular real number and use De Morgan's laws to write the negation for the statement. -7 < x < 3 -3 ≤ x ≤ 7 -7 > x or x > 3 -7 ≤ x ≤ 3 -7 < x < 3 -7 ≥ x or x ≥ 3
A sequence is defined recursively by the equation below. an+1 = (-1/2)an Write the first four terms of the sequence given that the 0th term is a0 = -1. You may either enter an exact answer or round to two decimal places.
Justify your answer by using the definitions of even, odd, prime, and composite numbers. Assume that r and s are particular integers. (a) Is 8rs even? Yes, because 8rs = 2(4rs) + 1 and 4rs is an integer. Yes, because 8rs = 2(4rs) and 4rs is an integer. No, because 8rs = 2(4rs) and 4rs is an integer. No, because 8rs = 2(4rs) + 1 and 4rs is an integer. (b) Is 2r + 6s 2 + 5 odd? Yes, because 2r + 6s 2 + 5 = 2(r + 3s 2 + 2) and r + 3s 2 + 2 is an integer. Yes, because 2r + 6s 2 + 5 = 2(r + 3s 2 + 2) + 1 and r + 3s 2 + 2 is an integer. No, because 2r + 6s 2 + 5 = 2(r + 3s 2 + 2) + 1 and r + 3s 2 + 2 is an integer. No, because 2r + 6s 2 + 5 = 2(r + 3s 2 + 2) and r + 3s 2 + 2 is an integer. (c) If r and s are both positive, is r 2 + 2rs + s 2 composite? Yes, because r 2 + 2rs + s 2 = (r + s) 2 and r + s is not an integer. Yes, because r 2 + 2rs + s 2 = (r + s) 2 and r + s is an integer. No, because r 2 + 2rs + s 2 = (r + s) 2 and r + s is not an integer No, because r 2 + 2rs + s 2 = (r + s) 2 and r + s is an integer.
Determine the explicit formula for the nth term of the sequence written below. Note that the first term is for n = 1. 1/4, 1/3, 2/5, 4/9, 8/17, 16/33, …
Consider the following statement. If 1 − 0.99999 … is less than every positive real number, then it equals zero. Use modus ponens or modus tollens to fill in the blanks in the argument so as to produce a valid inference. 1 − 0.99999 … is real number. ∴ The number 1 − 0.99999 … equals zero.
Determine the explicit formula for the sequence defined recursively that satisfies the following conditions. a0 = 2 an = an-1 / 2 , for n = 1,2,3, …
Determine whether the argument form is valid and explain how the truth table supports your answer. The argument is valid because all rows that have true premises have false conclusions. The argument is valid because all rows that have true premises have true conclusions. The argument is valid because all rows that have false premises have false conclusions. The argument is invalid because there exists a row that has true premises and a false conclusion. The argument is invalid because there exists a row that has false premises and a true conclusion.
Identify the explicit formula for the nth term of the sequence written below. Note that the first term is for n = 1. 1/3, 2/5, 4/9, 8/17, 16/33, 32/65, … Select the correct answer below: an = 2^n/2^(n+1) + 1 an = 2^n/2^(n+1) + 2 an = 2^n/2^(n+1) - 2 an = 2^n+1 / 2^n + 2
Justify your answers by using the definitions of even, odd, prime, and composite numbers. Assume that c is a particular integer. (a) Is -8c an even integer? Yes, because -8c = 2(-4c) and -4c is an integer. Yes, because -8c = 2(-4c) + 1 and -4c is an integer. No, because -8c = 2(-4c) and -4c is an integer. No, because -8c = 2(-4c) + 1 and -4c is an integer. (b) Is 8c + 7 an odd integer? Yes, because 8c + 7 = 2(4c + 3) + 1 and 4c + 3 is an integer. Yes, because 8c + 7 = 2(4c + 3) and 4c + 3 is an integer. No, because 8c + 7 = 2(4c + 3) + 1 and 4c + 3 is an integer. No, because 8c + 7 = 2(4c + 3) and 4c + 3 is an integer. (c) Is (c^2 + 1) - (c^2 - 1) - 2 an even integer? Yes, because (c^2 + 1) - (c^2 - 1) - 2 = 2(0) + 1 and 0 is an integer. Yes, because (c^2 + 1) - (c^2 - 1) - 2 = 2(0) and 0 is an integer. No, because (c^2 + 1) - (c^2 - 1) - 2 = 2(0) and 0 is an integer. No, because (c^2 + 1) - (c^2 - 1) - 2 = 2(0) + 1 and 0 is an integer.
Determine the limit of the sequence defined below. an = cos(16/n+6 ) If the limit does not exist, enter ∅. If the sequence approaches positive or negative infinity, enter ∞ or -∞, respectively.