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To complete the division algorithm equation, a = mq + r, using a = – 46 and m = 9, which of the following gives appropriate values for integers q and r, with r expressed as a non-negative integer between 0 and (m-1), inclusive.

For arbitrary positive integers a, b, and m with m>1, if a ≡ b (mod m), then a = b + km, for some integer k.

If the square root of every integer is an integer, then 2 is irrational.

For any predicates, P(x) and Q(x), ∀x [ P(x) ⋁ Q (x) ] ⟺ [ ( ∀x P(x) ) ⋁ ( ∀x Q(x) ) ].

For arbitrary positive integers a, b, c, and m with m>1, if (a + c) ≡ (b + d) (mod m), then a ≡ b (mod m) and c ≡ d (mod m).

Use the Euclidean algorithm to determine the GCD(268, 108).  Show your work. Then express the GCD(268, 108) value you identify as a linear combination of 268 and 108. Show your work.

Indicate which of these listed graphs are bipartite. Select ‘True’ if the graph is bipartite; otherwise select ‘False’.  There may be more than one or none. [A]   K4 [B]   C6 [C]   Q3 [D]   W5

For arbitrary positive integers a, b, c with a ≠ 0, if a | (b + c) then a | b or a | c.

Given relation R defined on the set { 2, 4, 6, 8 } as follows: (m, n) ∈ R if and only if m|n. Determine which properties relation R exhibits.  Select ‘True’ if the property does apply to relation R; otherwise select ‘False’.  There may be more than one or none. [A]   reflexive [B]   irreflexive [C]   symmetric [D]   antisymmetric [E]   asymmetric [F]   transitive

Prove the following statement using induction. “For all integers n ≥ 3, 2n + 1 ≤ 2n.” Use good proof technique.  Grading rubric:1 pt. State the basis step, then prove it.1 pt. State the inductive hypothesis.2 pt. Complete the proof of the inductive step.  [Hint:  The fact that 2k − 1 ≥ 0 when k ≥ 3 can be useful] 1 pt. State the final conclusion at the end of the proof.1 pt. Label each part: the basis step, inductive hypothesis, inductive step, and conclusion. Note: To avoid the need for typing superscript exponents, you may use the expression ‘2^n’ to represent 2n.  Also the ≥ symbol can be written as >=.