Name (printed):Student ID #:Section # (or TA’s:name and time)CMSC 250 Quiz #6 ANSWERS Wednesday, Mar. 3, 2004Write all answers legibly in the space provided. The number of points p os sible for each question is indicatedin square brackets – the total number of points on the quiz is 30, and you will have exactly 20 minutes tocomplete this quiz. You may not use calculators, textbooks or any other aids during this quiz.1. [24 pnts.] Disprove by counter example or Prove each of the following:a. The product of any rational number with an integer is a rational number.∀x ∈ Q, ∀y ∈ Z, xy ∈ QPROOF:Let x by arbitrary in Q and let y be arbitrary in Z.Since x is rational, ∃a, b ∈ Z, x =abwhere b 6= 0 by definition of rationalThe product of x and y can be written as y ∗abby substitutionAfter multiplying, xy =yabSince ya ∈ Z by closure of Z in multiplicationand b ∈ Z and b 6= 0 because it was defined as such above.Therefore xy ∈ Q by definition of rational. ∀x ∈ Q, ∀y ∈ Z, xy ∈ Q by generalizing from thegeneric particulab. For all integers n, if n is odd then n2is odd.∀n ∈ Z, n ∈ Zodd→ n2∈ ZoddPROOF: Let n be arbitrary in Z. |Assumen ∈ Zodd|Since n is an odd integer, ∃a ∈ Z n = 2a + 1 by the definition of odd.|Since n = 2a + 1, n2= (2a + 1)2by squaring both sides.|(2a + 1)2= 4a2+ 4a + 1 = 2(2a2+ 2a) + 1 by algebra.|Since 2a2+ 2a is an integer by closure of integers during addition and multiplication, n2is alsoeven by the definition of odd.n ∈ Zodd→ n2∈ Zoddby closing the conditional world∀n ∈ Z, n ∈ Zodd→ n2∈ Zoddby Generalizing from the Generic Particularc. For all integers n and m, (n + m) > m.False n = −2 and m = 5, (n + m) = −2 + 5 = 3 which is not grea ter than 5.2. [6 pnts.] State Yes or No for each of the following (only a small justification is needed for your answernot a complete proof). Assume a, b and c are integers and x, y and z are rationals for all of thefollowing questions.a. NO If a, b, and c are even,a+b+c2is also even.justification: a = 2, b = 2 and c = 2 then a+b+c = 6 and 6/2 is 3 which is odd.b. NO x + y ≤ x · yjustification: x = 2 and y = -2, x+y = 0 and xy = -4, but 0 6≤ −4c. NO If a > b and x > y, then a · x > b · y justification: a = 1 and b = -2and x = -4 and y = -5, while it is true that 1 ¿ -2 and -4 ¿ -5, it is not true that -4 ¿
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