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Proposition: Let π΄, π΅, and πΆ be subsets of some universal set U. Then π΄ β (π΅ β πΆ) β‘ (π΄ β© πΆ) βͺ (π΄ β π΅). PROOF: Let A, B, and C be subsets of some universal set U. Observe that: π΄ β (π΅ β πΆ) = π΄ β (π΅ β© πΆ π ) by the basic property, π π = π΄ β© (π΅ β© πΆ ) by the basic property, = π΄ β© (π΅ π βͺ (πΆ π )π ) by Dβmorgans law, π = π΄ β© (π΅ βͺ πΆ) by the basic property, π = (π΄ β© π΅ ) βͺ (π΄ β© πΆ) by the distributive property, = (π΄ β© πΆ) βͺ (π΄ β© π΅ π ) by the commutative property, = (π΄ β© πΆ) βͺ (π΄ β π΅) by the basic property. Thus we have proven π΄ β (π΅ β πΆ) β‘ (π΄ β© πΆ) βͺ (π΄ β π΅) by using the algebra of sets. Use induction to prove the following propostion: For each natural number π, π3 3 + π2 2 + 7π 6 is a natural number. Proof: To prove the proposition, we will use induction. Let π be a natural number. For our base case let π = 1. Observe, 1 1 7 + + = 2. 3 2 6 Thus our base case holds. π3 For our induction step, let π be a natural number, and assume that ππ = 3 + natural number. We will now prove that ππ+1 is a natural number. Observe, (π + 1)3 (π + 1)2 7(π + 1) ππ+1 = + + 3 2 6 3 2 π π 7π 1 1 7 = + + + π 2 + 2π + + + 3 32 2 6 3 2 6 π π 7π 2 = 3 + 2 + 6 + π + 2π + 2. π3 π2 7π π2 2 π3 + 7π 6 π2 is a 7π Since 3 + 2 + 6 is a natural number by closure properties, we obtain that = 3 + 2 + 6 + π 2 + 2π + 2 is a natural number. Thus ππ+1 is a natural number. Thus our original proposition is true. Proposition: The equation π7 β 3π4 β 9π β 7 = 0 has a natural number solution. Proof: Let π7 β 3π4 β 9π β 7 = 0. We will prove this by attempting to show that there is no natural number solution. Thus, using algebra, we get: π7 β 3π4 β 9π β 7 = 0 π7 β 3π4 β 9π = 7 π(π6 β 3π3 β 9) = 7. Now, let π = π6 β 3π3 β 9 where k is an integer. Thus, we get the equation ππ = 7. This can 7 also we written as π = π . Thus, for k to be a natural number, only the solutions π = 1 and π = 7 are possible. Given this, we will now test these two integers. So, using π = 7 we get: 7(76 β 3(7)3 β 9) = 7 7(117649 β 1029 β 9) = 7 7(116611) = 7 816277 β 7. Thus, π = 7 does not work. Now, testing π = 1, we get: 1(16 β 3(1)3 β 9) = 7 1(1 β 3 β 9) = 7 1(β11) = 7 β11 β 7. Because neither of these π values worked, we can conclude that there is no natural number solution to the equation π7 β 3π4 β 9π β 7 = 0. Thus, our proposition is true. Proposition: Prove the following proposition: For all real numbers π₯ and π¦, |π₯π¦| = |π₯||π¦|. Proof: Let π₯,π¦ β β. To prove this proposition, we must show it to be true for these cases: 1) +π₯, +π¦, 2) +π₯, -π¦, 3) βπ₯, +π¦, 4) βπ₯, -π¦, 5) either π₯ or π¦ is 0, and 6) both π₯ and π¦ are 0. The definition of absolute value is π₯ |π₯| = { π₯ > 0 }. π₯<0 βπ₯ 1) Assume π₯ and π¦ are greater than 0. Thus π₯π¦ is also positive and by definition of absolute value, |π₯π¦| = π₯π¦ |π₯||π¦| = π₯ β π¦ = π₯π¦ hence |π₯π¦| = |π₯||π¦|. 2) Assume π₯ is greater than 0 and π¦ is less than 0. Thus π₯π¦ is negative and by definition of absolute value, |π₯π¦| = βπ₯π¦ |π₯||π¦| = π₯ β βπ¦ = βπ₯π¦ hence |π₯π¦| = |π₯||π¦|. 3) Assume π₯ is less than 0 and π¦ is greater than 0. Thus π₯π¦ is negative and by definition of absolute value, |π₯π¦| = βπ₯π¦ |π₯||π¦| = βπ₯ β π¦ = βπ₯π¦ hence |π₯π¦| = |π₯||π¦|. 4) Assume π₯ and π¦ are less than 0. Thus π₯π¦ is positive and by definition of absolute value, |π₯π¦| = π₯π¦ |π₯||π¦| = βπ₯ β βπ¦ = π₯π¦ hence |π₯π¦| = |π₯||π¦|. 5) Assume either π₯ or π¦ is 0. Thus π₯π¦ is 0 and by definition of absolute value, |π₯π¦| = 0 |π₯||π¦| = 0 β π¦ = 0 hence |π₯π¦| = |π₯||π¦|. 6) Assume π₯ and π¦ are 0. Thus π₯π¦ is 0 and by definition of absolute value, |π₯π¦| = 0 |π₯||π¦| = 0 β 0 = 0 hence |π₯π¦| = |π₯||π¦|. Therefore, for all real numbers π₯ and π¦, |π₯π¦| = |π₯||π¦|.