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```Proposition: Let π΄, π΅, and πΆ be subsets of some universal set U. Then
π΄ β (π΅ β πΆ) β‘ (π΄ β© πΆ) βͺ (π΄ β π΅).
PROOF:
Let A, B, and C be subsets of some universal set U. Observe that:
π΄ β (π΅ β πΆ) = π΄ β (π΅ β© πΆ π )
by the basic property,
π
π
by the basic property,
= π΄ β© (π΅ π βͺ (πΆ π )π )
by Dβmorgans law,
π
= π΄ β© (π΅ βͺ πΆ)
by the basic property,
π
by the distributive
property,
by the commutative
property,
= (π΄ β© πΆ) βͺ (π΄ β π΅)
by the basic property.
Thus we have proven π΄ β (π΅ β πΆ) β‘ (π΄ β© πΆ) βͺ (π΄ β π΅) by using the algebra of sets.
Use induction to prove the following propostion:
For each natural number π,
π3
3
+
π2
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.
π3
For our induction step, let π be a natural number, and assume that ππ = 3 +
natural number. We will now prove that ππ+1 is a natural number. Observe,
(π + 1)3 (π + 1)2 7(π + 1)
ππ+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.
π3
π2
7π
π2
2
π3
+
7π
6
π2
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 ππ+1 is a natural number. Thus our original proposition is
true.
Proposition: The equation π7 β 3π4 β 9π β 7 = 0 has a natural number solution.
Proof: Let π7 β 3π4 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:
π7 β 3π4 β 9π β 7 = 0
π7 β 3π4 β 9π = 7
π(π6 β 3π3 β 9) = 7.
Now, let π = π6 β 3π3 β 9 where k is an integer. Thus, we get the equation ππ = 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(β11) = 7
β11 β  7.
Because neither of these π values worked, we can conclude that there is no natural number
solution to the equation π7 β 3π4 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) βπ₯, +π¦, 4) βπ₯, -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
βπ₯
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¦| = βπ₯π¦
|π₯||π¦| = π₯ β βπ¦ = βπ₯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¦| = βπ₯π¦
|π₯||π¦| = βπ₯ β π¦ = βπ₯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¦| = βπ₯ β βπ¦ = 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¦|.
```
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