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Proposition: Let 𝐴, 𝐡, and 𝐢 be subsets of some universal set U. Then
𝐴 βˆ’ (𝐡 βˆ’ 𝐢) ≑ (𝐴 ∩ 𝐢) βˆͺ (𝐴 βˆ’ 𝐡).
PROOF:
Let A, B, and C be subsets of some universal set U. Observe that:
𝐴 βˆ’ (𝐡 βˆ’ 𝐢) = 𝐴 βˆ’ (𝐡 ∩ 𝐢 𝑐 )
by the basic property,
𝑐
𝑐
= 𝐴 ∩ (𝐡 ∩ 𝐢 )
by the basic property,
= 𝐴 ∩ (𝐡 𝑐 βˆͺ (𝐢 𝑐 )𝑐 )
by D’morgans law,
𝑐
= 𝐴 ∩ (𝐡 βˆͺ 𝐢)
by the basic property,
𝑐
= (𝐴 ∩ 𝐡 ) βˆͺ (𝐴 ∩ 𝐢)
by the distributive
property,
= (𝐴 ∩ 𝐢) βˆͺ (𝐴 ∩ 𝐡 𝑐 )
by the commutative
property,
= (𝐴 ∩ 𝐢) βˆͺ (𝐴 βˆ’ 𝐡)
by the basic property.
Thus we have proven 𝐴 βˆ’ (𝐡 βˆ’ 𝐢) ≑ (𝐴 ∩ 𝐢) βˆͺ (𝐴 βˆ’ 𝐡) by using the algebra of sets.
Use induction to prove the following propostion:
For each natural number 𝑛,
𝑛3
3
+
𝑛2
2
+
7𝑛
6
is a natural number.
Proof: To prove the proposition, we will use induction. Let 𝑛 be a natural number. For our base
case let 𝑛 = 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 𝑛7 βˆ’ 3𝑛4 βˆ’ 9𝑛 βˆ’ 7 = 0 has a natural number solution.
Proof: Let 𝑛7 βˆ’ 3𝑛4 βˆ’ 9𝑛 βˆ’ 7 = 0. We will prove this by attempting to show that there is no
natural number solution. Thus, using algebra, we get:
𝑛7 βˆ’ 3𝑛4 βˆ’ 9𝑛 βˆ’ 7 = 0
𝑛7 βˆ’ 3𝑛4 βˆ’ 9𝑛 = 7
𝑛(𝑛6 βˆ’ 3𝑛3 βˆ’ 9) = 7.
Now, let π‘š = 𝑛6 βˆ’ 3𝑛3 βˆ’ 9 where k is an integer. Thus, we get the equation π‘›π‘š = 7. This can
7
also we written as π‘š = 𝑛 . Thus, for k to be a natural number, only the solutions 𝑛 = 1 and 𝑛 =
7 are possible. Given this, we will now test these two integers. So, using 𝑛 = 7 we get:
7(76 βˆ’ 3(7)3 βˆ’ 9) = 7
7(117649 βˆ’ 1029 βˆ’ 9) = 7
7(116611) = 7
816277 β‰  7.
Thus, 𝑛 = 7 does not work. Now, testing 𝑛 = 1, we get:
1(16 βˆ’ 3(1)3 βˆ’ 9) = 7
1(1 βˆ’ 3 βˆ’ 9) = 7
1(βˆ’11) = 7
βˆ’11 β‰  7.
Because neither of these 𝑛 values worked, we can conclude that there is no natural number
solution to the equation 𝑛7 βˆ’ 3𝑛4 βˆ’ 9𝑛 βˆ’ 7 = 0. Thus, our proposition is true.
Proposition:
Prove the following proposition:
For all real numbers π‘₯ and 𝑦, |π‘₯𝑦| = |π‘₯||𝑦|.
Proof:
Let π‘₯,𝑦 ∈ ℝ. To prove this proposition, we must show it to be true for these cases: 1) +π‘₯,
+𝑦, 2) +π‘₯, -𝑦, 3) –π‘₯, +𝑦, 4) –π‘₯, -𝑦, 5) either π‘₯ or 𝑦 is 0, and 6) both π‘₯ and 𝑦 are 0. The definition
of absolute value is
π‘₯
|π‘₯| = { π‘₯ > 0
}.
π‘₯<0
βˆ’π‘₯
1) Assume π‘₯ and 𝑦 are greater than 0. Thus π‘₯𝑦 is also positive and by definition of absolute
value,
|π‘₯𝑦| = π‘₯𝑦
|π‘₯||𝑦| = π‘₯ βˆ— 𝑦 = π‘₯𝑦
hence |π‘₯𝑦| = |π‘₯||𝑦|.
2) Assume π‘₯ is greater than 0 and 𝑦 is less than 0. Thus π‘₯𝑦 is negative and by definition of
absolute value,
|π‘₯𝑦| = βˆ’π‘₯𝑦
|π‘₯||𝑦| = π‘₯ βˆ— βˆ’π‘¦ = βˆ’π‘₯𝑦
hence |π‘₯𝑦| = |π‘₯||𝑦|.
3) Assume π‘₯ is less than 0 and 𝑦 is greater than 0. Thus π‘₯𝑦 is negative and by definition of
absolute value,
|π‘₯𝑦| = βˆ’π‘₯𝑦
|π‘₯||𝑦| = βˆ’π‘₯ βˆ— 𝑦 = βˆ’π‘₯𝑦
hence |π‘₯𝑦| = |π‘₯||𝑦|.
4) Assume π‘₯ and 𝑦 are less than 0. Thus π‘₯𝑦 is positive and by definition of absolute value,
|π‘₯𝑦| = π‘₯𝑦
|π‘₯||𝑦| = βˆ’π‘₯ βˆ— βˆ’π‘¦ = π‘₯𝑦
hence |π‘₯𝑦| = |π‘₯||𝑦|.
5) Assume either π‘₯ or 𝑦 is 0. Thus π‘₯𝑦 is 0 and by definition of absolute value,
|π‘₯𝑦| = 0
|π‘₯||𝑦| = 0 βˆ— 𝑦 = 0
hence |π‘₯𝑦| = |π‘₯||𝑦|.
6) Assume π‘₯ and 𝑦 are 0. Thus π‘₯𝑦 is 0 and by definition of absolute value,
|π‘₯𝑦| = 0
|π‘₯||𝑦| = 0 βˆ— 0 = 0
hence |π‘₯𝑦| = |π‘₯||𝑦|.
Therefore, for all real numbers π‘₯ and 𝑦, |π‘₯𝑦| = |π‘₯||𝑦|.