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Proposition: Let 𝐴, 𝐵, and 𝐶 be subsets of some universal set U. Then
𝐴 − (𝐵 − 𝐶) ≡ (𝐴 ∩ 𝐶) ∪ (𝐴 − 𝐵).
PROOF:
Let A, B, and C be subsets of some universal set U. Observe that:
𝐴 − (𝐵 − 𝐶) = 𝐴 − (𝐵 ∩ 𝐶 𝑐 )
by the basic property,
𝑐
𝑐
= 𝐴 ∩ (𝐵 ∩ 𝐶 )
by the basic property,
= 𝐴 ∩ (𝐵 𝑐 ∪ (𝐶 𝑐 )𝑐 )
by D’morgans law,
𝑐
= 𝐴 ∩ (𝐵 ∪ 𝐶)
by the basic property,
𝑐
= (𝐴 ∩ 𝐵 ) ∪ (𝐴 ∩ 𝐶)
by the distributive
property,
= (𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵 𝑐 )
by the commutative
property,
= (𝐴 ∩ 𝐶) ∪ (𝐴 − 𝐵)
by the basic property.
Thus we have proven 𝐴 − (𝐵 − 𝐶) ≡ (𝐴 ∩ 𝐶) ∪ (𝐴 − 𝐵) 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.
𝑘3
For our induction step, let 𝑘 be a natural number, and assume that 𝑓𝑘 = 3 +
natural number. We will now prove that 𝑓𝑘+1 is a natural number. Observe,
(𝑘 + 1)3 (𝑘 + 1)2 7(𝑘 + 1)
𝑓𝑘+1 =
+
+
3
2
6
3
2
𝑘
𝑘
7𝑘
1
1
7
=
+
+
+ 𝑘 2 + 2𝑘 + + +
3 32 2 6
3 2 6
𝑘
𝑘
7𝑘
2
= 3 + 2 + 6 + 𝑘 + 2𝑘 + 2.
𝑘3
𝑘2
7𝑘
𝑘2
2
𝑘3
+
7𝑘
6
𝑘2
is a
7𝑘
Since 3 + 2 + 6 is a natural number by closure properties, we obtain that = 3 + 2 + 6 +
𝑘 2 + 2𝑘 + 2 is a natural number. Thus 𝑓𝑘+1 is a natural number. Thus our original proposition is
true.
Proposition: The equation 𝑛7 − 3𝑛4 − 9𝑛 − 7 = 0 has a natural number solution.
Proof: Let 𝑛7 − 3𝑛4 − 9𝑛 − 7 = 0. We will prove this by attempting to show that there is no
natural number solution. Thus, using algebra, we get:
𝑛7 − 3𝑛4 − 9𝑛 − 7 = 0
𝑛7 − 3𝑛4 − 9𝑛 = 7
𝑛(𝑛6 − 3𝑛3 − 9) = 7.
Now, let 𝑚 = 𝑛6 − 3𝑛3 − 9 where k is an integer. Thus, we get the equation 𝑛𝑚 = 7. This can
7
also we written as 𝑚 = 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 − 3(7)3 − 9) = 7
7(117649 − 1029 − 9) = 7
7(116611) = 7
816277 ≠ 7.
Thus, 𝑛 = 7 does not work. Now, testing 𝑛 = 1, we get:
1(16 − 3(1)3 − 9) = 7
1(1 − 3 − 9) = 7
1(−11) = 7
−11 ≠ 7.
Because neither of these 𝑛 values worked, we can conclude that there is no natural number
solution to the equation 𝑛7 − 3𝑛4 − 9𝑛 − 7 = 0. Thus, our proposition is true.
Proposition:
Prove the following proposition:
For all real numbers 𝑥 and 𝑦, |𝑥𝑦| = |𝑥||𝑦|.
Proof:
Let 𝑥,𝑦 ∈ ℝ. To prove this proposition, we must show it to be true for these cases: 1) +𝑥,
+𝑦, 2) +𝑥, -𝑦, 3) –𝑥, +𝑦, 4) –𝑥, -𝑦, 5) either 𝑥 or 𝑦 is 0, and 6) both 𝑥 and 𝑦 are 0. The definition
of absolute value is
𝑥
|𝑥| = { 𝑥 > 0
}.
𝑥<0
−𝑥
1) Assume 𝑥 and 𝑦 are greater than 0. Thus 𝑥𝑦 is also positive and by definition of absolute
value,
|𝑥𝑦| = 𝑥𝑦
|𝑥||𝑦| = 𝑥 ∗ 𝑦 = 𝑥𝑦
hence |𝑥𝑦| = |𝑥||𝑦|.
2) Assume 𝑥 is greater than 0 and 𝑦 is less than 0. Thus 𝑥𝑦 is negative and by definition of
absolute value,
|𝑥𝑦| = −𝑥𝑦
|𝑥||𝑦| = 𝑥 ∗ −𝑦 = −𝑥𝑦
hence |𝑥𝑦| = |𝑥||𝑦|.
3) Assume 𝑥 is less than 0 and 𝑦 is greater than 0. Thus 𝑥𝑦 is negative and by definition of
absolute value,
|𝑥𝑦| = −𝑥𝑦
|𝑥||𝑦| = −𝑥 ∗ 𝑦 = −𝑥𝑦
hence |𝑥𝑦| = |𝑥||𝑦|.
4) Assume 𝑥 and 𝑦 are less than 0. Thus 𝑥𝑦 is positive and by definition of absolute value,
|𝑥𝑦| = 𝑥𝑦
|𝑥||𝑦| = −𝑥 ∗ −𝑦 = 𝑥𝑦
hence |𝑥𝑦| = |𝑥||𝑦|.
5) Assume either 𝑥 or 𝑦 is 0. Thus 𝑥𝑦 is 0 and by definition of absolute value,
|𝑥𝑦| = 0
|𝑥||𝑦| = 0 ∗ 𝑦 = 0
hence |𝑥𝑦| = |𝑥||𝑦|.
6) Assume 𝑥 and 𝑦 are 0. Thus 𝑥𝑦 is 0 and by definition of absolute value,
|𝑥𝑦| = 0
|𝑥||𝑦| = 0 ∗ 0 = 0
hence |𝑥𝑦| = |𝑥||𝑦|.
Therefore, for all real numbers 𝑥 and 𝑦, |𝑥𝑦| = |𝑥||𝑦|.