Ask a New Question
Menu

1. Prove that 12 + 22 + 32 + β‹― + 𝑛𝑛2 = 𝑛𝑛(𝑛𝑛+1)6(2𝑛𝑛+1) , βˆ€π‘›π‘› ∈ β„• by using the principle of

mathematical induction.
2. Determine the power set of 𝐴𝐴 = {{π‘Žπ‘Ž, 𝑏𝑏},{𝑐𝑐},{𝑑𝑑, 𝑒𝑒, 𝑓𝑓}}.
3. Suppose β„• = {1,2,3, … } is the universal set and 𝐴𝐴 = {π‘₯π‘₯: π‘₯π‘₯ ≀ 6}, 𝐡𝐡 = {π‘₯π‘₯: 4 ≀ π‘₯π‘₯ ≀ 9} and
𝐢𝐢 = {2,3,5,7,8}. Then, find 𝐴𝐴 ∩ (𝐡𝐡 βŠ• 𝐢𝐢).
4. Let 𝐴𝐴 = {1,2}, 𝐡𝐡 = {π‘Žπ‘Ž, 𝑏𝑏, 𝑐𝑐} π‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Ž 𝐢𝐢 = {𝑐𝑐, 𝑑𝑑}. Then show that (𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢) =
𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢).
5. Let 𝑅𝑅 be the equivalence relation on the set 𝐴𝐴 = {1, 2, 3, 4, 5, 6}, given by 𝑅𝑅 =
{(1,1), (1,5), (2,2), (2,3), (2,6), (3,2), (3,3), (3,6), (4,4), (5,1), (5,5), (6,2), (6,3), (6,6)}.
Then, find the partition of 𝐴𝐴 induced by 𝑅𝑅 (i.e., find the quotient set 𝐴𝐴/𝑅𝑅).
6. Show that 𝑓𝑓: ℝ β†’ ℝ, 𝑓𝑓(π‘₯π‘₯) = 3π‘₯π‘₯5+1
is a one-to-one correspondence.

1 answer

1. Base Case:
For n = 1, we have 12 = 1(1+1)/6(2(1)+1) which is true.

2. Inductive Hypothesis:
Assume that for some k ∈ β„•, 12 + 22 +...+ k^2 = k(k+1)/6(2k+1).

3. Inductive Step:
We need to show that (k+1)(k+1+1)/6(2(k+1)+1) = 1^2 + 2^2 +...+ k^2 + (k+1)^2.

Starting with the left side:
(k+1)(k+2)/6(2k+3) = (k(k+1) + 2(k+1))/6(2k+3)
= k(k+1)/6(2k+1) + (2(k+1))/(6(2k+3))
= 12 + 22 +...+ k^2 + (2(k+1))/(6(2k+3))
= 12 + 22 +...+ k^2 + (k+1)^2

Therefore, by the principle of mathematical induction, 12 + 22 +...+ n^2 = n(n+1)/6(2n+1) for all n ∈ β„•.

2. The power set of A = {{a, b}, {c}, {d, e, f}} is {{}, {a, b}, {c}, {d, e, f}, {a, b, c}, {a, b, d, e, f}, {c, d, e, f}, {a, b, c, d, e, f}}. It consists of all possible subsets of A, including the empty set and the set itself.

3. A ∩ (B βŠ• C) means the intersection of A with the symmetric difference of B and C.

B βŠ• C = (B βˆͺ C) \ (B ∩ C)
= ({x: 4 ≀ x ≀ 9} βˆͺ {2,3,5,7,8}) \ ({4,5,6,7,8,9} ∩ {2,3,5,7,8})
= {x: 4 ≀ x ≀ 9} \ {5,7,8}
= {4,6,9}

A ∩ (B βŠ• C) = {x: x ≀ 6} ∩ {4,6,9}
= {4,6}

4. (A Γ— B) ∩ (A Γ— C) is the intersection of A cross B and A cross C.

(A Γ— B) = {(1,a), (1,b), (1,c), (2,a), (2,b), (2,c)}
(A Γ— C) = {(1,c), (1,d), (2,c), (2,d)}

(A Γ— B) ∩ (A Γ— C) = {(1,c), (2,c)}

A Γ— (B ∩ C) is the cartesian product of A and the intersection of B and C.

(B ∩ C) = {c}

A Γ— (B ∩ C) = {(1,c), (2,c)}

Therefore, (A Γ— B) ∩ (A Γ— C) = A Γ— (B ∩ C).

5. The partition of A induced by R is the set of equivalence classes formed by R.

From the given relation R, we can find the equivalence classes as follows:
[1] = {1, 5}
[2] = {2, 3, 6}
[3] = {2, 3, 6}
[4] = {4}
[5] = {1, 5}
[6] = {2, 3, 6}

Therefore, the partition of A induced by R is {[1], [2], [3], [4], [5], [6]}.

6. To show that f: ℝ β†’ ℝ, f(x) = 3x^5+1 is a one-to-one correspondence, we need to show that it is both injective (one-to-one) and surjective (onto).

Injective:
Assume f(a) = f(b) for some a, b ∈ ℝ.
This means 3a^5+1 = 3b^5+1.
Subtracting 1 from both sides and dividing by 3, we get a^5 = b^5.
Taking the fifth root of both sides, we get a = b.
Therefore, f is injective.

Surjective:
For any y ∈ ℝ, we need to show that there exists an x ∈ ℝ such that f(x) = y.
Let x = ((y-1)/3)^(1/5), then f(x) = 3[((y-1)/3)^(1/5)]^5+1 = y.
Therefore, f is surjective.

Since f is both injective and surjective, it is a one-to-one correspondence.