Solution:
→ Theorem. For every natural number n, 6│n^3 – n + 6 or 3 ∣ (n^3 – n).
Proof. We use mathematical induction for this proof.
Base case: If n = 0, then we get n^3 – n + 6 = 6 = 6 • 1, so 6│n^3 – n + 6.
Induction Step: Let n be an arbitrary natural number and suppose 6│n^3 – n + 6.
Then we can choose an integer k such that 6k = n^3 – n + 6.
Thus, we plug n + 1 into the polynomial, n^3 – n + 6.
This gives us (n + 1)^3 – (n + 1) + 6 = (n + 1)^2 • (n +1) – (n + 1) + 6
= (n^2 + 2n + 1) • (n +1) – (n + 1) + 6 = (n^2 + 2n) • (n +1) + 6
= n^3 + 3n^2 + 2n + 6 = n^3 – n + 6 + 3n^2 + 3n
= 6k + 3n^2 + 3n = (3) • (2k + n^2 + n) = (3) • (2k + n • (n + 1)).
When n = 0, 6│n^3 + 3n^2 + 2n + 6 since 6 is divisible by itself.
When n = 1, 6│n^3 + 3n^2 + 2n + 6 since 12 is indeed divisible by 6.
When n = 2, 6│n^3 + 3n^2 + 2n + 6 since 24 is indeed divisible by 6.
→ Therefore, 6 ∣ ((n + 1)^3 – (n + 1) + 6), as required. ∴ It is true that 6│n^3 – n + 6 for any n ∈ ℕ by the principle of mathematical induction. ☐ㅁ◻ or Q.E.D
Is this true for the math proof problem?
#PROBLEM: Prove that if 6│n^3 – n + 6 for any n ∈ ℕ.
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Solution:
→ Theorem. For every natural number n, 6│n^3 – n + 6 or 3 ∣ (n^3 – n)...
I think you needed to mention that n(n + 1) is always an ever number
Showing the cases n = 0, 1, 2 doesn't complete the induction step
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(a) How many different relations are on X = {1, 2, 3}? (b) How many different equivalence relations are on this set?
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