If you mean (6^n) - 1 is divisible by 5 for all positive integers, then yes, it is true. We can prove it by a method known as Strong Induction. First, we show that the assertion is true for n = 1. This is called the Basis, or Base Step. Then we assume it is true for n = k, where k is any arbitrary positive integer k < (k + 1). This is called the Induction Step, or Inductive Hypothesis. Then we prove that is true for n = k + 1, based upon the last assumption. This is the Proof Step. If we can prove that the assertion is true for n = (k + 1), then it is true for all n, since we can let k equal any positive integer. Here's the proof:
Basis Step: Show assertion is true for n = 1.
(6^1) - 1 = 5m
6 - 1 = 5m
5 = 5(1), where m = 1.
Inductive Hypothesis: We assume the assertion is true for any n = k, where n is an integer.
Proof Step: Prove true for n = k + 1, for k any positive integer. This is the most rigorous part of the proof. The proof method involves a substitution based on our Inductive Hypothesis.
From the Inductive Hypothesis, (6^k) - 1 = 5m, where m is an integer.
Then 6^k = 5m + 1, and the steps below follow logically:
[6^(k + 1)] - 1 = 6 * (6^k) - 1
[6^(k + 1)] - 1 = 6 * (5m + 1) - 1
[6^(k + 1)] - 1 = 30m + 6 - 1
[6^(k + 1)] - 1 = 30m + 5
[6^(k + 1)] - 1 = 5 (6m + 1), where (6m + 1) is another integer.
So the assertion is proven, because we can let n equal any value of k, an integer.
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If you mean (6^n) - 1 is divisible by 5 for all positive integers, then yes, it is true. We can prove it by a method known as Strong Induction. First, we show that the assertion is true for n = 1. This is called the Basis, or Base Step. Then we assume it is true for n = k, where k is any arbitrary positive integer k < (k + 1). This is called the Induction Step, or Inductive Hypothesis. Then we prove that is true for n = k + 1, based upon the last assumption. This is the Proof Step. If we can prove that the assertion is true for n = (k + 1), then it is true for all n, since we can let k equal any positive integer. Here's the proof:
Basis Step: Show assertion is true for n = 1.
(6^1) - 1 = 5m
6 - 1 = 5m
5 = 5(1), where m = 1.
Inductive Hypothesis: We assume the assertion is true for any n = k, where n is an integer.
Proof Step: Prove true for n = k + 1, for k any positive integer. This is the most rigorous part of the proof. The proof method involves a substitution based on our Inductive Hypothesis.
From the Inductive Hypothesis, (6^k) - 1 = 5m, where m is an integer.
Then 6^k = 5m + 1, and the steps below follow logically:
[6^(k + 1)] - 1 = 6 * (6^k) - 1
[6^(k + 1)] - 1 = 6 * (5m + 1) - 1
[6^(k + 1)] - 1 = 30m + 6 - 1
[6^(k + 1)] - 1 = 30m + 5
[6^(k + 1)] - 1 = 5 (6m + 1), where (6m + 1) is another integer.
So the assertion is proven, because we can let n equal any value of k, an integer.
Technically, yes. You can divide, say, 17 by 5. You'd just get a fraction.
But the question probably wants if you can divide it evenly, to an integer, in which case no.