Mathematics, 12.03.2020 23:28 jmurguia888
In order for a proof by mathematical induction to be valid, the basis statement must be true for the initial value of n and the argument of the inductive step must be correct for every integer greater than or equal to the initial value.
Consider the following statement.
For every integer n โฅ 1, 3n โ 2 is even.
The following is a proposed proof by mathematical induction for the statement.
Since the property is true for n = 1, the basis step is true. Suppose the property is true for an integer k, where k โฅ 1.That is, suppose that 3k โ 2 is even. We must show that
3k + 1 โ 2 is even. Observe that 3k + 1 โ 2 = 3k ยท 3 โ 2 = 3k(1 + 2) โ 2
= (3k โ 2) + 3k ยท 2.
Now 3k โ 2 is even by inductive hypothesis and 3k ยท 2 is even by inspection. Hence the sum of the two quantities is even (by Theorem 4.1.1). It follows that
3k + 1 โ 2 is even, which is what we needed to show.
Identify the error(s) in the proof. (Select all that apply.)
3k + 1 โ 2 โ (3k โ 2) + 3k ยท 2
(3k โ 2) + 3k ยท 2 โ 3k(1 + 2) โ 2
3k โ 2 is odd by the inductive hypothesis.
The inductive hypothesis is assumed to be true.
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