Let S be a subset of the set of integers defined recursively as follows: Base case: 5 ES Recursive case: If a ES, then 3

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Let S Be A Subset Of The Set Of Integers Defined Recursively As Follows Base Case 5 Es Recursive Case If A Es Then 3 1 (47.3 KiB) Viewed 53 times

Let S be a subset of the set of integers defined recursively as follows: Base case: 5 ES Recursive case: If a ES, then 3a E S. we are using Structural Induction to show for all a ES that a = 5k for some k E N. Don't include zero in the natural number set. Evaluate if the following steps are correct or not. Type in "Correct" OR "Incorrect". (Don't include quotation mark.) i) Base case: 5 belongs to the set S and 5 is a multiple of 5 because 5 X k = 5 (where k = 1 and 1 is a natural number). (Correct/Incorrect?): ii) Inductive step: assume if'a' belongs to 'S' then a = 5k where k is a natural number. (Correct/Incorrect?): iii) Proof. 5 is a positive multiple of 5 because 5k = 5 where k = 1 and 1 is a natural number. (Correct/Incorrect?)