1
$\begingroup$

What is the consequence of not proving a base case in mathematical induction? For example:

prove that 2 + 4 + · · · + 2n = (n + 2)(n − 1) for any integer n ≥ 2

if a student jumped right into the inductive step without proving the base case We assume that P(n) : 2 + 4 + · · · + 2n = (n + 2)(n − 1), and we want to prove P(n + 1) : 2 + 4 + · · · + 2n + 2(n + 1) = (n + 3)n. 2 + 4 + · · · + 2n + 2(n + 1) = 2 + 4 + · · · + 2n + 2n + 2 = (n + 2)(n − 1) + 2n + 2 hypothesis = n(n + 2) − (n + 2) + 2n + 2 = n(n + 2) + n = n(n + 3)

What does this say?

$\endgroup$
  • $\begingroup$ here the base case is true but the student did not include that in their proof. From a markers point of view can we assume their proof is invalid/incomplete or just pass it? How important is that step @MauroALLEGRANZA $\endgroup$ – ekeith Oct 2 '16 at 17:25
  • $\begingroup$ I marked such problems on a scale of $10$. I’d have given $7$ for that if the intrusive hypothesis had not been (mis)placed in the middle of the chain of equalities and $6$ as it stands. In general $7$ meant that it was mostly right, but there was some non-trivial error or omission. I would probably also have written a note on the paper showing how essentially the same argument could be used to ‘prove’ that $\sum_{k=1}^n2k=(n+2)(n-1)+1$. $\endgroup$ – Brian M. Scott Oct 2 '16 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.