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I was solving the following question

Let $S(k) = 1+3+5+\cdots+[2k-1] = 3+k^2$. Then which of the following is true ?

  1. $S(1)$ is correct
  2. $S(k) \implies S(k+1)$
  3. $\neg (S(k) \implies S(k+1))$
  4. Principle of mathematical induction can be used to prove the formula.

Clearly the first option is incorrect since $S(1) \implies 1 = 4$, which is false. Then I worked on second option . For this I assumed $S(k)$ is true, i.e., $1+3+5+...+[2k-1] = 3 + k^2$. Adding $(2k+1)$ on both sides we get, $$1+3+5+...+[2k-1]+[2k+1] = 3+k^2+2k+1 \implies 1+3+5+...+[2k-1]+[2(k+1)-1] = 3+(k+1)^2 \implies S(k+1)$$ is true , so option $2$ is correct and simultaneously option $3$ is incorrect . But when I tried to find the least value of $k$ for which $S(k)$ is true I couldn't find a single value of $k$. So how is this possible ?

Community
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Sameer Nilkhan
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  • I am not entirely sure I understand what you mean by $=>$, $=/>$, etc. Also, what does $S(1) => 1 = 4$ mean? – an4s Jan 04 '20 at 11:36

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$S(k)$ is false for all $k$. So $S(k)\implies$ absolutely anything. Hence $2$ is the only correct option.

I suspect that whoever set this question wants you to suppose that $S(k)$ is true and use elementary algebra to show that $S(k+1)$ is true, but that is logically invalid.

Also, you can't use induction to prove it, because it's false.

PS Your title seems to be asking a different question.

TonyK
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  • Not quite. I think you have grasped the point of the question: false statements imply anything, and the claim is false. But (2) is a correct implication, – Ethan Bolker Jan 04 '20 at 11:48
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    @EthanBolker: This question is a logical minefield. Of course option $2$ is a correct implication, because $S(k)$ is false. It is not correct for any other reason, whatever the question setter intended. – TonyK Jan 04 '20 at 11:56
  • @TonyK as per your answer , can principle of mathematical induction be used only to prove a true statement and not to disprove a false statement ? If we know a statement is true , why will we prove it ? – Sameer Nilkhan Jan 04 '20 at 13:26
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    @TonyK Indeed it is a logical minefield, and a very bad question. I think it's intended to test the fact that the inductive step in a proof may be true but the proof still fail because there's no base case. Here the inductive step is true because the premise is false, but the instructor meant the student to prove it true with the usual crank-turning algebra. – Ethan Bolker Jan 04 '20 at 13:34
  • @Sameernilkhan: That's two questions again! And I don't really understand either of them. – TonyK Jan 04 '20 at 14:47